Laravel条件查询 [英] Laravel conditional query

问题描述

我有一个功能，我尝试在两列中搜索匹配的id，但是即使实际上没有结果，它总是返回 true .

I have function where I try to search in 2 columns for matched id but in all times it return true even where there is actually no result.

我想搜索用户以前的关系，但不确定如何使这种逻辑发生.

I want to search for users previous relations but not sure how to make this logic happen.

1. 我的表有2列 from_id 和 to_id (均会获取用户ID)
2. 我正在向后端发送2个ID(一个作为输入，另一个在请求中标头)
3. 我想说的是:其中输入的 id 是 from_id ，标头ID是 to_id ，或其中输入 id 是 to_id 和标头 id 是 from_id返回结果，否则显然为null.

1. My table has 2 columns from_id and to_id (both will get users ids)
2. I am sending 2 ids to back-end (one as input, another one in request header)
3. I want to say: where input id is from_id and header id is to_id , ORwhere input id is to_id and header id is from_id return the result otherwise is null obviously.

代码

public function check(Request $request) { // always return true!
        $user = $request->user();
        $userInput = $request->input('receiverId');
        $chat = PrivateChat::where(function ($query) use ($user) {
            $query->where('from_id', $user->id)->orWhere('to_id', $user->id);
        })->orWhere(function ($query) use ($userInput) {
            $query->where('from_id', $userInput)->orWhere('to_id', $userInput);
        })->first();

        $receiver = User::where('id', $request->input('receiverId'))->first();

        if($chat) {
            return response()->json([
                'data' => new PrivateChatResource($chat),
                'exist' => true,
                'receiver' => new UserResource($receiver),
                'message' => 'Chat data retrieved successfully.'
            ]);
        } else {
            return response()->json([
                'exist' => false,
                'receiver' => new UserResource($receiver),
                'message' => 'Chat data retrieved successfully.'
            ]);
        }
}

注意:这是我的旧问题的后续操作

知道是什么问题吗?

基于 Giacomo M 的评论，我认为这应该是最终代码

Based on Giacomo M comment i think this should be the final code

$chat = PrivateChat::where(function ($query) use ($user, $userInput) {
            $query
                ->where('from_id', $user->id)
                ->where('to_id', $userInput);
        })->orWhere(function ($query) use ($user, $userInput) {
            $query
                ->where('from_id', $userInput)
                ->where('to_id', $user->id);
        })->first();

有意义吗?

推荐答案

您也可以在子查询中使用 orWhere ，但是根据您的逻辑，您需要在子查询中使用AND.

You use orWhere also in the subqueries, but according to your logic you need AND in the subqueries.

您的查询将变为:

$chat = PrivateChat::where(function ($query) use ($user) {
    $query
        ->where('from_id', $user->id)
        ->where('to_id', $user->id); // where, NOT orWhere
})->orWhere(function ($query) use ($userInput) {
    $query
        ->where('from_id', $userInput)
        ->where('to_id', $userInput); // where, NOT orWhere
})->first();

这篇关于Laravel条件查询的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！