Related
SQL scheme:
bulletins
id increment
deals
id increment
seller_id
buyer_id
deals_items - items = bulletins
id increment
title
desc
bulletin_id
deal_id
How can I get deal row by bulletin id? In raw SQL it looks like:
select `deals`.* from `deals` inner join `deals_items` on `deals_items`.`deal_id` = `deals`.`id` where `deals_items`.`bulletin_id` = 10572
I tried:
public function deals()
{
return $this->hasManyThrough(DealItem::class,Deal::class, 'bulletin_id','dealid','id');
}
But it seems a wrong way. Can't find right way in laravel doc about relation.
#HCK shows right way.
but when I doing $bulletin->deals() in blade template I got empty collection of deals.
When just $bulletin->deal - all is fine, we have collection of deals.
I using protected $with = ['deals'] in my bulletin model, but what is different call method or property? Why with method empty result?
#Amarnasan was close, but the order of the foreign keys was wrong. Try this:
Deal.php
public function bulletins()
{
return $this
->belongsToMany(Bulletin::class, 'deals_items', 'deal_id', 'bulletin_id')
->withPivot('title','desc');
}
Bulletin.php
public function deals()
{
return $this
->belongsToMany(Deal::class, 'deals_items', 'bulletin_id', 'deal_id')
->withPivot('title','desc');
}
From the docs:
As mentioned previously, to determine the table name of the
relationship's joining table, Eloquent will join the two related model
names in alphabetical order. However, you are free to override this
convention. You may do so by passing a second argument to the
belongsToMany method:
return $this->belongsToMany('App\Role', 'role_user');
In addition to customizing the name of the joining table, you may also
customize the column names of the keys on the table by passing
additional arguments to the belongsToMany method. The third argument
is the foreign key name of the model on which you are defining the
relationship, while the fourth argument is the foreign key name of the
model that you are joining to:
return $this->belongsToMany('App\Role', 'role_user', 'user_id', 'role_id');
Update
When you access the relationship as a method: $bulletin->deals() you are accessing the relationship itself. This will return an instance of \Illuminate\Database\Eloquent\Relations\BelongsToMany (in your case). Here the query is not executed yet, so you could keep adding constrains to your query, for example:
$bulletin
->deals()
->where('seller_id', 45) // <---
->skip(5) // <---
-> ... (And so on)
When you access it as a dynamic property, you are already executing the query, so this will return a Collection instance. Is the same as calling the relationship as a method and then attach the ->get() at the end, so this two are equivalent:
$bulletin->deals()->get()
// equals to:
$bulletin->deals
Check this other answer, it answers your question.
DealClass:
public function bulletins()
return $this->belongsToMany('App\Bulletin', 'deals_items ', 'bulletin_id', 'deal_id')->withPivot('title','desc');
}
BulletinClass:
public function deals()
return $this->belongsToMany('App\Deal', 'deals_items ', 'deal_id', 'bulletin_id')->withPivot('title','desc');
}
deals model -
public function bulletins()
return $this->belongsToMany(Bulletin::class, 'deals_items ', 'bulletin_id', 'deal_id');
}
bulletin model:-
public function deals()
{
return $this
->belongsToMany(Deal::class, 'deals_items', 'deal_id', 'bulletin_id',);
}
I am trying to set up a conditional relationship in eloquent model based on the value of table column, but it does not works.
This is the code that I am using:
//RELATIONS
public function task_schedule()
{
if ($this->task_schedule_id === 0) {
return $this->belongsTo('TaskSchedule', 'hold_task_schedule_id', 'id');
} else {
return $this->belongsTo('TaskSchedule');
}
}
Basically I want to use different column to get my child model defined in belongs to relationship.
scopeProfile($query) would not work for me because I dont want to retrieve the child in every single query and also I am using Task->with('task_schedule') in a lot parts of code.
My User model have these fields :
user_id
username
name
family
supervisor
And in that model I defined an accesssor that same name as supervisor attribute like this (because I want to return supervisor user as an User object and not a simple id):
public function getSupervisorAttribute($value)
{
return is_null($value) ? null : User::select('user_id', 'name', 'family')->find($value);
}
In the other hand there is a OneToMany relationship like this:
public function child()
{
return $this->hasMany(self::class, 'supervisor', 'user_id');
}
Now each time I call child() relation it return Illegal offset type error. seems that supervisor field does not recognized in second argument of hasMany method.
There is any way to solve this problem Without having to change accessor name.
I think the problem comes when you try to retrieve the relationship child, why? Because you have an accessor on your supervisor which is a foreign key inside of child relationship, so what happens is when you ask for that relationship, Laravel will try to use your supervisor property, since it has an accessor, it will trigger and instead of getting a desired property (which i guess is an integer), you will either get NULL or a User. I hope this clarifies it for you.
One workaround for this is to add appends attribute to your Model and then put mutators and accessors on that attribute.
If a User has children then it's one to many(he/she can have many children or none)
Anyway,
Lets assume you have a table named Children make sure you change the table name in the model(laravel assumes it's childrens in the DB).
If public function child() {} is in the User model then,
/*
* children since he/she can have many children
* hasMany means this model has many of the other model by self::class
* it's as if you're saying this model has many of this model so change it
*/
public function children()
{
/* you're building a relationship between User('user_id' as local primary key)
* and Children('parent_id' as foreign key)
* means children table has foreign key parent_id(unsignedInt)
* it returns an array of all the children objects of this User row
*/
return $this->hasMany('Children', 'parent_id', 'user_id');
}
On the other hand the Children Model:
<?php
namespace App;
use Illuminate\Database\Eloquent\Model;
class Children extends Model
{
/**
* The table associated with the model.
*
* #var string
*/
protected $table = 'children';
public function parent()
{
// belongsTo means 'parent_id' in this model(Children) relates to 'user_id' on 'User' model
// it returns the User object which is the parent of this child row
return $this->belongsTo('User', 'user_id', 'parent_id');
}
}
This solution is for creating another table however it seems you want it with the same table(it's not very clear edit your post).
// this function makes no sense, it takes an integer and finds the parameter to was given
$userWithIdOne = $user->getSupervisorAttribute(1);
Give us the migrations of the table, show us the relationships.
I have two tables, User and Post. One User can have many posts and one post belongs to only one user.
In my User model I have a hasMany relation...
public function post(){
return $this->hasmany('post');
}
And in my post model I have a belongsTo relation...
public function user(){
return $this->belongsTo('user');
}
Now I want to join these two tables using Eloquent with() but want specific columns from the second table. I know I can use the Query Builder but I don't want to.
When in the Post model I write...
public function getAllPosts() {
return Post::with('user')->get();
}
It runs the following queries...
select * from `posts`
select * from `users` where `users`.`id` in (<1>, <2>)
But what I want is...
select * from `posts`
select id,username from `users` where `users`.`id` in (<1>, <2>)
When I use...
Post::with('user')->get(array('columns'....));
It only returns the column from the first table. I want specific columns using with() from the second table. How can I do that?
Well I found the solution. It can be done one by passing a closure function in with() as second index of array like
Post::query()
->with(['user' => function ($query) {
$query->select('id', 'username');
}])
->get()
It will only select id and username from other table. I hope this will help others.
Remember that the primary key (id in this case) needs to be the first param in the
$query->select() to actually retrieve the necessary results.*
You can do it like this since Laravel 5.5:
Post::with('user:id,username')->get();
Care for the id field and foreign keys as stated in the docs:
When using this feature, you should always include the id column and
any relevant foreign key columns in the list of columns you wish to
retrieve.
For example, if the user belongs to a team and has a team_id as a foreign key column, then $post->user->team is empty if you don't specifiy team_id
Post::with('user:id,username,team_id')->get();
Also, if the user belongs to the post (i.e. there is a column post_id in the users table), then you need to specify it like this:
Post::with('user:id,username,post_id')->get();
Otherwise $post->user will be empty.
For loading models with specific column, though not eager loading, you could:
In your Post model
public function user()
{
return $this->belongsTo('User')->select(['id', 'username']);
}
Original credit goes to Laravel Eager Loading - Load only specific columns
When going the other way (hasMany):
User::with(['post'=>function($query){
$query->select('id','user_id');
}])->get();
Don't forget to include the foreign key (assuming it is user_id in this example) to resolve the relationship, otherwise you'll get zero results for your relation.
In Laravel 5.7 you can call specific field like this
$users = App\Book::with('author:id,name')->get();
It is important to add foreign_key field in the selection.
If you want to get specific columns using with() in laravel eloquent then you can use code as below which is originally answered by #Adam in his answer here in response of this same question, the answer's main code is as below :
Post::with('user:id,username')->get();
So i have used it in my code but it was giving me error of 1052: Column 'id' in field list is ambiguous, so if you guys are also facing same problem
Then for solving it you have to specify table name before the id column in with() method as below code:
Post::with('user:user.id,username')->get();
I came across this issue but with a second layer of related objects. #Awais Qarni's answer holds up with the inclusion of the appropriate foreign key in the nested select statement. Just as an id is required in the first nested select statement to reference the related model, the foreign key is required to reference the second degree of related models; in this example the Company model.
Post::with(['user' => function ($query) {
$query->select('id','company_id', 'username');
}, 'user.company' => function ($query) {
$query->select('id', 'name');
}])->get();
Additionally, if you want to select specific columns from the Post model you would need to include the user_id column in the select statement in order to reference it.
Post::with(['user' => function ($query) {
$query->select('id', 'username');
}])
->select('title', 'content', 'user_id')
->get();
In your Post model:
public function userWithName()
{
return $this->belongsTo('User')->select(array('id', 'first_name', 'last_name'));
}
Now you can use $post->userWithName
There is another alternative you can eager load specific columns
public function show(Post $post)
{
$posts = $post->has('user')->with('user:id,name,email,picture')->findOrFail($post->id);
return view('your_blade_file_path',compact('posts);
}
In your Post model you should have user relationship also
public function user()
{
return $this->belongsTo( User::class, 'user_id')->withDefault();
}
Note: It is mentioned in Laravel docs.
https://laravel.com/docs/8.x/eloquent-relationships#eager-loading-specific-columns
Note that if you only need one column from the table then using 'lists' is quite nice. In my case i am retrieving a user's favourite articles but i only want the article id's:
$favourites = $user->favourites->lists('id');
Returns an array of ids, eg:
Array
(
[0] => 3
[1] => 7
[2] => 8
)
If you use PHP 7.4 or later you can also do it using arrow function so it looks cleaner:
Post::with(['user' => fn ($query) => $query->select('id','username')])->get();
I faced the same issue while using belongsToMany relationship with my user model (Laravel 8.x.x).
After a long search and trial and test method. I found out this answer
You have to make sure you are selecting the id's and any foreign keys that would be needed for the relationship from either side of that relationship. This allows Eloquent to match up parents to their children.
Original credit goes to https://stackoverflow.com/a/64233242/1551102
So I included
Groups::select('groupid')
...
And it worked like a charm. Although now I want to know how to hide the groupid field after fetching.
I know I can simply loop through the array and remove it. But is there any other method? potentially a simpler and better one.
You can also specify columns on related model at the time of accessing it.
Post::first()->user()->get(['columns....']);
You can try this code . It is tested in laravel 6 version.
Controller code
public function getSection(Request $request)
{
Section::with(['sectionType' => function($q) {
$q->select('id', 'name');
}])->where('position',1)->orderBy('serial_no', 'asc')->get(['id','name','','description']);
return response()->json($getSection);
}
Model code
public function sectionType(){
return $this->belongsTo(Section_Type::class, 'type_id');
}
Be careful that if you don't add the key column(s) it won't return anything. If you want to show only the username without the id you could instead define the $visible/$hidden properties within the Model, like so:
app/Models/User.php
protected $visible = ['username'];
Then it will retrieve only username column with:
Post::with('user')->get();
Hiding the key columns:
Alternatively you could hide the key column(s) and then retrieve only the columns you wish.
app/Models/User.php
protected $hidden = ['id'];
Specify which columns you want including the key or else it won't return anything, but this will actually only return the username, because id is $hidden.
Post::with('user:id,username')->get();
Now you can use the pluckmethod on a Collection instance:
This will return only the uuid attribute of the Post model
App\Models\User::find(2)->posts->pluck('uuid')
=> Illuminate\Support\Collection {#983
all: [
"1",
"2",
"3",
],
}
Try with conditions.
$id = 1;
Post::with(array('user'=>function($query) use ($id){
$query->where('id','=',$id);
$query->select('id','username');
}))->get();
So, similar to other solutions here is mine:
// For example you have this relation defined with "user()" method
public function user()
{
return $this->belongsTo('User');
}
// Just make another one defined with "user_frontend()" method
public function user_frontend()
{
return $this->belongsTo('User')->select(array('id', 'username'));
}
// Then use it later like this
$thing = new Thing();
$thing->with('user_frontend');
// This way, you get only id and username,
// and if you want all fields you can do this
$thing = new Thing();
$thing->with('user');
EmployeeGatePassStatus::with('user:id,name')->get();
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.