Question

我是laravel的新手。这是我的查询，过滤掉没有任何问题的问题（问题）或问题没有任何答案。但是我不确定如何使用laravel查询构建器正确编写它。考虑到我使用雄辩来建立关系如果我能够使用laravel eloquent得到相同的结果会更好。

SELECT DISTINCT q.id, q.content
FROM m_quiz q
RIGHT JOIN  (SELECT DISTINCT p.quiz_id as id
             FROM  m_problem p
             RIGHT JOIN m_answer_choice a
             ON p.id = a.problem_id) problem
ON q.id = problem.id
ORDER BY q.id;

Answer 1

未经测试，但我认为会是这样的。

 DB::table('problems')->selectRaw('m_problem.*, DISTINCT m_problem.quiz_id as id')
    ->join('m_answer_choice', 'm_problem.id', '=', 'm_answer_choice.problem_id')
    ->groupBy('m_problem.id')
    ->get();

但如果你可以做一些像

这样的事情会更好

namespace App;
use Illuminate\Database\Eloquent\Model;

    class Problem extends Model
    {
        /**
         * Get the comments for the blog post.
         */
        public function answerChoices()
        {
            return $this->hasMany('App\AnswerChoices');
        }
    }

并执行类似

的操作

$answerChoices= App\Problem::find(1)->answerChoices;

foreach ($answerChoices as $answerChoice) {
    //
}

Answer 2

SELECT 
    *, SUM(num_hr) AS total_hr, attendances.empID AS empid 
FROM attendances 
LEFT JOIN addsalaryemployees ON addsalaryemployees.empID=attendances.empID 
LEFT JOIN positions ON positions.id=addsalaryemployees.empPosition 
GROUP BY attendances.empID 
ORDER BY addsalaryemployees.empFirstName ASC

如何使用Laravel查询构建器编写此查询？

2 个答案: