如何与Laravel一起加入Left Outer?

问题描述

我想要一个表中的信息，如果还想要另一个表中的信息，则同样.

i want information from one table and if there is matched info from another table that as well.

这是我的代码

 $scoreObject = DB::table('responses')
        ->select('responses.id', 'responses.questions_id', 'responses.answer_id', 'responses.open_answer', 'responses.user_id',  'responses.scan_id',
             'questions.question', 'questions.question_nr', 'questions.type', 'questions.totalsection_id',
            'answers.id as answerID', 'answers.answer', 'answers.questions_id', 'answers.points'
        )
        ->Join('answers as answers', 'responses.answer_id', '=', 'answers.id')
        ->Join('questions as questions', 'answers.questions_id', '=', 'questions.id')
        ->orderBy('questions.id', 'ASC')
        ->where('responses.scan_id', $scanid)
        ->where('responses.user_id', $userid)
        ->groupBy('questions.id')
        ->get();

它返回所有与答案匹配的答案(answers.questions_idquestions.id').一些响应不匹配(因为没有response.answer_id)，但是我仍然想要响应信息.

It returns all responses that have matches with answers (answers.questions_id questions.id'). some responses don't have matched (because there is no responses.answer_id) but i still want the responses info then.

我如何在laravel中获得这种左外部连接?

how can i get such a left outer join in laravel ?

推荐答案

您可以尝试将连接指定为左外部连接:

You could try specifying the join as being a left outer join:

->join('answers as answers', 'responses.answer_id', '=', 'answers.id', 'left outer')

join方法的第四个参数是$type，如果未指定，则默认为值inner.但是由于左连接和左外部连接是同一件事 ，您可以改为使用leftJoin方法，以使其更具可读性:

The fourth parameter of the join method is $type, which when not specified, defaults to the value inner. But since left join and left outer join are the same thing, you could just use the leftJoin method instead, to make it more readable:

->leftJoin('answers as answers', 'responses.answer_id', '=', 'answers.id')

这篇关于如何与Laravel一起加入Left Outer?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！