Question 17800 in Laravel Question posted in
The official Laravel Documentation can be found here.

Get users that are only subscribed in Laravel Cashier – all users not a user

Here is the query I tried to write and gives an error

$users =User::has('subscriptions', function (Builder $q) {
 $q->whereNotNull('ends_at');
})->get();

Getting this error

SQLSTATE[42601]: Syntax error: 7 ERROR: SELECT * with no tables specified is not valid LINE 1: ...sers"."id" = "subscriptions"."user_id") = (select * where "e... ^ (SQL: select * from "users" where (select count(*) from "subscriptions" where "users"."id" = "subscriptions"."user_id") = (select * where "ends_at" > now) and "users"."deleted_at" is null)

When I write this code I get results but need to filter result to get a list of subscribed users without calling User::all() then loop to filter.

User::has('subscriptions')->get();
Tags:

4

Answers


  1. 0

    Solution is here 😁

    $users =  User::whereHas('subscriptions', function (Builder $q) {
        return $q->active();
    })->get()->toArray();

    Have a good day

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  2. 0

    use

    $users = User::with(['subscriptions' => static function ($query) {
    $query->whereNotNull('ends_at');
}])->get();

    Read Constraining Eager Loads

    To query, you need to load the subscriptions relationship first.

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  3. 0

    Do it like this

    $users = User::with(['subscriptions' => static function ($query) { $query->whereNotNull('ends_at'); }])->get();

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  4. 0 
    $users = User::with(['subscriptions' => function ($query) {
    return $query->whereNotNull('ends_at');
}])->get();

    Or,

    $users = User::with('subscriptions')->whereHas('subscriptions', function ($query) {
    return $query->whereNotNull('ends_at');
})->get();

    This will give you only subscribed users. But if you fetch all the users and then apply filter to the fetched result, then every row will be fetched from the database, which is not a good practice.

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