Question

我有一个SQL查询，我想将其转换为查询构建器或Laravel ORM。它工作正常。但我希望它成为查询构建器表单或ORM。是否可以在查询构建器或ORM中编写？。

我有四个表答案，问题，用户和upvote_answers。查询有三个'连接'和一个'左连接'只是为了检查当前登录用户upvoted（boolean）答案与否以及其他属性

   SELECT answers.answer_content as answer_content,
          answers.id as answer_id,
          answers.created_at as created_at,
          answers.created_at as answer_upvote,
          answers.created_at as answer_downvote,
          questions.id as question_id, 
          questions.question_title as question_title, 
          questions.question_slug as question_slug, 
          users.id as user_id, 
          users.name as user_name, 
          users.user_slug as user_slug,
          upvote_answers.upvote as upvote
                                        FROM 
                                            answers
                                            JOIN questions on questions.id = answers.question_id
                                            JOIN users on users.id = answers.user_id
                                            LEFT JOIN upvote_answers ON
                                                upvote_answers.answer_id = answers.id AND
                                                upvote_answers.user_id = '2'
                                        WHERE 
                                            questions.question_active = 1 and 
                                            answers.answer_active = 1

Answer 1

使用查询构建器，您可以将查询编写为

DB::table('answers as a')
    ->join('questions as q', 'q.id', '=', 'a.question_id')
    ->join('users as u', 'u.id', '=', 'a.user_id')
    ->leftJoin('upvote_answers as ua', function ($join) {
        $join->on('ua.answer_id', '=', 'a.id')
             ->where('ua.user_id', '=', 2);
    })
    ->where('q.question_active', '=', 1)
    ->where('a.answer_active', '=', 1)
    ->select(DB::raw('a.answer_content as answer_content,a.id as answer_id,a.created_at as created_at,a.created_at as answer_upvote,a.created_at as answer_downvote,q.id as question_id, q.question_title as question_title, q.question_slug as question_slug, u.id as user_id, u.name as user_name, u.user_slug as user_slug,ua.upvote as upvote'))
    ->get();

Answer 2

使用加入条款

https://laravel.com/docs/master/queries#joins

简单示例

$users = DB::table('users')
            ->join('contacts', 'users.id', '=', 'contacts.user_id')
            ->join('orders', 'users.id', '=', 'orders.user_id')
            ->select('users.*', 'contacts.phone', 'orders.price')
            ->where('users.id', '=', 100)
            ->get();

您也可以使用Left Join Clause。

<强>更新

$yourQuery = \DB::table('answers')
    ->join('questions', 'questions.id', '=', 'answers.question_id')
    ->join('users', 'users.id', '=', 'answers.user_id')
    ->leftJoin('upvote_answers', 'upvote_answers.answer_id', '=', 'answers.id')
    ->where('questions.question_active', '=', '1')
    ->where('answers.answer_active', '=', '1')
    ->selectRaw('answers.answer_content as answer_content,
                  answers.id as answer_id,
                  answers.created_at as created_at,
                  answers.created_at as answer_upvote,
                  answers.created_at as answer_downvote,
                  questions.id as question_id, 
                  questions.question_title as question_title, 
                  questions.question_slug as question_slug, 
                  users.id as user_id, 
                  users.name as user_name, 
                  users.user_slug as user_slug,
                  upvote_answers.upvote as upvote');

echo $yourQuery->toSql(); //generated sql query as string
//$yourQuery->get(); //for get data from db

结果：

选择answers.answer_content为answer_content，answers.id为 answer_id，answers.created_at as created_at，answers.created_at as answer_upvote，answers.created_at为answer_downvote，questions.id为 question_id，questions.question_title as question_title， questions.question_slug为question_slug，users.id为user_id， users.name为user_name，users.user_slug为user_slug， upvote_answers.upvote作为upvote从答案内部加入问题 questions.id = answers.question_id在users.id =上的内部加入用户 answers.user_id在upvote_answers.answer_id =上加入upvote_answers answers.id where questions.question_active =？和 answers.answer_active =？

如果你需要upvote_answers.user_id = '2'在leftJoin必须使用提前左连接：

->leftJoin('upvote_answers', function($advancedLeftJoin){
    $advancedLeftJoin->on('users.id', '=', 'contacts.user_id')
        ->where('upvote_answers.user_id', '=', 2);
})

最后，您的示例的答案是：

$yourQuery = \DB::table('answers')
    ->join('questions', 'questions.id', '=', 'answers.question_id')
    ->join('users', 'users.id', '=', 'answers.user_id')
    ->leftJoin('upvote_answers', function($advancedLeftJoin){
        $advancedLeftJoin->on('users.id', '=', 'contacts.user_id')
            ->where('upvote_answers.user_id', '=', 2);
    })
    ->where('questions.question_active', '=', '1')
    ->where('answers.answer_active', '=', '1')
    ->selectRaw('answers.answer_content as answer_content,
                  answers.id as answer_id,
                  answers.created_at as created_at,
                  answers.created_at as answer_upvote,
                  answers.created_at as answer_downvote,
                  questions.id as question_id, 
                  questions.question_title as question_title, 
                  questions.question_slug as question_slug, 
                  users.id as user_id, 
                  users.name as user_name, 
                  users.user_slug as user_slug,
                  upvote_answers.upvote as upvote');

echo $yourQuery->toSql(); //generated sql query as string
//dd($yourQuery->get()); //result as collection

结果：

选择answers.answer_content为answer_content，answers.id为 answer_id，answers.created_at as created_at，answers.created_at as answer_upvote，answers.created_at为answer_downvote，questions.id为 question_id，questions.question_title as question_title， questions.question_slug为question_slug，users.id为user_id， users.name为user_name，users.user_slug为user_slug， upvote_answers.upvote作为upvote从答案内部加入问题 questions.id = answers.question_id在users.id =上的内部加入用户 answers.user_id在users.id =上加入upvote_answers contacts.user_id和upvote_answers.user_id =？哪里 questions.question_active =？和answers.answer_active =？

使用?

后，

->get()值绑定

Laravel加入并在查询中保持联接

2 个答案: