I have a users table, roles table and a user_role table.
User
id | name
Roles
id | title
User_Role
user_id | role_id
I need an eloquent query that finds if a user has all roles.
So far I have:
$userId = 1;
$role = array('admin', 'public');
$data = User::whereHas('roles', function($q) use ($role){
$q->whereRaw($q->whereIn('title', $role)->count() .' = '.count($role));
})->find($userId);
But I get the error:
Unknown column user.id in where clause.
Where am I going wrong?
This should work.
The query should return null if the user doesn't have the specified roles.
$userId = 1;
$role = array('admin', 'public');
$data = User::whereHas('roles', function($q) use ($role){
$q->whereIn('title', $role);
}, '>=', count($role))->find($userId);
Related
I have the relationship below, an application belongs to a user.
Users table
ID | Name
1 | Danny
2 | Mark
Applications table
ID | user_id
1 | 1
2 | 2
Application.php
public function user()
{
return $this->belongsTo(User::class);
}
I'm trying to sort the application by user's name and paginate the result. I tried the code below, the pagination is working but the order by doesn't have any effect.
$query = Application::query();
$query = $query->with(['user' => function ($query){
$query->orderBy('name', 'DESC');
}]);
$query = $query->paginate(10);
Try using inner join and order by relation column:
$applications = Application::select('*')
->join('authors', 'applications.user_id', '=', 'users.id')
->orderBy('authors.name', 'DESC')
->paginate(10);
Try this:
Application::join('users', 'applications.user_id', '=', 'users.id')
->orderBy('users.name','desc')
->with('users')
->get();
I have three tables as follows
Base Table
users
id - integer
name - string
Target Table
roles
id - integer
name - string
Pivot Table
role_user
user_id - integer
role_id - integer
Many to many relationship is exist between users and roles table
If I want all the users with role_id of '1' so I can simply do this as
$result = User::whereHas('roles', function ($q){
$q->where('roles.id', 1);
});
but I also want the role name correspond to role_id '1' which is name column of roles table.
Is there any way I can append role name into $result collection.
I think you might want to use map(). The eager loading (with()) is very important though, to prevent the N+1 issue in your query. Without eager loading one additional query would be executed for each user in your result.
$result = App\User::query()
->with([
'roles' => function ($q) {
$q->whereId(1);
}
])
->whereHas('roles', function ($q) {
$q->whereId(1);
})
->get()
->map(function ($user) {
$user['name'] = $user->roles->first()->name;
return $user;
});
You can Eagerload the relationship to get the column you want:
User::whereHas('roles', function ($q){
$q->where('roles.id', 1);
})->with(['roles' => function ($q){
$q->select('name')->where('roles.id', 1);
}])->get();
I am learning larvel 5.6 so I am trying to retrieve number of messages that have id larger than last_seen_id in pivot table
I have user table which have the default columns generated by:
php artisan make:auth
and messages tables which have the following columns:
id, from, message_content, group_id
and the group table have the columns:
id,type
now there is many to many relation between the users and groups table through the custom pivot table which have the columns:
group_id,user_id,last_id_seen
now if I want to retrieve the messages which belong to same group and have larger id than last_id_seen in the pivot table how to do it?
I think you are looking for something like this:
$groupId = 1; // set to whatever you want
$lastSeenId = \Auth::user()->groups()
->where('group_id', $groupId)
->first()->pivot->last_id_seen;
$messages = Message::where('id', '>', $lastSeenId)->get();
A more robust version, which does not fail when your user does not have an entry for the group yet, would be:
$groupId = 1; // set to whatever you want
$group = \Auth::user()->groups()->where('group_id', $groupId)->first();
$lastSeenId = $group ? $group->pivot->last_id_seen : null;
$messages = Message::when($lastSeenId, function($query, $id) {
$query->where('id', '>', $id);
})->get();
Note: normally you'd use optional() in the second snippet, but Laravel 5.2 does not come with this helper...
If you want both the count() of the results and the results themselves, you can store the query in a variable and perform two queries with it. You then don't have to rewrite the same query twice:
$groupId = 1; // set to whatever you want
$group = \Auth::user()->groups()->where('group_id', $groupId)->first();
$lastSeenId = $group ? $group->pivot->last_id_seen : null;
$query = Message::when($lastSeenId, function($query, $id) {
$query->where('id', '>', $id);
});
$count = $query->count();
$messages = $query->get();
Using Laravel eloquent how do I make a query like this:
select * from branches where user_id =(select id from users where name ='sara' )
Assuming that you have a user relationship in your Branch model you could use whereHas:
$branches = Branch::whereHas('user', function ($query) {
$query->where('name', 'sara');
})->get();
Update
If you're using v8.57.0 or above, you can now use the whereRelation() method instead:
Branch::whereRelation('user', 'name', 'sara')->get();
$id = Users::select('id')->where('name','sara')->first();
$barnches = branches::where('id',$id)->get();
Here Users and branches are models , first is using for 1 row and get for many rows
I would split it into two queries. First getting the id, then getting the list. Expecting your models to be called "User" and "Branches"
$user = User::where('name', 'sara');
$id = $user->id;
$branches = Branch::where('id', $id);
This site may help you Link
Try this.
$name = 'sara';
$results = BranchModel::whereIn("user_id", function ($query) use ($name) {
$query->select("id")
->from((new UserModel)->getTable())
->where("name", $name);
})->get();
You can use this:
$users = User::whereName("sara")->get()->pluck('id');
Branch::whereIn('user_id',$users)->get();
i have three table in mysql:
1-users
table of users
2-projects
table of project
3-project_user
there is id and project_id and user_id for relation
there is two model : user and project
the relation between these table are belongsToMany
when a project create maybe one project define for two person
NOW how can i show the project of each person?
Assuming you properly defined your relationships, do
$users = User::with('projects')->get();
//show each user projects
foreach($users as $user) {
echo $user->projects;
}
//Getting project users
$projects = Project::with('users')->get();
This will give you projects list, in each list you can see which user has access or not.
Updates
Taking project with specified user_id
$projects = Project::with(['users' => function ($query) use ($user_id) {
$query->where('user_id', '=', $user_id);
}])->get();
New code
Or try to to get the user
$user = User::with('projects')->where('id', $user_id)->first();
Or via projects, constraining by user_id
$projects = Project::with('users')->whereHas('users', function ($query) use ($user_id) {
$query->where($user_id);
})->get();
the way i solve it:
//save user_id
$user_id = $request->id;
//select in project_user where user_id = $user_id
$project_user = DB::table('project_user')->where('user_id', '=', $user_id)->get();
//create an empty array
$p_id = array();
//fill empty array with project_id
foreach($project_user as $pro) {
$p_id[] = $pro->project_id;
}
//select code and id in project table where id = one of that [$p_id] array
$data = Project::select('code','id')->whereIn('id', $p_id)->get();
//return your data in json
return response()->json($data);
As far as I understood your question, you want to query Project as per user_id. Here is how you do it.
Project::with('users')->whereHas('users', function ($query) use ($user_id_array) {
$query->whereIn('id',$user_id_array);
})->get();
This will give you projects whose user_id are from $user_id_array. Let me know if you need any further help.