I have 3 models
MasterParts hasmany Parts
Parts hasmany PickLines
MasterPart.php:
public $displayField = 'mp_part_nr';
Part.php:
Parts display field is another foreign key:
public $displayField = 'master_part_id';
In Picklines I have a dropdown for partnr's referencing the available parts, so I need to show a grouped list of all the partnr.'s of the Parts table.
Unfortunately is shows a list of id's. (note: I created the models, controllers and views using cake bake.)
I would imagine Cake would link the functionality so I do not need to write something like:
public $displayField = 'MasterPart.mp_part_nr';
(which doesn't work anyways)
How can I get a list of partnr's instead of id's?
Figured it out myself, I just need to change the code in my PickLinesController by adding the MasterParts class:
$parts = $this->PickLine->Part->MasterPart->find('list');
Related
I'm trying to create a connection between a JSON field in my database and a table which stores music by ID. So, I have a table called "playlists" which has a field called "songs". In this "songs" field I have a array[] of song ID's e.g. [1,2]. I tried the following code to make a relationship between these two tables:
class Playlist extends Model
{
protected $table = 'playlists';
public function songs()
{
return $this->hasMany('App\Music', 'id');
}
}
I used the foreign_key id because of the songs table which has a id field.
The code I used to retrieve the playlist from the controller is as follows:
$playlist = Playlist::find($id)->songs;
print_r($playlist);
Which outputs:
[1,2]
I most probably did something wrong, not understanding the relationships correctly. Could someone explain how this works? I looked up the documentation but did not get any wiser.
Laravel has no native support for JSON relationships.
I created a package for this: https://github.com/staudenmeir/eloquent-json-relations
If you rename the songs column to song_ids, you can define a many-to-many relationship like this:
class Playlist extends Model
{
use \Staudenmeir\EloquentJsonRelations\HasJsonRelationships;
protected $casts = [
'song_ids' => 'json',
];
public function songs()
{
return $this->belongsToJson('App\Music', 'song_ids');
}
}
class Music extends Model
{
use \Staudenmeir\EloquentJsonRelations\HasJsonRelationships;
public function playlists()
{
return $this->hasManyJson('App\Playlist', 'song_ids');
}
}
Although this is a very old post but I will go ahead and drop my own opinion for my future self and fellow googlers.....
So, If I got this question correctly, you are trying to use a JSON field for a relationship query. This issue I have stumbled across a couple of times, at different occasions for different use-cases. With the most recent being for the purpose of saving a couple of Ids belonging to different tables, in a single JSON field on a given table (While I keep pondering on why the Laravel guy won't just add this functionality already! I Know Pivots, Data Normalization etc....But I'm pleading for the 1%). Until I came across this post on Laracast that worked like a charm.
Apologies for the long intro, let me get right into it....
On your Playlist model (in Laravel 8.0 and a few older versions I can't really keep track of) you can do something like so;
public function songs()
{
$related = $this->hasMany(Song::class);
$related->setQuery(
Song::whereIn('id', $this->song_ids)->getQuery()
);
return $related;
}
I have the really good solution for keeping data in column on json format. It help me on previous project online shop
https://scotch.io/tutorials/working-with-json-in-mysql
So, going into the problem straight away. someone told me that we dont need to make a pivot table if we only want to have ids of the table. laravel can itself handle this situation. I dont know how this works. I have a table community and another table idea. relation is like this;
One community can contain many ideas and an idea can be found in many
communities.
Relation in idea Model:
public function community() {
return $this->belongsToMany('App\Community')->withTimestamps();
}
Relation in community Model:
public function idea() {
return $this->belongsToMany('App\idea');
}
Now i want to fetch all the records related to a single community to show on its page Let's say the community is Arts.
Here is Controller function:
public function showCommunities($id) {
$community = Community::findOrFail($id)->community()->get();
return view('publicPages.ideas_in_community', compact('community'));
}
When i attach ->community()->get() to the Community::findOrFail($id) Then it throws the error
SQLSTATE[42S02]: Base table or view not found laravel
Any help would be appreciated.
Edit:
Logically, this piece of code Community::findOrFail($id)->community()->get() should be like this Community::findOrFail($id)->idea()->get(). Now it is true but it has little issue. it throws an error
Fatal error: Class 'App\idea' not found
The way you define the many-to-many relation looks ok - I'd just call them communities() and ideas(), as they'll return a collection of objects, not a single object.
Make sure you use correct class names - I can see you refering to your model classes using different case - see App\Community and App\idea.
In order to find related models, Eloquent will look for matching rows in the pivot table - in your case it should be named community_idea and have 3 fields: community_id, idea_id and autoincrement primary key id.
With that in place, you should be able to get all ideas linked to given community with:
$ideas = Community::findOrFail($communityId)->ideas;
If you need communities linked to given idea, just do:
$communities = Idea::findOrFail($ideaId)->communities;
You can read more about how to use many-to-many relationships here: https://laravel.com/docs/5.1/eloquent-relationships#many-to-many
someone told me that we dont need to make a pivot table if we only want to have ids of the table
The above is not true (unless I've just misunderstood).
For a many-to-many (belongsToMany) their must be the two related table and then an intermediate (pivot) table. The intermediate table will contain the primary key for table 1 and the primary key for table 2.
In laravel, the convention for naming tables is plural for your main tables i.e. Community = 'communities' and Idea = 'ideas'. The pivot table name will be derived from the alphabetical order of the related model names i.e.
community_idea.
Now, if you don't want/can't to follow these conventions that's absolutely fine. For more information you can refer to the documentation: https://laravel.com/docs/5.2/eloquent-relationships#many-to-many
Once you're happy that you have the necessary tables with the necessary fields you can access the relationship by:
$ideas = $community->ideas()->get();
//or
$ideas = $community->ideas;
So you controller would look something like:
public function showCommunities($id)
{
$community = Community::findOrFail($id);
//The below isn't necessary as you're passing the Model to a view
// but it's good for self documentation
$community->load('ideas');
return view('publicPages.ideas_in_community', compact('community'));
}
Alternatively, you could add the ideas to the array of data passed to the view to be a bit more verbose:
public function showCommunities($id)
{
$community = Community::findOrFail($id);
$ideas = $community->ideas
return view('publicPages.ideas_in_community', compact('community', 'ideas));
}
Hope this helps!
UPDATE
I would imagine the reason that you're receiving the App\idea not found is because the model names don't match. It's good practice (and in certain environments essential) to Capitalise you class names so make sure of the following:
Your class name is Idea and it's file is called Idea.php
The class has it's namespace declared i.e. namespace App;
If you've added a new class and it's not being found you might need to run composer dump-autoload from the command line to update the autoloader.
How to implement information and information_description tables in laravel Eloquent Model ? Some how it need to set language, because a title should be a one record.
$information = App\Information::find(1);
$information->title
tables structure
You can review how to do this, and plenty more, by reading the documents by Laravel. They are a great help and this particular question has an example and everything. Having said that, I'll help with getting you started.
Define a relationship in either model, information or information_description, or both. For brevity, I'll use information only.
Pass the foreign_key and local_key in the hasMany() relationship since it differs from Laravel's default behavior.
So we have a model that now looks like:
class Information extends Model
{
/**
* Get the descriptions for the Information model.
* Note the 2nd and 3rd arguments in the method
* which define foreign_key and local_key.
*/
public function description()
{
return $this->hasMany('App\InformationDescription', 'information_id', 'information_id');
}
}
Now that you have the relationship defined, you can perform your query.
// Get the description for the information
$information = App\Information::find(1)->description;
// Iterate over the results
foreach ($information as $description) {
$description->title;
}
The table naming convention used is a little odd, but if I understand it correctly, this will work. Hope it helps.
how can I choose which fields I want to get from the with ORM eloquent. For example
$tourTeams = Tournament::with('teams')->where('id', $tourId)->first();
From the teams relation I want only to get the name (without the id and timestamps).
I didn't it in the documentation. For the Tournament eloquent I can do it via the get function while passing it an array of fields names, like this: get(array('name', 'id')). But how do I do this on the Team eloquent?
Note: here is how Team related to Tournament, this code taken from the Tournament eloquent file:
public function teams()
{
return $this->belongsToMany('Team', 'Tournament_Team');
}
You can get specific columns from the relation like this:
$tourTeams = Tournament::with(['teams'=>function($q){
$q->select('id','name');
}])->where('id', $tourId)->first();
how can I access any table from database in my model?
For example, I have Indexcontroller and code inside it:
$results = $this->Index->query("SELECT COUNT(*) FROM my_own_table");
Error: Database table indices for model Index was not found.
So, as I understand, I can access only table with naming related to model/controller name. But what to do if I can't modify the table naming and I want to access it's data?
You're not limited to using a model that's directly associated with your controller (this is just default behaviour); you can use any model.
To achieve what you want, create a new model for this table, eg. MyOwnTable, and in your controller, you can add this property to the class:
public $uses = array('Index', 'MyOwnTable');
Now you can access MyOwnTable using CakePHP's built in ActiveRecord functionality:
$results = $this->MyOwnTable->find('count');
If you have other tables you want to access, simply create models for those and add them to the $uses property. (You can also use $this->loadModel('Model') inside the action if you prefer).
If you have a table name that isn't very readable (eg. my_tb_own_1_x or some such), you can call the model class something human readable (eg. MyTable), and add the $useTable property to the model:
public $useTable = 'my_tb_own_1_x';
/* and change the default primary key if you have an unusual one */
public $primaryKey = 'my_tb_own_1_x_idx_pk';
See the CakePHP manual for more info on how to change default model and controller behaviour:
1.3 - Model Attributes
2.0 - Model Attributes
1.3 - Controller Attributes
2.0 - Controller Attributes
Nope. You can access different tables. However, CakePHP stumbles over the fact that the table that is associated by default to the Index model doesn't exist.
In other words, the model Index expects a table 'indices' to exist (and an error is thrown when it doesn't). You can do one of two things:
Create a table indices
Add the following to your Index model: var $useTable = false;
If you have any use for an indices table I'd go with option 1. If you're not going to use the indices table, go with option 2.
If you go with either step 1 or 2, your example should start working.