Question

我需要使用数据库查询构建器将以下原始查询转换为Laravel查询的帮助。该查询必须使用DB（外观）查询构建器构建，而不是Eloquent。

SELECT distinct p.client_id
  FROM people p
JOIN (SELECT client_id, COUNT(*)
  FROM people
GROUP BY client_id
HAVING count(client_id) > 1 ) b
ON p.client_id = b.client_id
ORDER BY p.client_id')

到目前为止，这是我的解决方案：

$subQuery = DB::table('people')
            ->select('client_id', DB::raw('count(*)'))
            ->groupBy('client_id')
            ->havingRaw('count(client_id) > ?', [1]);

$mainQuery = DB::table('people')
            ->distinct('client_id')
            ->joinSub($subQuery, 'b', function ($join) {
                $join->on('people.client_id', '=', 'b.client_id');
            })
            ->orderBy('client_id')
            ->get();

Answer 1

像这样尝试DB::raw()：

DB::select(DB::raw("Your query here"));

DB::raw()用于生成未解析的任意SQL命令查询构建器可以进行任何进一步的处理。

Answer 2

我找到了所需的解决方案：

$subQuery = DB::table('people')
   ->select('client_id', DB::raw('count(*)'))
   ->groupBy('client_id')
   ->havingRaw('count(client_id) > ?', [1]);

$peopleWithSameClient = DB::table('people')
   ->distinct('people.client_id')
   ->joinSub($subQuery, 'b', function ($join) {
          $join->on('people.client_id', '=', 'b.client_id');
   })
   ->orderBy('people.client_id')
   ->get();

将原始SQL查询转换为Laravel DB构建器查询

2 个答案: