Laravel - 停止多个查询链接

Question

我目前有使用查询生成器而不是 Eloquent 从我的数据库中获取一些统计信息的功能。

假设我想计算拥有 iOS 令牌的用户数量，然后是在我的一张表中拥有 android 令牌的用户。

class Connection {
    public function __construct()
    {
        return $this->db = DB::connection('database_one');
    }
}

class Statistics extends Connection {
    public function devices()
    {
        $this->devices = $this->db->table('users_devices');

        $arr = [
            'ios' => $this->devices->where('ios_push_token', '!=', '')->count(),
            'android' => $this->devices->where('android_push_token', '!=', '')->count(),
        ];

        return $arr;        
    }
}

获取ios设备的查询正确：

select count(*) as aggregate from `users_devices` where `ios_push_token` != ?
array (size=1)
      0 => string '' (length=0)

但是，然后我遇到了 android 值的问题，查询尝试执行：

select count(*) as aggregate from `users_devices` where `ios_push_token` != ? and `android_push_token` != ?
array (size=2)
  0 => string '' (length=0)
  1 => string '' (length=0)

似乎将第一个查询中的 where 子句链接到第二个查询上，以此类推，这给了我不正确的数据。

我认为这与使用DB::connection 的一个实例有关，但我不确定？

Answer 1

怎么样：

class Statistics {
    public function devices()
    {
        $arr = [
            'ios' => DB::table('users_devices')->where('ios_push_token', '!=', '')->count(),
            'android' => DB::table('users_devices')->where('android_push_token', '!=', '')->count(),
        ];

        return $arr;        
    }
}

或者克隆对象：

class Connection {
    public function __construct()
    {
        return $this->db = DB::connection('database_one');
    }
}

class Statistics extends Connection {
    public function devices()
    {
        $this->devices = $this->db->table('users_devices')->remember(30); // etc.

        $ios = clone $this->devices;
        $android= clone $this->devices;

        $arr = [
            'ios' => $ios->where('ios_push_token', '!=', '')->count(),
            'android' => $android->where('android_push_token', '!=', '')->count(),
        ];

        return $arr;        
    }
}