ErrorException:
stripos() expects parameter 1 to be string, object given
For the groupBy() call in the with() method
$user = User::with([
'pricelists' => function($query) {
$query->groupBy(function($var) {
return Carbon::parse($var->pivot->created_at)->format('m');
});
}
])->where('id', $id)->get();
I already saw a few posts talking about how to manage this problem and that it shall not be possible to use groupBy() in eloquent but I do not really understand why...
To be clear:
User and Pricelist model got a many-to-many relationship with the default timestamps() method. I am trying to get the downloaded pricelists grouped by their months they were downloaded from the current user.
After a few attempts I just deleted the above shown => function($query... statement from the with() method and just left the with(['pricelist']) to fetch all datasets and tried this:
$user->pricelists = $user->pricelists->groupBy(function($var) {
return Carbon::parse($var->pivot->created_at)->format('m');
});
return $user->pricelists;
And it works fine and returns an array with multiple arrays for each month... But returning it like this:
return $user;
returns just 1 array with all entries... I do not really get the sense behind it right now...
The two groupBy() method that you are using in the two code you provide are totally different methods.
The first groupBy() where you use it in the callback is actually being called by $query which is a query builder object. The groupBy() here is used to add SQL GROUP BY Statement into the query. And as per the documentation, it only take string variables as parameter.
The groupBy() in your second code is being called by $user->pricelists which is a laravel eloquent collection. The groupBy() method here is actually from the base collection class and is used to group the items inside the collection into multiple collections under the different key defined by the parameter passed to the function. Please read the documentation here.
For your case, the second groupBy() is the one you should be using since you plan to use a callback and will allow you to use more complicated logic.
If you have a query that uses eager loading like this:
Brand::with('tags')
->where('id', $id)
->get();
A brand can have many tags.
I then also have an array of tag ids like this [2,4]. How do I add a condition to this query where it returns only those brands whose tags are in the array?
I tried the eager load constraints but that condition is then placed on the tags model, not the Brand.
I tried this also but it returns an unknown method error:
public function tagsIn($allTags){
return $this->belongsToMany('App\Tag', 'brand_tags')
->whereIn('tags.id', $allTags);
}
Brand::with('tags')
->tagsIn('[2,4]')
->get();
I suspect a possible limitation to getting it to work is the fact that Eloquent makes two separate database calls. But is there a way nevertheless?
DB::table('Brands')
->join('brand_tag','brands.id','=','brand_tag.brand_id')
->join('tags','brand_tag.tag_id','=','tags.id')
->whereIn('tags.id',$allTags)
->get();
Try this DB::table('name')->whereIn('column', array(1, 2, 3))->get();
I think you should use this package to handle tag. I used it in my projects. laravel-tagging
I am fairly new to laravel and I built a little "similar posts" section. So every post has a tag and I query all the id's from the current tag. And then I find all the posts with thoses id's. Now my problem is that the current post is always included. Is there an easy way to exclude the current id when querying?
I can't seem to find anything in the helper function on the laravel docs site
this is my function:
public function show($id)
{
$project = Project::findOrFail($id);
foreach ($project->tags as $tag){
$theTag = $tag->name;
}
$tag_ids = DB::table('tags')
->where('name', "=", $theTag)
->value('id');
$similarProjects = Tag::find($tag_ids)->projects;
return view('projects.show', ['project' => $project, 'similarProjects' => $similarProjects]);
}
An easy way to solve your issue would be to use the Relationship method directly instead of referring to it by property, which you can add additional filters just like any eloquent transaction.
In other words, you would need to replace this:
Tag::find($tag_ids)->projects
With this:
Tag::find($tag_ids)->projects()->where('id', '!=', $id)->get()
Where $id is the current project's id. The reason behind this is that by using the method projects(), you are referring your model's defined Relationship directly (most probably a BelongsToMany, judging by your code) which can be used as a Query Builder (just as any model instance extending laravel's own Eloquent\Model).
You can find more information about laravel relationships and how the Query Builder works here:
https://laravel.com/docs/5.1/eloquent-relationships
https://laravel.com/docs/5.1/queries
However, the way you are handling it might cause some issues along the way.
From your code i can assume that the relationship between Project and Tag is a many to many relationship, which can cause duplicate results for projects sharing more than 1 tag (just as stated by user Ohgodwhy).
In this type of cases is better to use laravel's whereHas() method, which lets you filter your results based on a condition from your model's relation directly (you can find more info on how it works on the link i provided for eloquent-relationships). You would have to do the following:
// Array containing the current post tags
$tagIds = [...];
// Fetch all Projects that have tags corresponding to the defined array
Project::whereHas('tags', function($query) use ($tagIds) {
$query->whereIn('id', $tagIds);
})->where('id', !=, $postId)->get();
That way you can exclude your current Project while avoiding any duplicates in your result.
I don't think that Tag::find($tag_ids)->projects is a good way to go about this. The reason being is that multiple tags may belong to a project and you will end up getting back tons of project queries that are duplicates, resulting in poor performance.
Instead, you should be finding all projects that are not the existing project. That's easy.
$related_projects = Project::whereNotIn('id', [$project->id])->with('tags')->get();
Also you could improve your code by using Dependency Injection and Route Model Binding to ensure that the Model is provided to you automagically, instead of querying for it yourself.
public function show(Project $project)
Then change your route to something like this (replacing your controller name with whatever your controller is:
Route::get('/projects/{project}', 'ProjectController#show');
Now your $project will always be available within the show function and you only need to include tags (which was performed in the "with" statement above)
This should be simple but for some reason I am unable to get the expected response. I have a model (App\Product) that has many permissions (App\ProductPermission) and my objective is to check if a product holds a particular permission described by a text string. I can access these fine like..
\App\Product::find($id)->permissions
This will give me the collection of ProductPermission objects as expected. One of the attributes of the ProductPermission model is 'permission' which is the text string. For example "users*".
I want to be able to do something similar to this
\App\Product::find(3)->permissions->search($permission) //$permission = "users*"
But this returns false despite a ProductPermission object with the attribute permission=users* existing. How can I search the attributes (or the specific attribute) of all the objects in the relationship collection?
Thanks in advance..
This works...
\App\Product::find($id)->permissions->where('permission',$permission)->count();
But its not as pretty
You may use whereHas like follows:
$productsWithPermissionsCount = Product::whereHas('permissions', function ($query) {
$query->where('permission', '=', $permission);
})->count();
I have got 2 joined tables in Eloquent namely themes and users.
theme model:
public function user() {
return $this->belongs_to('User');
}
user model:
public function themes() {
return $this->has_many('Theme');
}
My Eloquent api call looks as below:
return Response::eloquent(Theme::with('user')->get());
Which returns all columns from theme (that's fine), and all columns from user (not fine). I only need the 'username' column from the user model, how can I limit the query to that?
Change your model to specify what columns you want selected:
public function user() {
return $this->belongs_to('User')->select(array('id', 'username'));
}
And don't forget to include the column you're joining on.
For Laravel >= 5.2
Use the ->pluck() method
$roles = DB::table('roles')->pluck('title');
If you would like to retrieve an array containing the values of a single column, you may use the pluck method
For Laravel <= 5.1
Use the ->lists() method
$roles = DB::table('roles')->lists('title');
This method will return an array of role titles. You may also specify a custom key column for the returned array:
You can supply an array of fields in the get parameter like so:
return Response::eloquent(Theme::with('user')->get(array('user.username'));
UPDATE (for Laravel 5.2)
From the docs, you can do this:
$response = DB::table('themes')
->select('themes.*', 'users.username')
->join('users', 'users.id', '=', 'themes.user_id')
->get();
I know, you ask for Eloquent but you can do it with Fluent Query Builder
$data = DB::table('themes')
->join('users', 'users.id', '=', 'themes.user_id')
->get(array('themes.*', 'users.username'));
This is how i do it
$posts = Post::with(['category' => function($query){
$query->select('id', 'name');
}])->get();
First answer by user2317976 did not work for me, i am using laravel 5.1
Using with pagination
$data = DB::table('themes')
->join('users', 'users.id', '=', 'themes.user_id')
->select('themes.*', 'users.username')
->paginate(6);
Another option is to make use of the $hidden property on the model to hide the columns you don't want to display. You can define this property on the fly or set defaults on your model.
public static $hidden = array('password');
Now the users password will be hidden when you return the JSON response.
You can also set it on the fly in a similar manner.
User::$hidden = array('password');
user2317976 has introduced a great static way of selecting related tables' columns.
Here is a dynamic trick I've found so you can get whatever you want when using the model:
return Response::eloquent(Theme::with(array('user' => function ($q) {
$q->addSelect(array('id','username'))
}))->get();
I just found this trick also works well with load() too. This is very convenient.
$queriedTheme->load(array('user'=>function($q){$q->addSelect(..)});
Make sure you also include target table's key otherwise it won't be able to find it.
This Way:
Post::with(array('user'=>function($query){
$query->select('id','username');
}))->get();
I know that this is an old question, but if you are building an API, as the author of the question does, use output transformers to perform such tasks.
Transofrmer is a layer between your actual database query result and a controller. It allows to easily control and modify what is going to be output to a user or an API consumer.
I recommend Fractal as a solid foundation of your output transformation layer. You can read the documentation here.
In Laravel 4 you can hide certain fields from being returned by adding the following in your model.
protected $hidden = array('password','secret_field');
http://laravel.com/docs/eloquent#converting-to-arrays-or-json
On Laravel 5.5, the cleanest way to do this is:
Theme::with('user:userid,name,address')->get()
You add a colon and the fields you wish to select separated by a comma and without a space between them.
Using Model:
Model::where('column','value')->get(['column1','column2','column3',...]);
Using Query Builder:
DB::table('table_name')->where('column','value')->get(['column1','column2','column3',...]);
If I good understood this what is returned is fine except you want to see only one column. If so this below should be much simpler:
return Response::eloquent(Theme::with('user')->get(['username']));
#You can get selected columns from two or three different tables
$users= DB::Table('profiles')->select('users.name','users.status','users.avatar','users.phone','profiles.user_id','profiles.full_name','profiles.email','profiles.experience','profiles.gender','profiles.profession','profiles.dob',)->join('users','profiles.user_id','=','users.id')
->paginate(10);
Check out, http://laravel.com/docs/database/eloquent#to-array
You should be able to define which columns you do not want displayed in your api.