Question

我在痛苦中

这是我的帮手

$friend_querry = Friend::where([
        'user_id' => Auth::id(),
        'friend_id' => $friend_id,
        ])->orWhere([
        'user_id' => $friend_id,
        'friend_id' => Auth::id(),
        ])->first();




        $friendship = new stdClass();
        $friendship->exist = false;
        $friendship->accepted = false;

        if (! is_null($friend_querry))
        {
        $friendship->accepted = $friend_querry->accepted;
        $friendship->exist = true;
    }
        var_dump($friendship->exist);
        exit;
        return $friendship;

当我记录为'user_id'=＆gt; auth :: id（） - 一切都很好但是当我被记录为'friend_id'时，请求总是var_dump返回我的确...这是第三天我找不到解决方案。 DB的照片包含在帖子中。

Answer 1

试试这样：

$friend_querry = Friend::where([
                                'user_id' => Auth::id(),
                                'friend_id' => $friend_id,
                                ])
                        ->orWhere(function ($query) use($friend_id) {
                                    $query->where([
                                            'user_id' => $friend_id,
                                            'friend_id' => Auth::id(),
                                            ]);
                                })
                        ->first();

Answer 2

谢谢你，只需要很少的语法错误

->orWhere(function ($query) {

但有一个问题 - 未定义的变量：friend_id

就像这样

function friendship($friend_id){
friend_querry = Friend::where([
        'user_id' => Auth::id(),
        'friend_id' => $friend_id,
        ])->orWhere(function ($query) {
                                $query->where([
                                        'user_id' => $friend_id,
                                        'friend_id' => Auth::id(),
                                        ]);
                            })
                    ->first();
... }

Answer 3

当你想＆＃34; group＆＃34;您的where条款，您应该传递closure，然后在其中执行您的条件。比如说，您希望查询看起来像这样

SELECT * FROM friends WHERE（user_id =？AND friend_id =？）OR（user_id =？AND friend_id =？）LIMIT 1

然后，这个雄辩的版本将是

$friend_query = Friend::where(function ($query) use ($friend_id) {
    $query->where('user_id', Auth::id())
        ->where('friend_id', $friend_id);
})->orWhere(function ($query) use ($friend_id) {
    $query->where('friend_id', Auth::id())
        ->where('user_id', $friend_id);
})->first();

其中orWhere Laravel 5.5逻辑DB问题

3 个答案: