Question

这是我的聊天表

的表格结构

    Schema::create('chats', function (Blueprint $table) {
                $table->bigIncrements('id');
                $table->enum('block', ['no', 'yes'])->default('no');
                $table->timestamps();

                $table->bigInteger('user_one')->unsigned();
                $table->foreign('user_one')
                    ->references('id')
                    ->on('users')
                    ->onDelete('cascade');

                $table->bigInteger('user_two')->unsigned();
                $table->foreign('user_two')
                    ->references('id')
                    ->on('users')
                    ->onDelete('cascade');
            });

这是我的chat_messages表格的一个

Schema::create('chat_messages', function (Blueprint $table) {
                $table->increments('id');
                $table->enum('viewed', ['no', 'yes'])->default('no');
                $table->string('msg', 1000);
                $table->timestamps();

                $table->bigInteger('chat_id')->unsigned();
                $table->foreign('chat_id')
                        ->references('id')
                        ->on('chats')
                        ->onDelete('cascade');

                $table->bigInteger('user_id')->unsigned();
                $table->foreign('user_id')
                    ->references('id')
                    ->on('users')
                    ->onDelete('cascade');
            });

我现在的问题是我想运行这个在phpmyadmin中工作正常的查询

SELECT U.id,C.id,U.name,U.email FROM users as U, chats as C, chat_messages as M WHERE CASE WHEN C.user_one = '2' THEN C.user_two = U.id WHEN C.user_two = '2' THEN C.user_one= U.id END AND C.id = M.chat_id AND (C.user_one ='2' OR C.user_two ='2') ORDER BY C.id DESC

以下是使用Laravel查询构建器的上述查询的版本，似乎无法正常工作

$authUser = Auth::user()->id;
    $chatList = DB::select(DB::raw('U.id, C.id, U.name'))
                ->from(DB::raw('users as U, chats as C, chat_messages as M'))
                ->where(DB::raw('CASE
                    WHEN C.user_one = '.Auth::user()->id.'
                    THEN C.user_two = U.id
                    WHEN C.user_two = '.Auth::user()->id.'
                    THEN C.user_one= U.id END'))
                ->where('C.id', '=', 'M.chat_id')
                ->where(function($query) use ($authUser) {
                    $query->where('C.user_one', '=', $authUser)
                            ->orWhere('C.user_two', '=', $authUser);
                })
                ->orderBy('C.id', 'desc')
                 ->get();

它出了这个错误

SQLSTATE[42000]: Syntax error or access violation: 1064 You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'U.id, C.id, U.name' at line 1 (SQL: U.id, C.id, U.name)

我已经尝试了很多方法，但似乎没有用......我需要你的帮助才能弄清楚如何解决这个问题...谢谢

Answer 1

这里是

    $authUser = Auth::user()->id;
    $chatList = DB::table(DB::raw('users as U, chats as C, chat_messages as M'))
        ->select(DB::raw('U.id, C.id, U.name, U.email'))
        ->whereRaw('CASE
                WHEN C.user_one = '.Auth::user()->id.'
                THEN C.user_two = U.id
                WHEN C.user_two = '.Auth::user()->id.'
                THEN C.user_one= U.id END
                AND C.id = M.chat_id AND 
                (C.user_one = '.$authUser.' OR C.user_two = '.$authUser.')')
        ->orderBy('C.id', 'DESC')
        ->get();

如何将此SQL查询转换为laravel eloquent或查询生成器

1 个答案: