Let's say I have a model with 2 tables : Owner (int: id) and Car (int: id, int:owner_id).
I'm trying to build a validation rule on Car in order to avoid non existing owner_ids to be bound to the Car.owner_id field. I'd like to have this validation rule in the code instead of just using the DB foreign key check, because it allows me to easily display error messages on the form instead on having to process DB exceptions later.
Thus, in my model, I'd like to have something like :
public function rules() {
return array(
array('owner_id', 'in', 'range' => array(11, 12, 13)),
);
}
where 11, 12, 13 are the existing Owner ids.
I can find those ids through code like :
$ars = Yii::app()->db->createCommand("SELECT id FROM owner")->queryAll();
$ids = array();
foreach($ars as $ar) {
$ids[] = $ar['id'];
}
However, I wondered if there is any built in method in Yii allowing to get this array in a more lazy way like "$ids = Owner::model()->findIdsAsArray()" or something similar.
You dont want to do validate a range. Yii provides CExistValidator with alias exist exactly for foreign key check. The range validator may fail if the owner ids are not continuous ie if we delete an intermediate user. So You can use a validation rule as follows
array('owner_id', 'exist', 'attributeName'=>'id', 'className'=>'Owner'),
The above code validates that there exist an owner_id in the model Owner and raises an error if its not exist in the DB.
Related
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?!?
in Yii framework, can I use unique validation rule to check uniqueness of an field within some condition (I know there is criteria, but this condition is a bit tricky)? Ie, i want to check num_days unique by property_id.
table:
NUM PROP_ID
3 4
3 5
validation should pass in case i try insert 3, 6, but fail in case of 3, 4
Check out UniqueAttributesValidator, and also this answer. In the links you'll see that they have used $this->attributename for the params array of the criteria option of CUniqueValidator, but for some reason $this->attributename was null for me. I believe that this is because the $this is not being passed correctly to the validator, so anyway it would be best to use the UniqueAttributesValidator class, because it has been made just for these kinds of situations.
The resulting sql will have a WHERE clause like this:
SELECT ... WHERE (`num`=:value) AND (`prop_id`=:prop_id) ...
which will easily fail for 3, 4 and pass for 3, 6. So it should work for your situation.
First Create unique field in you table
and in model add this to your rules()
array('field_name', 'unique'),
For combination of two fields unique use this code
public function rules() {
return array(
array('firstKey', 'unique', 'criteria'=>array(
'condition'=>'`secondKey`=:secondKey',
'params'=>array(
':secondKey'=>$this->secondKey
)
)),
);
}
Create your custom Validator or validation function, it's simple.
I've this
$reviewModel=Review::model()->findAll();
Now I want to perform a search operation on $reviewModel,say I want to search if a user of id 1 has posted a review or not. so is there any function provided by yii.
Note: I dont want to use
$reviewModel=Review::model()->findAll(array('condition'=>'')); As I
need all the reviews & then perform a search.
This is what I would recommend and is good design practice, also it will allow optimal performance from your database, + reduce the amount of queries. I would make your reviews have a relation with the user. As a user will post many reviews. so user HAS_MANY reviews. and Review and HAS_ONE user. so use that and create a foreign key to build the relationship.
So inside your user model (Relations) id have something like this:
'reviews' => array(self::HAS_MANY, 'Review', 'originator),
now inside your review model id have something like this:
'user' => array(self::BELONGS_TO, 'User', 'originator'),
Once you have done that all you would have to do is loop through all the users reviews. If its null, they don't have any reviews.
$user = User::model()->findbyPk(1);
and user->reviews would contain all the reviews for that user.
Not that I am aware of (if I understand your question correctly). Yii provides various ways to ask for what you want from the DB but not from an array of active record models in memory.
If you want to do it this way then get the array of data as you already did and then use PHP to iterate through the array of models looking for what you want.
You can filter through them manually. For example:
$reviewModel=Review::model()->findAll();
$matched = false;
foreach($reviewModel as $r) //cycle through each review
{
if($r->user == 1) //check if user is desired
{
$matched = true; //set matched as true
break; //stop searching, as we already found one
}
}
if you want to search for multiple things, you can use an array for matched and then check that array.
$reviewModel=Review::model()->findAll();
$wanted = array(0=>false,3=>false,6=>false,8=>false);
foreach($reviewModel as $r) //cycle through each review
{
if(isset($wanted[$r->user]) and !$wanted[$r->user]) //check if user is in wanted list and still false
{
$wanted[$r->user] == true; //set appropiate user to true
}
}
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.