I have 2 models, Building and Apartment. Building has a column named address and apartment has a column named addresss
I want to add relation like this:
public function building() {
return $this->belongsTo(Building::class, 'addresss', 'address');
}
But then this gives null:
Apartment::first()->building
Even if this works:
Apartment::first()->building()->first()
So what should I do to get it to work using only this:
Apartment::first()->building
If i kept it right the Apartment is part of the building?
That is why you set the Relation in Model Apartment:
public function building() {
return $this->belongsTo(Building::class, 'addresss', 'address');
}
When i use the belongsTo Relation, I just mention the class and do not append anymore inside the brackets like in this case:
public function user()
{
return $this->belongsTo(User::class);
}
What is the purpose of mentioning appartment's addresss aswell in the relation ? Have you set the foreign key's propely & set a hasMany Relationship in the Building-Model ?
Regards.
The issue was I had a column in apartments table named building, removing this fixed it
Related
If I have a table called Master Index which has for example : Country Name = " USA ", and I want to have several models linked to it (GDP,Population,Inequality,etc) , how do I define that list of models in a field so that I can know which properties does that Country has?
Let me know if its possible, thanks!
You can read HERE
First you need to create table but make it in migration that the table you will create is base on the hierarchy example:
If you have Master Index as your general root of relationship young need to make it first in the migration. It will look like this on your Database > Migrations folder.
2020_07_14_0000001_create_master_indexs_table.php
2020_07_14_0000001_create_gdps_table.php
2020_07_14_0000001_create_populations_table.php
2020_07_14_0000001_create_inequalities_table.php
Master Index Model
You will specify the relationship it should look like this:
public function gdps() {
return $this->hasMany(Gdp::class); // if you have different foreign key you can specify it in the next argument return [$this->hasMany(Gdp::class, 'gdp_id');] like this
}
public function populations() {
return $this->hasMany(Population::class);
}
public function inequalities() {
return $this->hasMany(Inequality::class);
}
GDPS Model / Populations Model / Inequality Model
You need to specify where it belongs. It should be like this.
public function master_index() {
return $this->belongsTo(MasterIndex::class);
}
GDPS Migration / Population Migration / Inequality Migration
In your migration you should specify the foreign key.
If you're using Laravel 7.x you can do like this.
$table->foreignId('master_index_id')->constrained()->cascadeOnDelete();
If you're not familiar with the above code you can do also like this:
$table->unsignedBigInteger('master_index_id');
$table->foreign('master_index_id')->references('id')->on('master_indexs')->onDelete('cascade');
I have 2 tables called jobs & job_records. Its relationship as below:
JobRecords Model
public function job()
{
return $this->belongsTo(Job::class);
}
Job Model:
public function jobRecord()
{
return $this->hasOne(JobRecord::class);
}
jobs table has 2 columns that I need to display alongside my job_records table view. It's total_pges & status.
In my JobRecords Controller, I have tried the following method. It throws me an error of Call to undefined relationship.
JobRecordController:
$job_records = JobRecord::whereStatus('In Progress')
->with('jobs', 'jobs.status', 'jobs.total_pges')
->get();
return DataTables::of($job_records)
I am still beginning with Laravel and PHP. I can sense that there is something wrong with the relationship. But I couldn't figure out what it is exactly. Can anyone help me out with this matter?
In your JobRecord model change the relation ship as
public function job()
{
return $this->hasOne('App\Models\Job','foreign_key','local_key');
}
Similarly, in Job model
public function job()
{
return $this->belongsTo('App\Models\JobRecord','foreign_key','local_key');
}
Replace foreign_key and local_key with appropriate values...
I deleted my previous answer. What are you trying to do exactly? You can't use "jobs" in the "with function" without to define "jobs" as function in the model.
If you change it to "job" (instead of "jobs), then it would work, but I don't know if you want this. With your query you saying that a record have many jobs? But your model doesn't define that.
I'm new to laravel relationship so many apologizes if it's just dumb question. I'm using a pivot table named users_email on the project to get Emails of users. Pivot table contains the foreign key Uid and Email_id. Uid references users table
primary key and the same as Email_id. I can get the result while joining them using QueryBuilder.
$recent_inbox_email=DB::table('users_email')->
join('email','users_email.email_id','=','email.Id')->
join('users','users_email.Uid','=','users.Id')->
where('users_email.Uid','=',$Uid)->
where('email.draft','<>','true')->
where('email.trash','<>','true')->
where('email.status','=','unread')->count();
here's how I define the relationship in my models
public function getUid()//User Model
{
return $this->hasMany("User_Email",'Uid');
}
public function getEmId()//Email Model
{
return $this->hasMany("User_Email",'email_id');
}
//User_Email Model
public function email()
{
return $this->belongsTo('Email','Id','email_id');
}
public function user()
{
return $this->belongsTo('User','Id','Uid');
}
Now I want to query something like this using Eloquent
$query= select * from users_email inner join
email on users_email.email_id=email.Id
inner join users on users_email.Uid=users.Id
where users.Id=users_email.Uid limit 0,10
foreach($query as $emails)
{
echo $emails->f_name;
echo $emails->Message
}
DB designer Pic
Link to image
Thanks
There are no dumb questions. I'll try to give you an explanation! I'm not a pro, but maybe I can help.
Laravel uses some conventions that are not mandatory, but if you use them, things work like a charm.
For example, as a general recommendation, tables should be named in plural (your table users is ok. Your "email" table should be "emails"). The model, should be named in singular. This is User.php for table users, Email.php for table emails.
"The pivot table is derived from the alphabetical order of the related model names...", in this case "email_user". I repeat, you are not obliged to name them like this, as you can specify the table for the model setting the $table property in the model.
Once you have set up things like this, you only have to add this to your User model:
public function emails()
{
return $this->belongsToMany('Email');
}
And in your Email model:
public function users()
{
return $this->belongsToMany('User');
}
The "User" and "Email" between parentheses is the name of the related model.
And that's it. You can now do this:
$user = User::find(1);
foreach($user->emails as $email) {
echo $email->subject . '<br>';
echo $email->message . '<br>';
}
If you decide not to follow conventions, you can still use Eloquent relationships. You have to set up the relationship like this:
public function nameOfRelation()
{
return $this->belongsToMany('NameOfRelatedModel', 'name_of_table', 'foreign_key', 'other_key');
}
In the case of the User model for example:
public function emails()
{
return $this->belongsToMany('Email', 'users_email', 'Uid', 'email_id');
}
And in the email model, the other way round.
The answer got long! I didn't test the code, but this should give you an idea!
You can always check the official Laravel documentation, it is really helpful!
http://laravel.com/docs/4.2/eloquent
Hope I helped
I'm creating an online tool for companies that each have a set of users in Laravel.
When a user is connected, he has a $connected_company_id variable
For every SELECT request (called by ::all(), find(), ...), i would like to add the condition: where company_id = $connected_company_id. I have found this post: laravel set an automatic where clause, but it doesn't work by overriding newQuery().
For every INSERT request, i would like to add the company_id.
Is this possible without changing my code inside all the controllers ?
I thought about extending Eloquent with customEloquent, and then make my models extend customEloquent, but I don't know how to write the code for customEloquent and if it could work.
Well, you could make use of the Eloquent Model Events. I assume you have the connected_company_id stored in the Session company_id
class BaseModel extends Eloquent{
public static function boot(){
parent::boot();
//Column to inject when inserting
static::creating(function ($obj){
$obj->company_id = Session::get('company_id');
});
//Column to inject when updating
static::updating(function ($obj){
$obj->company_id = Session::get('company_id');
});
}
}
You can extend the BaseModel class on all the models that you want the company_id to be inserted or updated. Take a look at Eloquent Model Events for more information.
The above code will automatically insert or update the company_id to the model that you extend the BaseModel to. When you do a Model::all() or Model::get(), you automatically get the company_id on that Model and you can also perform searches as you requested on Point `
Hope this helps.
well, you can just add the company id to the find query.
Model::where("company_id","=",$company_id):
Or you can create a scope:
class theModel extends Eloquent {
static $company_id;
static for_company($company_id){
self::company_id=$company_id;
return __CLASS__;
}
public function scopeCompany($query)
{
return $query->where('company_id', '=', self::company_id);
}
}
//And later
$scope=theModel::for_company($company_id);
$res=$scope::company->where(...);
Disclaimer: I haven't tried this. Just a solution I constructed. Let me know if this works. This will not work under PHP 5.3
Table.linkedIndex is related to LinkedIndex.ID. The value of the field LinkedIndex.TableName is either Linked1 or Linked2 and defines which of these tables is related to a row in Table.
Now i want to make a dynamical link with Yii models so that i can easily get from a Table row to the corresponding Linked1 or Linked2 row:
Table.linkedID = [LinkedIndex.TableName].ID
Example
Table values:
LinkedIndex values:
Now I should get the row from Linked2 where ID=2:
$model = Table::model()->findByPk(0);
$row = $model->linked;
Model
In the model Table, I tried to make the relation to the table with the name of the value of linkedIndex.TableName:
public function relations()
{
return array(
'linkedIndex' => array(self::HAS_ONE, 'LinkedIndex', array('ID' => 'linkedIndex')),
'linked' => array(
self::HAS_ONE,
'linkedIndex.TableName',
array('ID' => 'linkedID'),
)
)
}
But then I get the error:
include(linkedIndex.TableName.php) [function.include]: failed to open stream: No such file or directory
Is there any way to make a dynamic relation Table.linkedID -> [LinkedIndex.TableName].ID with Yii Models?
Per the Yii docs here:
http://www.yiiframework.com/doc/api/1.1/CActiveRecord#relations-detail
I'd suggest using self::HAS_ONE instead (unless there can be multiple rows in LinkedIndex with the same ID - although from the looks of above, I doubt that's the case).
You can link tables together that have different keys by following the schema:
foreign_key => primary_key
In case you need to specify custom PK->FK association you can define it as array('fk'=>'pk'). For composite keys it will be array('fk_c1'=>'pk_с1','fk_c2'=>'pk_c2').
so in your case:
public function relations(){
return array(
'linkedIndex' => array(self::HAS_ONE, 'LinkedIndex', array('ID' => 'linkedIndex')),
);
}
where LinkedIndex is the class name for the LinkedIndex model (relative to your Table model - i.e. same folder. You could change that, of course) and array('ID' => 'linkedIndex') specifies the relationship as LinkedIndex.ID = Table.linkedIndex.
Edit
Looking at your updated example, I think you're misunderstanding how the relations function works. You're getting the error
include(linkedIndex.TableName.php) [function.include]: failed to open stream: No such file or directory
because you're trying to create another relation here:
'linked' => array(
self::BELONGS_TO,
'linkedIndex.TableName',
array('ID' => 'linkedID'),
)
This part: linkedIndex.TableName refers to a new model class linkedIndex.TableName, so Yii attempts to load that class' file linkedIndex.TableName.php and throws an error since it doesn't exist.
I think what you're looking for is to be able to access the value TableName within the table LinkedIndex, correct? If so, that's accessible from within the Table model via:
$this->linkedIndex->TableName
This is made possible by the relation we set up above. $this refers to the Table model, linkedIndex refers to the LinkedIndex relation we made above, and TableName is an attribute of that LinkedIndex model.
Edit 2
Per your comments, it looks like you're trying to make a more complex relationship. I'll be honest that this isn't really the way you should be using linking tables (ideally you should have a linking table between two tables, not a linking table that says which 3rd table to link to) but I'll try and answer your question as best as possible within Yii.
Ideally, this relationship should be made from within the LinkedIndex model, since that's where the relationship lies.
Since you're using the table name as the linking factor, you'll need to create a way to dynamically pass in the table you want to use after the record is found.
You can use the LinkedIndex model's afterFind function to create the secondary link after the model is created within Yii, and instantiate the new linked model there.
Something like this for your LinkedIndex model:
class LinkedIndex extends CActiveRecord{
public $linked;
public static function model($className = __CLASS__){
return parent::model($className);
}
public function tableName(){
return 'LinkedIndex';
}
public function afterFind(){
$this->linked = new Linked($this->TableName);
parent::afterFind();
}
//...etc.
}
The afterFind instantiates a new Linked model, and passes in the table name to use. That allows us to do something like this from within the Linked model:
class Linked extends CActiveRecord{
private $table_name;
public function __construct($table_name){
$this->table_name = $table_name;
}
public static function model($className = __CLASS__){
return parent::model($className);
}
public function tableName(){
return $this->table_name;
}
//...etc.
}
which is how we dynamically create a class with interchangeable table names. Of course, this fails of the classes need to have separate operations done per-method, but you could check what the table_name is and act accordingly (that's pretty janky, but would work).
All of this would result in being to access a property of the linked table via (from within the Table model):
$this->linkedIndex->linked->foo;
Because the value of LinkedIndex.TableName and Table.linkedID is needed to get the values, I moved the afterFind, suggested by M Sost, directly into the Table-Class and changed its content accordingly. No more need for a virtual model.
class Table extends CActiveRecord {
public $linked; // Needs to be public, to be accessible
// ...etc.
public function afterFind() {
$model = new $this->linkedIndex->TableName;
$this->linked = $model::model()->findByPk( $this->linkedID );
parent::afterFind();
}
// ...
}
Now I get the row from Linked2 where ID=2:
$model = Table::model()->findByPk(0);
$row = $model->linked;