I can get the results using this query in mysql.
SELECT group_concat(im.instrument_name) AS instrument
FROM ot_note otn
LEFT JOIN ot_instrument_entry oie ON otn.id=oie.ot_note_id
LEFT JOIN instrument_master im ON oie.instrument_name=im.id
GROUP BY otn.id
Now I have a model OTNote, InstrumentMaster and OTInstrumentEntry.
I have tried to use a relation using via table, but it looks like I am missing something.
I have tried to create the relation like this, but it is throwing error;
public function getInstrumentMaster()
{
return $this
->hasMany(OtInstrumentEntry::className(), ['ot_note_id' => 'id'])
->viaTable('instrument_master', ['id'=>'instrument_name']);
}
How can I can create a relation so I can access the column instrument_master.instrument_name column which is related to ot_instrument_entry?
ot_instrument_entry is the linking table. In class OTNote you should have:
// class OTNote
public function getInstrumentMaster()
{
return $this
->hasMany(InstrumentMaster::className(), ['id' => 'instrument_name'])
->viaTable('ot_instrument_entry', ['ot_note_id' => 'id']);
}
In class InstrumentMaster you may want to have this for vice versa access:
// class InstrumentMaster
public function getOTNotes()
{
return $this
->hasMany(OTNote::className(), ['id' => 'ot_note_id'])
->viaTable('ot_instrument_entry', ['instrument_name' => 'id']);
}
Details can be found here.
Related
When creating an entry using create() - Why does it return a relationship of pilot_id table instead of just showing the value of pilot_id?
For example, in the repository class:
public function createFlight()
$flight = App\Flight::create([
'name' => 'Flight 10'
'pilot_id' => 4
]);
return $flight;
}
In the controller:
public function show()
{
$data = $this->flight->createFlight();
return $data
}
It would return json on the screen but it is not showing the relationship table (pilot) of pilot_id.
Try adding the following to your Flight model, like so. By default you need to tell Laravel to include any relationships.
protected $with = ['pilot'];
That will make it so everytime it includes the relationship. If this is not desirable, then you will want to load the relationships when you return the flight, like so.
return $flight->load(['pilot']);
It shows pilot_id is 4 because that's what its value is. Did you create a relationship on the Flight so that Laravel knows how to retrieve the model for Pilot? It should look something like this:
public function pilot()
{
return $this->hasOne('App\Pilot');
}
When you return a model directly from the controller, it invokes the toJson() method to convert the object to a string. If you want to append the contents of a related model you can do so by adding the relationship to the $with variable on the Flight model.
protected $with = ['pilot']
I am new to yii2. I have read the documentation and some answers on sof but still I cant get to work with relations in yii2. I am able to create raw mysql query for the problem but I dont know how to create the same query using yii2 relations. I am confused with via, joinWith and some key concepts. I will make the problem as descriptive as possible.
I have four models.
Category, CategoryNews, NewsTags, Tags
category table - cat_id, cat_name
news_category table - nc_id, nc_cat_id, nc_news_id
news_tags table - nt_id, nt_news_id, nt_tag_id
tags table - tag_id, tag_name
What I need is tags model object for each category, that is for each category i need all news tags belonging to that category. Request is from gridview.
The generated relations are:
Category Model:
public function getNewsCategory()
{
return $this->hasMany(NewsCategory::className(), ['nc_cat_id' => 'cat_id']);
}
NewsCategory Model:
public function getNcNews()
{
return $this->hasOne(News::className(), ['news_id' => 'nc_news_id']);
}
public function getNcCat()
{
return $this->hasOne(Category::className(), ['cat_id' => 'nc_cat_id']);
}
NewsTags Model:
public function getNtNews()
{
return $this->hasOne(News::className(), ['news_id' => 'nt_news_id']);
}
public function getNtTag()
{
return $this->hasOne(Tags::className(), ['tag_id' => 'nt_tag_id']);
}
News Model:
public function getNewsCategory()
{
return $this->hasMany(NewsCategory::className(), ['nc_news_id' => 'news_id']);
}
public function getNewsTags()
{
return $this->hasMany(NewsTags::className(), ['nt_news_id' => 'news_id']);
}
Tags Model:
public function getNewsTags()
{
return $this->hasMany(NewsTags::className(), ['nt_tag_id' => 'tag_id']);
}
ie. each category contains multiple news and each news contain mutiple tags and I need all tags related to each category.
More precisely, on the gridview I need all categories and a column displaying all tags related to these categories.
Please help!!
You can avoid declaration of models for junction tables, using viaTable syntax for many-to-many relations. Then your code will contain only three models (Category, News and Tag) and everything will be much simplier.
Your code for AR models and relations could looks as follows:
public class Category extends ActiveRecord
{
public function getNews()
{
return $this->hasMany(News::className(), ['id' => 'news_id'])
->viaTable('news_category_table', ['category_id' => 'id']);
}
}
public class News extends ActiveRecord
{
public function getCategories()
{
return $this->hasMany(Category::className(), ['id' => 'category_id'])
->viaTable('news_category_table', ['news_id' => 'id']);
}
public function getTags()
{
return $this->hasMany(Tags::className(), ['id' => 'tag_id'])
->viaTable('news_tags_table', ['news_id' => 'id']);
}
}
public class Tag extends ActiveRecord
{
public function getNews()
{
return $this->hasMany(News::className(), ['id' => 'news_id'])
->viaTable('news_tags_table', ['tag_id' => 'id']);
}
}
These relations you can use in link and unlink functions (rows in junction tables will be managed by Yii in backround). But keep in mind that you should use TRUE as second param in unlink() to remove row in junction table:
$article = new News();
$tag = new Tag();
$tag->save();
$article->link('tags', $tag);
$article->link('caterories', $category);
OR vice versa
$tag->link('news', $article);
$category->link('news', $article);
To get all tags in given category you can declare following function in Category class:
public function getTags()
{
return Tags::find()
->joinWith(['news', 'news.categories C'])
->where(['C.id' => $this->id])
->distinct();
}
This will work as relation query and you can use it as $category->tags or as $category->getTags()->count() or any other way (but not in link and unlink functions).
P.S. To use provided example in your code You should first change names, because I used singular form for AR classes names (Tag) and short notation for primary and foreign keys (id, tag_id etc). And I'd recommend you also to use such naming approach in your code and DB structure.
P.P.S. This example code wasn't tested so be careful :)
I have three models: Person, Feature and PersonFeature. PersonFeature is a junction table with two foreign keys, person_id referencing id in the person table and feature_id referencing id in the feature table.
My question is does Gii in Yii2 generate all the relevant relations. These are the relation in each of the three models
Person:
public function getPersonfeatures()
{
return $this->hasMany(Personfeature::className(), ['personid' => 'id']);
}
Feature:
public function getPersonfeatures()
{
return $this->hasMany(Personfeature::className(), ['featureid' => 'id']);
}
PersonFeature:
public function getPerson()
{
return $this->hasOne(Person::className(), ['id' => 'personid']);
}
public function getFeature()
{
return $this->hasOne(Feature::className(), ['id' => 'featureid']);
}
But when I browse other posts I see that there exists a 'viaTable' operation for example:
public function getPerson() {
return $this->hasMany(Person::className(), ['id' => 'personid'])
->viaTable('personfeature', ['featureid' => 'id']);}
So basically my question is, is Yii supposed to generate this for me? Or can I manually add it in?
Cheers
The last function (with viaTable) is a many to many relationship, that function can be used just as any other relational function (for instance in a ->with() query).
You don't need a model for your join table, unless you want to use it for something else.
Hope this helps.
I'm trying to get data from a join table in Yii2 without an additional query. I have 2 models (User, Group) associated via the junction table (user_group). In the user_group table, I want to store extra data (admin flag, ...) for this relation.
What's the best way to add data to the junction table? The link method accepts a parameter extraColumns but I can't figure out how this works.
What's the best way to retrieve this data? I wrote an additional query to get the values out of the junction table. There must be a cleaner way to do this?!
FYI, this is how I defined the relation in the models:
Group.php
public function getUsers() {
return $this->hasMany(User::className(), ['id' => 'user_id'])
->viaTable('user_group', ['group_id' => 'id']);
}
User.php
public function getGroups() {
return $this->hasMany(Group::className(), ['id' => 'group_id'])
->viaTable('user_group', ['user_id' => 'id']);
}
In short: Using an ActiveRecord for the junction table like you suggested is IMHO the right way because you can set up via() to use that existing ActiveRecord. This allows you to use Yii's link() method to create items in the junction table while adding data (like your admin flag) at the same time.
The official Yii Guide 2.0 states two ways of using a junction table: using viaTable() and using via() (see here). While the former expects the name of the junction table as parameter the latter expects a relation name as parameter.
If you need access to the data inside the junction table I would use an ActiveRecord for the junction table as you suggested and use via():
class User extends ActiveRecord
{
public function getUserGroups() {
// one-to-many
return $this->hasMany(UserGroup::className(), ['user_id' => 'id']);
}
}
class Group extends ActiveRecord
{
public function getUserGroups() {
// one-to-many
return $this->hasMany(UserGroup::className(), ['group_id' => 'id']);
}
public function getUsers()
{
// many-to-many: uses userGroups relation above which uses an ActiveRecord class
return $this->hasMany(User::className(), ['id' => 'user_id'])
->via('userGroups');
}
}
class UserGroup extends ActiveRecord
{
public function getUser() {
// one-to-one
return $this->hasOne(User::className(), ['id' => 'user_id']);
}
public function getGroup() {
// one-to-one
return $this->hasOne(Group::className(), ['id' => 'userh_id']);
}
}
This way you can get the data of the junction table without additional queries using the userGroups relation (like with any other one-to-many relation):
$group = Group::find()->where(['id' => $id])->with('userGroups.user')->one();
// --> 3 queries: find group, find user_group, find user
// $group->userGroups contains data of the junction table, for example:
$isAdmin = $group->userGroups[0]->adminFlag
// and the user is also fetched:
$userName = $group->userGroups[0]->user->name
This all can be done using the hasMany relation. So you may ask why you should declare the many-to-many relation using via(): Because you can use Yii's link() method to create items in the junction table:
$userGroup = new UserGroup();
// load data from form into $userGroup and validate
if ($userGroup->load(Yii::$app->request->post()) && $userGroup->validate()) {
// all data in $userGroup is valid
// --> create item in junction table incl. additional data
$group->link('users', $user, $userGroup->getDirtyAttributes())
}
I don't know for sure it is best solution. But for my project it will be good for now :)
1) Left join
Add new class attribute in User model public $flag;.
Append two lines to your basic relation but don't remove viaTable this can (and should) stay.
public function getUsers()
{
return $this->hasMany(User::className(), ['id' => 'user_id'])
->viaTable('user_group', ['group_id' => 'id'])
->leftJoin('user_group', '{{user}}.id=user_id')
->select('{{user}}.*, flag') //or all ->select('*');
}
leftJoin makes possible to select data from junction table and with select to customize your return columns.
Remember that viaTable must stay because link() relies on it.
2) sub-select query
Add new class attribute in User model public $flag;
And in Group model modified getUsers() relation:
public function getUsers()
{
return $this->hasMany(User::className(), ['id' => 'user_id'])
->viaTable('user_group', ['group_id' => 'id'])
->select('*, (SELECT flag FROM user_group WHERE group_id='.$this->id.' AND user_id=user.id LIMIT 1) as flag');
}
As you can see i added sub-select for default select list. This select is for users not group model. Yes, i agree this is litle bit ugly but does the job.
3) Condition relations
Different option is to create one more relation for admins only:
// Select all users
public function getUsers() { .. }
// Select only admins (users with specific flag )
public function getAdmins()
{
return $this->hasMany(User::className(), ['id' => 'user_id'])
->viaTable('user_group', ['group_id' => 'id'],
function($q){
return $q->andWhere([ 'flag' => 'ADMIN' ]);
});
}
$Group->admins - get users with specific admin flag. But this solution doesn't add attribute $flag. You need to know when you select only admins and when all users. Downside: you need to create separate relation for every flag value.
Your solution with using separate model UserGroup still is more flexible and universal for all cases. Like you can add validation and basic ActiveRecord stuff. These solutions are more one way direction - to get stuff out.
Since I have received no answer for almost 14 days, I'll post how I solved this problem. This is not exactly what I had in mind but it works, that's enough for now. So... this is what I did:
Added a model UserGroup for the junction table
Added a relation to Group
public function getUserGroups()
{
return $this->hasMany(UserGroup::className(), ['user_id' => 'id']);
}
Joined UserGroup in my search model function
$query = Group::find()->where('id =' . $id)->with('users')->with('userGroups');
This get's me what I wanted, the Group with all Users and, represented by my new model UserGroup, the data from the junction table.
I thought about extending the query building Yii2 function first - this might be a better way to solve this. But since I don't know Yii2 very well yet, I decided not to do for now.
Please let me know if you have a better solution.
For that purpose I've created a simple extension, that allows to attach columns in junction table to child model in relation as properties.
So after setting up this extension you will be able to access junction table attributes like
foreach ($parentModel->relatedModels as $childModel)
{
$childModel->junction_table_column1;
$childModel->junction_table_column2;
....
}
For more info please have look at
Yii2 junction table attributes extension
Thanks.
I have a model Listing that inherits through its belongsTo('Model') relationship should inherently belong to the Manufacturer that its corresponding Model belongs to.
Here's from my Listing model:
public function model()
{
return $this->belongsTo('Model', 'model_id');
}
public function manufacturer()
{
return $this->belongsTo('Manufacturer', 'models.manufacturer_id');
/*
$manufacturer_id = $this->model->manufacturer_id;
return Manufacturer::find($manufacturer_id)->name;*/
}
and my Manufacturer model:
public function listings()
{
return $this->hasManyThrough('Listing', 'Model', 'manufacturer_id', 'model_id');
}
public function models()
{
return $this->hasMany('Model', 'manufacturer_id');
}
I am able to echo $listing->model->name in a view, but not $listing->manufacturer->name. That throws an error. I tried the commented out 2 lines in the Listing model just to get the effect so then I could echo $listing->manufacturer() and that would work, but that doesn't properly establish their relationship. How do I do this? Thanks.
Revised Listing model (thanks to answerer):
public function model()
{
return $this->belongsTo('Model', 'model_id');
}
public function manufacturer()
{
return $this->belongsTo('Model', 'model_id')
->join('manufacturers', 'manufacturers.id', '=', 'models.manufacturer_id');
}
I found a solution, but it's not extremely straight forward. I've posted it below, but I posted what I think is the better solution first.
You shouldn't be able to access manufacturer directly from the listing, since manufacturer applies to the Model only. Though you can eager-load the manufacturer relationships from the listing object, see below.
class Listing extends Eloquent
{
public function model()
{
return $this->belongsTo('Model', 'model_id');
}
}
class Model extends Eloquent
{
public function manufacturer()
{
return $this->belongsTo('manufacturer');
}
}
class Manufacturer extends Eloquent
{
}
$listings = Listing::with('model.manufacturer')->all();
foreach($listings as $listing) {
echo $listing->model->name . ' by ' . $listing->model->manufacturer->name;
}
It took a bit of finagling, to get your requested solution working. The solution looks like this:
public function manufacturer()
{
$instance = new Manufacturer();
$instance->setTable('models');
$query = $instance->newQuery();
return (new BelongsTo($query, $this, 'model_id', $instance->getKeyName(), 'manufacturer'))
->join('manufacturers', 'manufacturers.id', '=', 'models.manufacturer_id')
->select(DB::raw('manufacturers.*'));
}
I started off by working with the query and building the response from that. The query I was looking to create was something along the lines of:
SELECT * FROM manufacturers ma
JOIN models m on m.manufacturer_id = ma.id
WHERE m.id in (?)
The query that would be normally created by doing return $this->belongsTo('Manufacturer');
select * from `manufacturers` where `manufacturers`.`id` in (?)
The ? would be replaced by the value of manufacturer_id columns from the listings table. This column doesn't exist, so a single 0 would be inserted and you'd never return a manufacturer.
In the query I wanted to recreate I was constraining by models.id. I could easily access that value in my relationship by defining the foreign key. So the relationship became
return $this->belongsTo('Manufacturer', 'model_id');
This produces the same query as it did before, but populates the ? with the model_ids. So this returns results, but generally incorrect results. Then I aimed to change the base table that I was selecting from. This value is derived from the model, so I changed the passed in model to Model.
return $this->belongsTo('Model', 'model_id');
We've now mimic the model relationship, so that's great I hadn't really got anywhere. But at least now, I could make the join to the manufacturers table. So again I updated the relationship:
return $this->belongsTo('Model', 'model_id')
->join('manufacturers', 'manufacturers.id', '=', 'models.manufacturer_id');
This got us one step closer, generating the following query:
select * from `models`
inner join `manufacturers` on `manufacturers`.`id` = `models`.`manufacturer_id`
where `models`.`id` in (?)
From here, I wanted to limit the columns I was querying for to just the manufacturer columns, to do this I added the select specification. This brought the relationship to:
return $this->belongsTo('Model', 'model_id')
->join('manufacturers', 'manufacturers.id', '=', 'models.manufacturer_id')
->select(DB::raw('manufacturers.*'));
And got the query to
select manufacturers.* from `models`
inner join `manufacturers` on `manufacturers`.`id` = `models`.`manufacturer_id`
where `models`.`id` in (?)
Now we have a 100% valid query, but the objects being returned from the relationship are of type Model not Manufacturer. And that's where the last bit of trickery came in. I needed to return a Manufacturer, but wanted it to constrain by themodelstable in the where clause. I created a new instance of Manufacturer and set the table tomodels` and manually create the relationship.
It is important to note, that saving will not work.
$listing = Listing::find(1);
$listing->manufacturer()->associate(Manufacturer::create([]));
$listing->save();
This will create a new Manufacturer and then update listings.model_id to the new manufacturer's id.
I guess that this could help, it helped me:
class Car extends Model
{
public function mechanical()
{
return $this->belongsTo(Mechanical::class);
}
}
class CarPiece extends Model
{
public function car()
{
return $this->belongsTo(Car::class);
}
public function mechanical()
{
return $this->car->mechanical();
}
}
At least, it was this need that made me think of the existence of a belongsToThrough
You can do something like this (Student Group -> Users -> Poll results):
// poll result
public function studentGroup(): HasOneDeep
{
return $this->hasOneDeepFromRelations($this->user(), (new User())->studentGroup());
}