MySQL选择不在where子句中具有相同表的行

Question

我有2张桌子，像这样：

表格1

+----+-----------+--------------------+
| ID | Postcode4 |     woonplaats     |
+----+-----------+--------------------+
|  1 |      9044 | Beetgum            |
|  2 |      9045 | Beetgummermole     |
|  3 |      1028 | Amsterdam          |
|  4 |      1029 | Amsterdam          |
|  5 |      1030 | Amsterdam          |
+----+-----------+--------------------+

表2

+----+------+-------+
| ID | 4PP  | Regio |
+----+------+-------+
|  1 | 9044 |     2 |
|  2 | 9045 |     2 |
|  3 | 1028 |     4 |
|  4 | 1029 |     4 |
|  5 | 1030 |     4 |
+----+------+-------+

我首先要选择Beetgum中的区域，在这种情况下为区域2。然后，我希望表2中的所有其他4PP具有相同的区域（2）。 因此查询结果必须为9044,9045

Answer 1

这是一个加入的:)

SELECT t2.*
FROM Table1 
INNER JOIN Table2 ON Table2.4PP = Table1.Postcode4
INNER JOIN Table2 t2 ON t2.Regio = Table2.regio
WHERE Table1.woonplaats = 'Beetgum'

小提琴： http ://sqlfiddle.com/#!2/347f3/3

Answer 2

这是一个非常糟糕的查询，直到有人给您看中了加入的查询（然后您应该使用）

select t.4PP from Table2 t
WHERE t.Regio in
(
  select Regio
  from Table2 t2
  join Table1 t1 on t2.4PP=t1.Postcode4
  where t1.woonplats='Beetgum'
)

Answer 3

这应该做

select 4PP from table2 where Regio in (select Regio from table2 where 4PP in (select PostCode4 from table1 where woonplaats = "Beetgum"))

Answer 4

似乎这是您要寻找的：

SELECT GROUP_CONCAT(Table3.4PP) 4PP FROM Table1
INNER JOIN Table2 ON(Table1.PostCode4 = Table2.4PP) 
INNER JOIN Table2 AS Table3 ON(Table2.Regio = Table3.Regio) 
WHERE woonplaats = 'Beetgum';

演示