Laravel：将结果与UnionAll和条件

Question

让我们说我有2张桌子

resume_profiles表：

   user_id  current_location      city
   | 3 |     | Chennai  |    | Kolkatta   |
   | 4 |     | Mumbai   |    | Ahmaedabad |
   | 5 |     | Pune     |    | Kolkatta   |
   | 6 |     | Kolkatta |    | Pune       |

resume_locations表：

   user_id     location
   | 2 |     | Chennai  |
   | 2 |     | Mumbai   |
   | 3 |     | Pune     |
   | 4 |     | Kolkatta |

我需要将这些结果表位置组合到一个SET中，其中resume_profiles.user_id = resume_locations.user_id使用Union ALL，所以我有这样的事情：

   City        Aggregate
| Chennai    |    | 1 |
| Mumbai     |    | 1 |
| Pune       |    | 2 |
| Kolkatta   |    | 3 |
| Ahmaedabad |    | 1 |

工作查询没有user_id：

$preferred_locations = ResumeLocation::selectRaw('location as city');

$current_locations = ResumeProfile::selectRaw('current_location as city');

$subquery = ResumeProfile::selectRaw('city')
            ->unionAll($current_locations)
            ->unionAll($preferred_locations);

$locations = DB::table(DB::raw("({$subquery->toSql()}) AS s"))
            ->selectRaw('s.city,count(*) as aggregate')
            ->groupBy('s.city')
            ->get();

**在使用条件时在查询中获取异常：**

$preferred_locations = ResumeLocation::selectRaw('location as city')
            ->where('resume_locations.user_id','resume_profiles.user_id');

$current_locations = ResumeProfile::selectRaw('current_location as city');

$subquery = ResumeProfile::selectRaw('city')
            ->unionAll($current_locations)
            ->unionAll($preferred_locations);

$locations = DB::table(DB::raw("({$subquery->toSql()}) AS s"))
            ->selectRaw('s.city,count(*) as aggregate')
            ->groupBy('s.city')
            ->get();

我不知道如何在表格中实现此where条件。

Answer 1

<强>其中：

$preferred_locations = ResumeLocation::selectRaw('location as city')
->whereIn('user_id', function($q) {
  $q->select('user_id')->from('resume_profiles');
});