Question

我已经安装了用于关注管理的包，但我想更多地将其集成到我的应用程序中，以便只显示关注您的用户的内容，例如instagram，以便更好地了解我们。

https://github.com/overtrue/laravel-follow#api

我现在的代码是这样的，怎么集成？

$items = Items::whereHas('category', function ($query) {
            $query->where('status', 1); // only where the category it belongs to is active
        })->whereHas('user', function ($query) {
            $query->where('access', 1); // only by users with a public profile
        })->where('visible', 1)
            ->where('status', 1)
            ->orderByDesc('id')
            ->paginate(25);

表：用户、项目、User_follower

Answer 1

我留下了我正在尝试做的事情的小更新，但有错误：

Integrity constraint violation: 1052 Column 'id' in where clause is ambiguous (SQL: select count(*) as aggregate from `items` where exists (select * from `users` where `items`.`user_id` = `users`.`id` and exists (select * from `users` as `laravel_reserved_0` inner join `user_follower` on `laravel_reserved_0`.`id` = `user_follower`.`following_id` where `users`.`id` = `user_follower`.`follower_id` and `id` = 1)) and exists (select * from `categories` where `items`.`category_id` = `categories`.`id` and `status` = 1) and exists (select * from `users` where `items`.`user_id` = `users`.`id` and `access` = 1) and `visible` = 1 and `status` = 1)




$items = Items::whereHas('user.followings', function ($query) use ($user_id) {
            $query->where('id', $user_id);
        })->whereHas('category', function ($query) {
            $query->where('status', 1); // only where the category it belongs to is active
        })->whereHas('user', function ($query) {
            $query->where('access', 1); // only by users with a public profile
        })->where('visible', 1)
            ->where('status', 1)
            ->orderByDesc('id')
            ->paginate(25);



public function user(){
    return $this->belongsTo(User::class);
}

public function followings(){
    return $this->belongsTo(Follow::class);
}

显示您关注的用户的帖子 [overtrue/laravel-follow]

1 个答案: