我要在 Laravel 中实现 SQL 查询。我的查询是:
SELECT * FROM officer_ranks ra
LEFT JOIN (SELECT * FROM officer_rank_details WHERE void = 0) AS r ON ra.rank = r.present_rank
LEFT JOIN officer_languages l ON r.officer_id = l.officer_id
GROUP BY(ra.rank)
ORDER BY ra.id ASC;
为了实现这个查询,我写了:
DB::table('officer_ranks as ra')
->leftJoin(DB::raw('SELECT * FROM officer_rank_details WHERE void = 0'), 'ra.rank', '=', 'present_rank')
->leftJoin('officer_languages as l', 'officer_id', '=', 'l.officer_id')
->groupBy('ra.rank')
->get();
我的问题是如何获取 (DB::raw(...))
结果表作为命名 r
因为我需要 JOIN
表。
我试过用
leftJoin(DB::raw('SELECT * FROM officer_rank_details WHERE void = 0') as r, 'ra.rank', '=', 'present_rank')
但是没有成功。
最佳答案
我在评论中写了如何修复您的代码。但是你原来的 SQL 应该是:
SELECT * FROM officer_ranks ra
LEFT JOIN officer_rank_details AS r
ON ra.rank = r.present_rank
AND r.void = 0
LEFT JOIN officer_languages l ON r.officer_id = l.officer_id
ORDER BY ra.id ASC;
在 Laravel 中:
DB::table('officer_ranks as ra')
->leftJoin('officer_rank_details as r', function($join){
$join->on('ra.rank', '=', 'r.present_rank');
$join->on('r.void', '=', 0);
})
->leftJoin('officer_languages as l', 'r.officer_id', '=', 'l.officer_id')
->get();
我还删除了 GROUP BY
子句,因为它的使用方式没有意义。
关于mysql - Laravel 查询生成器 - 获取表结果 `as`,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/41524103/