I have table user which have fields username,password, and type. The type can be any or combination of these employee,vendor and client i.e a user can be vendor or client both or some another combination. For type field I have used the multiple checkbox, see the code below. This is the views/users/add.ctp file
Form->create('User');?>
Form->input('username');
echo $this->Form->input('password');
echo $this->Form->input('type', array('type' => 'select', 'multiple' => 'checkbox','options' => array(
'client' => 'Client',
'vendor' => 'Vendor',
'employee' => 'Employee'
)
));
?>
Form->end(__('Submit', true));?>
This is the code I have used in the model file. A callback method beforeSave
app/models/user.php
function beforeSave() {
if(!empty($this->data['User']['type'])) {
$this->data['User']['type'] = join(',', $this->data['User']['type']);
}
return true;
}
This code saves the multiple values as comma separated value in db.
The main problem comes when Im editing a user. If a user has selected multiple types during user creation I can't find the checkbox checked for that user types.
you should never be saving serialized data, json or csv in a field. This makes your life real hard later on down the line.
While habtm is one way to do things, if your binary maths is reasonable you might want to checkout bitmasks for this. here is a great post http://mark-story.com/posts/view/using-bitmasks-to-indicate-status
basics would be
1 = employee
2 = vendor
4 = client
// 8 = next_type
then, if the user was type employee & vendor the type would be 3 (1 + 2) and if it was a vendor & client the type would be 6 (2 + 4)
as you can see there is no way to mix it up, and bitwise works pretty good in mysql aswell so finds are pretty easy. See the post for much more detailed information
You should have a table types and a join table users_types.
What you're looking at is a HABTM relationship, so you should handle it like one.
In the joining UsersType model you should add a custom validation rule that checks if the current combination of types is allowed.
If you want to modify data after it's been found in the database, you can use the afterFind() callback in your model.
So in your case, put something like this is your user model:
function afterFind($results) {
$results['User']['type'] = explode(',', $results['User']['type']);
return $results;
}
There's more info on afterFind in the CakePHP manual.
That being said, it might be worth considering another approach, like a HABTM relationship as deceze first suggested above.
Related
Apologies if this has been asked before. All of the examples I can find are old or apply to legacy versions of CakePHP, e.g. cakephp: saving to multiple models using one form is 7 years old.
I have an application in CakePHP 4.1.6. Two of the tables in the database are called tbl_users and tbl_orgs ("orgs" in this case means "Organisations").
When I add an Organisation I also want to create a User who is the main contact within the Organisation. This involves saving to both the tbl_orgs and tbl_users tables when the form is submitted.
The problem I'm experiencing is how to get the form working in a way where it will run the validation rules for both tbl_users and tbl_orgs when submitted.
This is how our application is currently structured:
There is a Controller method called add() in src/Controller/TblOrgsController.php. This was generated by bake and was initially used to insert a new Organisation into the tbl_orgs table. At this point it didn't do anything in terms of tbl_users however it worked in terms of saving a new Organisation and running the appropriate validation rules.
One validation rule is that every companyname record in tbl_orgs must be unique. If you try to insert more than 1 company with the name "My Company Limited" it would give the validation error "This company name already exists":
// src/Model/Table/TblOrgsTable.php
public function buildRules(RulesChecker $rules): RulesChecker
{
$rules->add(
$rules->isUnique(['companyname']),
[
'errorField' => 'companyname',
'message' => 'This company name already exists',
]
);
return $rules;
}
Whilst the above applies to TblOrgs we also have an buildRules() in TblUsers which applies similar logic on an email field to make sure that all email addresses are unique per user.
In the add() Controller method we start by specifying a new empty entity for TblOrgs:
// src/Controller/TblOrgsController.php
public function add()
{
$org = $this->TblOrgs->newEmptyEntity();
// ...
$this->set(compact('org'));
}
When the form is created we pass $org:
// templates/TblOrgs/add.php
<?= $this->Form->create($org) ?>
<?= $this->Form->control('companyname') ?>
<?= $this->Form->end() ?>
When the TblOrgs fields are rendered by the browser we can inspect the HTML and see these are obeying the corresponding Model. This is clear because of things such as required="required" and maxlength="100" which correspond to the fact that field is not allowed to be empty and is a VARCHAR(100) field in the database:
<input type="text" name="companyname" required="required" id="companyname" maxlength="100">
It also works in terms of the rules specified in buildRules for TblOrgs. For example if I enter the same company name twice it shows the appropriate error in-line:
I then tried to introduce fields for TblUsers. I prefixed the form fields with dot notation, e.g. this is intended to correspond to tbl_users.email input field:
<?= $this->Form->control('TblUser.email') ?>
When inspecting the HTML it doesn't do the equivalent as for TblOrgs. For example things like maxlength or required are not present. It effectively isn't aware of TblUsers. I understand that $org in my Controller method is specifying a new entity for TblOrgs and not TblUsers. I reviewed the CakePHP documentation on Saving With Associations which says
The save() method is also able to create new records for associations
However, in the documentation the example it gives:
$firstComment = $articlesTable->Comments->newEmptyEntity();
// ...
$tag2 = $articlesTable->Tags->newEmptyEntity();
In this case Tags is a different Model to Comments but newEmtpyEntity() works for both. With this in mind I adapted my add() method to become:
$org = $this->TblOrgs->TblUsers->newEmptyEntity();
But this now gives an Entity for TblUsers. It seems you can have either one or the other, but not both.
The reason this doesn't work for my use-case is that I can either run my Validation Rules for TblOrgs (but not TblUsers) or vice-versa.
How do you set this up in a way where it will run the validation rules for both Models? It doesn't seem to be an unreasonable requirement that a form may need to save data to multiple tables and you'd want the validation rules for each of them to run. I get the impression from the documentation that it is possible, but it's unclear how.
For reference there is an appropriate relationship between the two tables:
// src/Model/Table/TblOrgsTable.php
public function initialize(array $config): void
{
$this->hasMany('TblUsers', [
'foreignKey' => 'o_id',
'joinType' => 'INNER',
]);
}
and
// src/Model/Table/TblUsersTable.php
public function initialize(array $config): void
{
$this->belongsTo('TblOrgs', [
'foreignKey' => 'o_id',
'joinType' => 'INNER',
]);
}
Okay, lots of confusion to clear up here. :-) My assumption here, based on what you've written, is that you're trying to use a single form to add a new organization, and the first user in it, and then maybe later you'll add more users to the org.
First, $this->TblOrgs->TblUsers is your users table object, so when you use
$org = $this->TblOrgs->TblUsers->newEmptyEntity();
what you're doing is creating a new user entity. The fact that you got to that table object through the orgs table, and that you're calling it $org doesn't change that. It doesn't somehow magically create a blank org entity with a blank user entity in it. But you don't need that entity structure here at all here, just the empty org entity. Go back to simply:
$org = $this->TblOrgs->newEmptyEntity();
Now, in your form, you'll want something like this:
<?= $this->Form->create($org) ?>
<?= $this->Form->control('companyname') ?>
<?= $this->Form->control('tbl_users.0.email') ?>
<?= $this->Form->end() ?>
The field is called tbl_users.0.email because:
The table name gets converted to lower case underscore format.
It's a hasMany relation from orgs to users, so it's expecting an array of users; we have to give a numeric index into that array, and 0 is a great place to start. If you were going to add a second user at the same time, the field for that would be tbl_users.1.email.
Note: A great way to figure out what format the form helper is expecting you to create your field names in is to read an existing set of records from the database (in this case, an org and its users), and then just dump that data, with something like debug($org);. You'll see that $org has a property called tbl_users, which is an array, and that will point straight to this structure I've described above.
With the fields set up like this, you should be able to patch the resulting data directly into your $org entity in your controller, and save that without any other work. The patch will created the entire structure, with a entity of class TblOrg, with a tbl_users property which is an array containing a single entity of class TblUser, and validation will have been done on both of them. (At least it should; you can use debug($org); as mentioned above to confirm it.) And when you save this entity, it will first save the TblOrg entity, then add that new ID into the TblUser entity before saving it, as well as checking the rules for both and making sure that nothing gets saved to the database if it can't all be saved. That all happens automatically for you with the single save call!
If your association was a hasOne or belongsTo relation (for example if you were adding a new user and also the org that they're in, instead of the other way around), you could dump a sample $user, and see that it has a property called tbl_org which is just a straight-up entity, not an array of entities, and note that tbl_org is now singular, because it's just one entity instead of a bunch. In this case, the field name to use would be tbl_org.companyname, no array index in there at all.
I'm working on a user-generated content blog that allows a user to go through the whole upload process before being prompted to sign up. Basic flow: fill out form to pick username/basic info->upload blog post->prompt to sign up with email/password. The purpose of reversing the normal flow is to increase the UX and conversion rate and avoid a wall in the beginning.
Instead of migrating, I've just created the tables manually in PHPmyAdmin. I have 3 relational models: Usermeta->hasOne(App\Mopdels\Post), Post->belongsTo(App\Models\Usermeta), and User->belongsTo(App\Models\Usermeta).
What I'm having trouble with is once the user has created a username and submits the first form to the usermeta table, and then submits the second form to upload their blog post to the post table, it doesn't seem to be attaching the usermeta.id to posts.usermeta_id linking them together. I must be missing something or not attaching it correctly. Here's my StoryController:
<?php
namespace App\Controllers\Story;
use App\Models\Post;
use App\Models\User;
use App\Models\Usermeta;
use App\Controllers\Controller;
use Respect\Validation\Validator as v;
class StoryUploadController extends Controller
{
public function guidance($request, $response)
{
return $this->view->render($response, 'storyupload/guidance.twig');
}
//set up our the Upload Story class so the user can upload their story
//render the view 'uploadstory.twig'
public function getStoryUpload($request, $response)
{
return $this->view->render($response, 'storyupload/upload.twig');
}
// This method is called when the user submits the final form
public function postStoryUpload($request, $response, $id)
{
//set up our validation rules for our complete sign up form
$validation = $this->validator->validate($request, [
'title' => v::stringType()->notEmpty()->length(1, 80),
'body' => v::stringType()->notEmpty()->length(1, 2500),
]);
//if validation fails, stay on story upload page
if ($validation->failed()) {
return $response->withRedirect($this->router>pathFor('storyupload.upload'));
}
$user = Usermeta::find($id)->first();
//We can use our Post Model to send the form data to the database
$post = Post::create([
'title' => $request->getParam('title'),
'body' => $request->getParam('body'),
'category' => $request->getParam('category'),
'files' => $request->getParam('img_path'),
'usermeta_id' => usermeta()->attach($user->id),
]);
//after submit, redirect to completesignup page
return $response->withRedirect($this->router->pathFor('auth.completesignup'));
}
}
I continue to get the error 'usermeta_id cannot be null' so it's definitely not pulling the id from the usermeta table correctly.
I've used the create() method to send the usermeta data to the table in my Auth controller.
Would it be better to have all of my form submissions in the Auth controller and what is the proper way using my example to make sure that my posts.usermeta_id is linked to my usermeta.id?
The usermeta form is taken care of by my Auth Controller:
//render the view 'signup.twig'
public function getSignUp($request, $response)
{
return $this->view->render($response, 'auth/signup.twig');
}
// This method is called when the user submits the form
public function postSignUp($request, $response)
{
$validation = $this->validator->validate($request, [
'name' => v::notEmpty()->alpha(),
'username' => v::noWhitespace()->notEmpty()->UsernameAvailable(),
'city' => v::notEmpty()->alpha(),
'country' => v::notEmpty()->alpha(),
]);
//if validation fails, stay on signup page
if ($validation->failed()) {
return $response->withRedirect($this->router->pathFor('auth.signup'));
}
$usermeta = Usermeta::create([
'name' => $request->getParam('name'),
'username' => $request->getParam('username'),
'city' => $request->getParam('city'),
'country' => $request->getParam('country'),
'share_location' => $request->getParam('share_location'),
]);
//after submit, redirect to storyupload/guidance
return $response->withRedirect($this->router>pathFor('storyupload.guidance'));
}
I wrote quite a bit here. To jump directly to what I believe will solve your problem, see the "Your Issue" section. The rest is here as an educational exercise.
A Quick Intro to Laravel Relations
As you probably already know, "relations" in Laravel are virtual concepts that are derived from the hard data in the database. Because they are virtual, there is some overlap in the definition of relations.
When you say "Usermeta has one Post" - what this means is that the posts table will have a usermeta_id field.
When you say "Post belongs to Usermeta" - what this means is that the posts table will have a usermeta_id field.
Notice that these two relations map to the exact same field in the exact same table. Declaring one relation will declare the other by simple congruence. "Usermeta has one Post" and "Post belongs to Usermeta" are identical relations.
A Tweak to Your Relations
There's one other relation that share this same schema (the posts table have a usermeta_id field). That is "Usermeta has many Posts". The difference here is not in how the relations are stored to the database, but in how Laravel interprets the relations and in what queries Laravel will run.
When you say "Usermeta has one Post", Laravel will scan the database for the first Post with a matching usermeta_id and return that as an instance of the Usermeta model.
When you say "Usermeta has many Posts", Laravel will scan the database for all matching usermeta_ids and return them as a Collection of Usermeta models. You likely want this second behavior -- otherwise users won't be able to make a second post after they sign up.
Setting the usermeta_id Field
Laravel allows you to set database fields directly through a relationship. See their documentation on inserting related models for details.
Because many relationships are just ciphers for the same underlying schema, there's no need to insert or update a related model both ways. For instance, suppose we had the following two models:
class User extends Eloquent {
public function posts() {
return $this->hasMany("App\Post");
}
}
class Post extends Eloquent {
public function user() {
return $this->belongsTo("App\User");
}
}
In this case, the following two lines of code are identical and you only need to use one of them:
$post->user()->associate($user);
$user->posts()->save($post);
Both of these will have the same effect (setting the user_id field on the posts table)
The reason I mention this is that it looks like you're trying to double-dip in your code. You're using attach() (conceivably to set the usermeta_id) and you're also setting the usermeta_id directly. I've added a side-note on the attach method below - as I don't believe it's the right method, anyway.
To use Laravel's relations, you would want code like the following to set this field:
public function postStoryUpload($request, $response, $id)
{
//set up our validation rules for our complete sign up form
$validation = $this->validator->validate($request, [
'title' => v::stringType()->notEmpty()->length(1, 80),
'body' => v::stringType()->notEmpty()->length(1, 2500),
]);
//if validation fails, stay on story upload page
if ($validation->failed()) {
return $response->withRedirect($this->router>pathFor('storyupload.upload'));
}
$user = Usermeta::find($id)->first();
//We can use our Post Model to send the form data to the database
$post = Post::create([
'title' => $request->getParam('title'),
'body' => $request->getParam('body'),
'category' => $request->getParam('category'),
'files' => $request->getParam('img_path'),
]);
// Set the usermeta_id field
$post->usermeta()->associate($user);
// Save the model so we write changes to the database
$post->save();
//after submit, redirect to completesignup page
return $response->withRedirect($this->router->pathFor('auth.completesignup'));
}
Manually Setting the usermeta_id Field
Instead of using Laravel's relations to set this field, you can set the field manually. This can sometimes be cleaner, but it's less explicit and can lead to minor bugs if you aren't careful. To do this, you need to treat the usermeta_id field like any other field on your model.
$post->usermeta_id = $user->id;
This also works when mass assigning attributes using fill or create like so:
$post = \App\Post::create([
'title' => $title,
'body' => $body,
'usermeta_id' => $user->id
]);
$post->fill([
'title' => $title,
'body' => $body,
'usermeta_id' => $user->id
]);
Note that when manually setting the usermeta_id like this, you do not need to use any relationship methods. The following code is redundant:
$post->usermeta_id = $user->id;
$post->usermeta()->associate($user);
Your Issue (I Believe)
There's a caveat to mass assignment, however. Per the Laravel documentation, mass assignment requires you to fill out the model's fillable or guarded attributes.
This is one of the most common bugs, if not the most common bug, in any Laravel code - and it doesn't throw an obvious error so it's easy to miss. Consider the following model:
class Post extends Eloquent {
private $fillable = ["title", "body"];
}
If you attempt to mass assign the usermeta_id field like so:
$post = \App\Post::create([
'title' => $title,
'body' => $body,
'usermeta_id' => $user->id
]);
Then it will silently fail. No error is thrown and the Post is created but the usermeta_id field will be NULL - because it's not mass assignable. This is fixed by updating your model like so:
class Post extends Eloquent {
private $fillable = ["title", "body", "usermeta_id"];
}
I will repeat again, as I did above, that if using mass assignment like this you do not not need to use the associate or save relationship methods. This would be redundant. Therefore you can just set usermeta_id directly to $user->id without any of the usermeta()->associate() shenanigans.
The Bugs I Mentioned
I mentioned that manually setting the field like this can cause bugs. So let's actually discuss what some of those bugs are now instead of glossing over them.
If you update the relationship field manually, Laravel will be unaware that the two models are related until it reloads the model from the database. Consider the following two chunks of code:
$post = new Post();
$post->usermeta_id = $user->id;
dd( $post->usermeta->name );
$post = new Post();
$post->usermeta()->associate($user);
dd( $post->usermeta->name );
The first code block will fail, throwing the error "cannot read attribute of null object" -- because as far as Laravel is aware, $post->usermeta is NULL. You set $post->usermeta_id, but you didn't set $post->usermeta.
The second code block will work as expected, because by running the associate function it sets both usermeta_id and usermeta.
95% of the time this doesn't really cause any issues, however. If you're using an asynchronous API call to save the post and then a separate asynchronous API call to read the post at a later time, then Laravel will read the post from the database and properly set up the relation automatically when we sees the usermeta_id field is filled out.
Side-note On the attach() Method
Laravel uses different methods for saving different types of relations - because the different relations imply different underlying database fields.
associate: This sets the *_id field on the current model's table. For instance: $post->user()->associate($user) will set the user_id on the posts table
save: This sets the *_id field on the other model's table. For instance: $post->comments()->save($comment) will set the post_id on the comments table
attach: This sets both *_id fields on a linking table for many to many relationships. For instance, if you had a tag system then $post->tags()->attach($tag) would set post_id and tag_id on the post_tags table
It can be a bit tricky to remember which of these three functions you need. In general, there's a direct mapping from relation to function:
hasOne, hasMany --> save
belongsTo --> associate
belongsToMany --> attach
I assume that this should all be in one query in order to prevent duplicate data in the database. Is this correct?
How do I simplify this code into one Eloquent query?
$user = User::where( 'id', '=', $otherID )->first();
if( $user != null )
{
if( $user->requestReceived() )
accept_friend( $otherID );
else if( !$user->requestSent() )
{
$friend = new Friend;
$friend->user_1= $myID;
$friend->user_2 = $otherID;
$friend->accepted = 0;
$friend->save();
}
}
I assume that this should all be in one query in order to prevent
duplicate data in the database. Is this correct?
It's not correct. You prevent duplication by placing unique constraints on database level.
There's literally nothing you can do in php or any other language for that matter, that will prevent duplicates, if you don't have unique keys on your table(s). That's a simple fact, and if anyone tells you anything different - that person is blatantly wrong. I can explain why, but the explanation would be a lengthy one so I'll skip it.
Your code should be quite simple - just insert the data. Since it's not exactly clear how uniqueness is handled (it appears to be user_2, accepted, but there's an edge case), without a bit more data form you - it's not possible to suggest a complete solution.
You can always disregard what I wrote and try to go with suggested solutions, but they will fail miserably and you'll end up with duplicates.
I would say if there is a relationship between User and Friend you can simply employ Laravel's model relationship, such as:
$status = User::find($id)->friends()->updateOrCreate(['user_id' => $id], $attributes_to_update));
Thats what I would do to ensure that the new data is updated or a new one is created.
PS: I have used updateOrCreate() on Laravel 5.2.* only. And also it would be nice to actually do some check on user existence before updating else some errors might be thrown for null.
UPDATE
I'm not sure what to do. Could you explain a bit more what I should do? What about $attributes_to_update ?
Okay. Depending on what fields in the friends table marks the two friends, now using your example user_1 and user_2. By the example I gave, the $attributes_to_update would be (assuming otherID is the new friend's id):
$attributes_to_update = ['user_2' => otherID, 'accepted' => 0 ];
If your relationship between User and Friend is set properly, then the user_1 would already included in the insertion.
Furthermore,on this updateOrCreate function:
updateOrCreate($attributes_to_check, $attributes_to_update);
$attributes_to_check would mean those fields you want to check if they already exists before you create/update new one so if I want to ensure, the check is made when accepted is 0 then I can pass both say `['user_1' => 1, 'accepted' => 0]
Hope this is clearer now.
I'm assuming "friends" here represents a many-to-many relation between users. Apparently friend requests from one user (myID) to another (otherId).
You can represent that with Eloquent as:
class User extends Model
{
//...
public function friends()
{
return $this->belongsToMany(User::class, 'friends', 'myId', 'otherId')->withPivot('accepted');
}
}
That is, no need for Friend model.
Then, I think this is equivalent to what you want to accomplish (if not, please update with clarification):
$me = User::find($myId);
$me->friends()->syncWithoutDetaching([$otherId => ['accepted' => 0]]);
(accepted 0 or 1, according to your business logic).
This sync method prevents duplicate inserts, and updates or creates any row for the given pair of "myId - otherId". You can set any number of additional fields in the pivot table with this method.
However, I agree with #Mjh about setting unique constraints at database level as well.
For this kind of issue, First of all, you have to enjoy the code and database if you are working in laravel. For this first you create realtionship between both table friend and user in database as well as in Models . Also you have to use unique in database .
$data= array('accepted' => 0);
User::find($otherID)->friends()->updateOrCreate(['user_id', $otherID], $data));
This is query you can work with this . Also you can pass multiple condition here. Thanks
You can use firstOrCreate/ firstOrNew methods (https://laravel.com/docs/5.3/eloquent)
Example (from docs) :
// Retrieve the flight by the attributes, or create it if it doesn't exist...
$flight = App\Flight::firstOrCreate(['name' => 'Flight 10']);
// Retrieve the flight by the attributes, or instantiate a new instance...
$flight = App\Flight::firstOrNew(['name' => 'Flight 10']);
use `firstOrCreate' it will do same as you did manually.
Definition of FirstOrCreate copied from the Laravel Manual.
The firstOrCreate method will attempt to locate a database record using the given column / value pairs. If the model can not be found in the database, a record will be inserted with the given attributes.
So according to that you should try :
$user = User::where( 'id', '=', $otherID )->first();
$friend=Friend::firstOrCreate(['user_id' => $myId], ['user_2' => $otherId]);
It will check with both IDs if not exists then create record in friends table.
I am using PHP Yii framework's Active Records to model a relation between two tables. The join involves a column and a literal, and could match 2+ rows but must be limited to only ever return 1 row.
I'm using Yii version 1.1.13, and MySQL 5.1.something.
My problem isn't the SQL, but how to configure the Yii model classes to work in all cases. I can get the classes to work sometimes (simple eager loading) but not always (never for lazy loading).
First I will describe the database. Then the goal. Then I will include examples of code I've tried and why it failed.
Sorry for the length, this is complex and examples are necessary.
The database:
TABLE sites
columns:
id INT
name VARCHAR
type VARCHAR
rows:
id name type
-- ------- -----
1 Site A foo
2 Site B bar
3 Site C bar
TABLE field_options
columns:
id INT
field VARCHAR
option_value VARCHAR
option_label VARCHAR
rows:
id field option_value option_label
-- ----------- ------------- -------------
1 sites.type foo Foo Style Site
2 sites.type bar Bar-Like Site
3 sites.type bar Bar Site
So sites has an informal a reference to field_options where:
field_options.field = 'sites.type' and
field_options.option_value = sites.type
The goal:
The goal is for sites to look up the relevant field_options.option_label to go with its type value. If there happens to be more than one matching row, pick only one (any one, doesn't matter which).
Using SQL this is easy, I can do it 2 ways:
I can join using a subquery:
SELECT
sites.id,
f1.option_label AS type_label
FROM sites
LEFT JOIN field_options AS f1 ON f1.id = (
SELECT id FROM field_options
WHERE
field_options.field = 'sites.type'
AND field_options.option_value = sites.type
LIMIT 1
)
Or I can use a subquery as a column reference in the select clause:
SELECT
sites.id,
(
SELECT id FROM field_options
WHERE
field_options.field = 'sites.type'
AND field_options.option_value = sites.type
LIMIT 1
) AS type_label
FROM sites
Either way works great. So how do I model this in Yii??
What I've tried so far:
1. Use "on" array key in relation
I can get a simple eager lookup to work with this code:
class Sites extends CActiveRecord
{
...
public function relations()
{
return array(
'type_option' => array(
self::BELONGS_TO,
'FieldOptions', // that's the class for field_options
'', // no normal foreign key
'on' => "type_option.id = (SELECT id FROM field_options WHERE field = 'sites.type' AND option_value = t.type LIMIT 1)",
),
);
}
}
This works when I load a set of Sites objects and force it to eager load type_label, e.g. Sites::model()->with('type_label')->findByPk(1).
It does not work if type_label is lazy-loaded.
$site = Sites::model()->findByPk(1);
$label = $site->type_option->option_label; // ERROR: column t.type doesn't exist
2. Force eager loading always
Building on #1 above, I tried forcing Yii to always to eager loading, never lazy loading:
class Sites extends CActiveRecord
{
public function relations()
{
....
}
public function defaultScope()
{
return array(
'with' => array( 'type_option' ),
);
}
}
Now everything always works when I load Sites, but it's no good because there are other models (not pictured here) that have relations that point to Sites, and those result in errors:
$site = Sites::model()->findByPk(1);
$label = $site->type_option->option_label; // works now
$other = OtherModel::model()->with('site_relation')->findByPk(1); // ERROR: column t.type doesn't exist, because 't' refers to OtherModel now
3. Make the reference to the base table somehow relative
If there was a way that I could refer to the base table, other than "t", that was guaranteed to point to the correct alias, that would work, e.g.
'on' => "type_option.id = (SELECT id FROM field_options WHERE field = 'sites.type' AND option_value = %%BASE_TABLE%%.type LIMIT 1)",
where %%BASE_TABLE%% always refers to the correct alias for table sites. But I know of no such token.
4. Add a true virtual database column
This way would be the best, if I could convince Yii that the table has an extra column, which should be loaded just like every other column, except the SQL is a subquery -- that would be awesome. But again, I don't see any way to mess with the column list, it's all done automatically.
So, after all that... does anyone have any ideas?
EDIT Mar 21/15: I just spent a long time investigating the possibility of subclassing parts of Yii to get the job done. No luck.
I tried creating a new type of relation based on BELONGS_TO (class CBelongsToRelation), to see if I could somehow add in context sensitivity so it could react differently depending on whether it was being lazy-loaded or not. But Yii isn't built that way. There is no place where I can hook in code during query buiding from inside a relation object. And there is also no way I can tell even what the base class is, relation objects have no link back to the parent model.
All of the code that assembles these queries for active records and their relations is locked up in a separate set of classes (CActiveFinder, CJoinQuery, etc.) that cannot be extended or replaced without replacing the entire AR system pretty much. So that's out.
I then tried to see if I can create "fake" database column entries that would actually be a subquery. Answer: no. I figured out how I could add additional columns to Yii's automatically generated schema data. But,
a) there's no way to define a column in such a way that it can be a derived value, Yii assumes it's a column name in way too many places for that; and
b) there also doesn't appear to be any way to avoid having it try to insert/update to those columns on save.
So it really is looking like Yii (1.x) just does not have any way to make this happen.
Limited solution provided by #eggyal in comments: #eggyal has a suggestion that will meet my needs. He suggests creating a MySQL view table to add extra columns for each label, using a subquery to look up the value. To allow editing, the view would have to be tied to a separate Yii class, so the downside is everywhere in my code I need to be aware of whether I'm loading a record for reading only (must use the view's class) or read/write (must use the base table's class, does not have the extra columns). That said, it is a workable solution for my particular case, maybe even the only solution -- although not an answer to this question as written, so I'm not going to put it in as an answer.
OK, after a lot of attempts, I have found a solution. Thanks to #eggyal for making me think about database views.
As a quick recap, my goal was:
link one Yii model (CActiveRecord) to another using a relation()
the table join is complex and could match more than one row
the relation must never join more than one row (i.e. LIMIT 1)
I got it to work by:
creating a view from the field_options base table, using SQL GROUP BY to eliminate duplicate rows
creating a separate Yii model (CActiveRecord class) for the view
using the new model/view for the relation(), not the original table
Even then there were some wrinkles (maybe a Yii bug?) I had to work around.
Here are all the details:
The SQL view:
CREATE VIEW field_options_distinct AS
SELECT
field,
option_value,
option_label
FROM
field_options
GROUP BY
field,
option_value
;
This view contains only the columns I care about, and only ever one row per field/option_value pair.
The Yii model class:
class FieldOptionsDistinct extends CActiveRecord
{
public function tableName()
{
return 'field_options_distinct'; // the view
}
/*
I found I needed the following to override Yii's default table data.
The view doesn't have a primary key, and that confused Yii's AR finding system
and resulted in a PHP "invalid foreach()" error.
So the code below works around it by diving into the Yii table metadata object
and manually setting the primary key column list.
*/
private $bMetaDataSet = FALSE;
public function getMetaData()
{
$oMetaData = parent::getMetaData();
if (!$this->bMetaDataSet) {
$oMetaData->tableSchema->primaryKey = array( 'field', 'option_value' );
$this->bMetaDataSet = TRUE;
}
return $oMetaData;
}
}
The Yii relation():
class Sites extends CActiveRecord
{
// ...
public function relations()
{
return (
'type_option' => array(
self::BELONGS_TO,
'FieldOptionsDistinct',
array(
'type' => 'option_value',
),
'on' => "type_option.field = 'sites.type'",
),
);
}
}
And all that does the trick. Easy, right?!?
I new in yii framework. i create an application in yii framework. i created model, controller, views using gii. After that i alter database table. I deleted 2 column and add 3 new columns. After that overwrite the model using the gii. But when i am trying to save into that table it show property(which was old column that I deleted) is not defined. Plz provide me a solution for this.
You need to define all columns in the validation rules() method in your model, have a look and make sure that you have defined a rule for every column in the table there, for example (if it's a string with max length 128):
public function rules()
{
return array(
...
array('myField', 'length', 'max'=>128),
...
);
}
See some info about validation rules.
Also, for forms if you're using CActiveForm widget and calling fields like so:
echo $form->labelEx($model,'myField');
echo $form->textField($model,'myField');
Then you'll need to make sure that a label is defined in the model too, in the attributeLabels() method, for example:
public function attributeLabels()
{
return array(
...
'myField'=>'My Field',
...
);
}
Lastly, if you want the field to be searchable, you'll need to add a statement to the search() method in the model, for example:
public function search()
{
...
$criteria->compare('myField',$this->myField);
...
}
Make sure you have all of those elements present and you shouldn't get the '* is not defined' error.
Also, if you're using schema caching in your main config file, you'll have to clear your cache before the app will see your new database structure.
Your changes should also be set at the Views since there are forms, widgets using the old properties !! (for this exact save issue, you will need to fix _form.php which is the partial responsible from your model Save & Update actions.
You can either do the same as you did with the model: (regenerate it using gii) or you can edit it manually (i recommend you get used to this since in the future you will have code you don't want to loose just because of altering a column name. simple Find & edit in most of the text editors will do the job).
May be you need to read a bit more about how MVC works in general & in Yii in special
This is because you are using schema-cache. Your table schema is cached in Yii. You need to flush AR cache. Either flush full schema cache or use
Yii::app()->db->schema->getTable('tablename', true); in start of your action. This will update model schema-cache.