Question

我有表A

============
| id | val |
=====+======
| 1  | abc |
| 1  | def |
| 2  | ghi |
| 2  | jkl |
============

我有表B

============
| id | val2 |
=====+======
| 1  | rty |
| 1  | vbn |
| 2  | uio |
| 2  | zxc |
============

我想像这样显示两个表..

===================
| id | val | val2 |
=====+=====+=======
| 1  | abc | rty  |
| 1  | def | vbn  |
| 2  | ghi | uio  |
| 2  | jkl | zxc  |
===================

我的问题是我有裁员..

Answer 1

是的，您遇到问题，因为您没有合适的join密钥。您可以使用变量创建一个变量来完成此操作。这样的东西适用于您提供的数据：

select min(id), max(aval), max(bval)
from ((select id, val as aval, NULL as bval, @rna := @rna + 1 as seqnum
       from tablea a cross join (select @rna := 0)
      ) union all
      (select id, NULL val, @rnb := @rnb + 1 as seqnum
       from tableb b cross join (select @rnb := 0)
      )
     ) ab
group by seqnum;

Answer 2

您可以使用INNER JOIN执行此操作。请参阅下面的查询：

SELECT A.id,val,val2 FROM
(SELECT
     @row_number:=@row_number+1 AS RowNumber,
     id,
     val
FROM TableA, (SELECT @row_number:=0) AS t ORDER BY val) AS A
INNER JOIN
(SELECT
     @row_number:=@row_number+1 AS RowNumber,
     id,
     val2
FROM TableB, (SELECT @row_number:=0) AS t ORDER BY val2) AS B
ON A.RowNumber=B.RowNumber

Answer 3

我喜欢Gordon的方法，因为它不假设每个表中的行数相同，但这里是JOIN版本：

SELECT a.id,a.val,b.val2
FROM (SELECT @row_number:=@row_number+1 AS row_number
            ,id,val
      FROM Table1 a cross join (select @row_number := 0) b
      )a 
JOIN (SELECT @row_number2:=@row_number2+1 AS row_number2
            ,id,val2
      FROM Table2 a cross join (select @row_number2 := 0) b
      )b
 ON a.Row_Number = b.Row_Number2
  AND a.id = b.id

这是他的UNION版本的工作版本：

SELECT Row_Number,ID,MAX(Val) AS Val,MAX(Val2) AS Val2
FROM (SELECT @row_number:=@row_number+1 AS row_number
            ,id,val,NULL as Val2
      FROM Table1 a cross join (select @row_number := 0) b      
      UNION ALL
      SELECT @row_number2:=@row_number2+1 AS row_number
            ,id,NULL,val2
      FROM Table2 a cross join (select @row_number2 := 0) b
      )sub
GROUP BY Row_Number,ID

两者的演示：SQL Fiddle

使用select sql将两个表显示为一个

3 个答案: