Question

我有两个表A，B和我的查询显示:(加入A和B）

select A.i,A.j,B.x,B.y,B.z from A inner join B on A.id=B.id_B order by A.i,A.j

+-----+-----+-----+-----+-----+
| A.i | A.j | B.x | B.y | B.z |
+=====+=====+=====+=====+=====+
|  bk | bg  |  1  | inf1| bar | 
+-----+-----+-----+-----+-----+
|  bk | bg  |  2  | inf2| bar | 
+-----+-----+-----+-----+-----+
|  bk | bg  |  3  | inf3|  y  | 
+-----+-----+-----+-----+-----+
|  ro | fn  |  5  | enf1| bar | 
+-----+-----+-----+-----+-----+
|  ro | fn  |  3  | enf2| bar | 
+-----+-----+-----+-----+-----+
|  st | st  |  3  | onf1| bar | 
+-----+-----+-----+-----+-----+

现在我想知道是多少次

A.I

重复，如：

    +-----+-----+-----+-----+-----+
    | A.i | A.j | B.x | B.y | B.z |RepeatColumn
    +=====+=====+=====+=====+=====+---
    |  bk | bg  |  1  | inf1| bar | 3 |(bk is repeated 3 times)
    +-----+-----+-----+-----+-----+---
    |  bk | bg  |  2  | inf2| bar | 3 |(bk is repeated 3 times)
    +-----+-----+-----+-----+-----+---
    |  bk | bg  |  3  | inf3|  y  | 3 |(bk is repeated 3 times)
    +-----+-----+-----+-----+-----+---
    |  ro | fn  |  5  | enf1| bar | 2 |(ro is repeated 2 times)
    +-----+-----+-----+-----+-----+---
    |  ro | fn  |  3  | enf2| bar | 2 |(ro is repeated 2 times)
    +-----+-----+-----+-----+-----+---
    |  st | st  |  3  | onf1| bar | 1 |(st is repeated once)
    +-----+-----+-----+-----+-----+---

我如何做这个查询，任何人都有任何想法？

Answer 1

您需要subquery：

select A.i,A.j,
       B.x,B.y,B.z,
       (select count(*) from b where b.id = a.id) RepeatColumn
from A 
inner join B 
order by A.i,A.j;

但是，我在o / p中的表格（A，B）之间找不到任何关系。所以，我只是在飞行中使用。

Answer 2

编辑：不幸的是（对你而言）看起来MySQL并不支持简单的窗口函数。

像这样：

COUNT(1) OVER
  ( PARTITION BY A.i
  ) AS cnt

这取决于您的用例是否比您所显示的更复杂，如果值得拼凑group_concat变通方法，或者只是使用子查询来获取计数。

Answer 3

从tablename中选择count（A.I）;

Answer 4

xcrun simctl addmedia

Answer 5

尝试使用subquery

SELECT 
A.i,
A.j,
B.x,
B.y,
B.z,
(SELECT COUNT(1) FROM A as WHERE i = A.i) AS 'RepeatColumn'
FROM A 
INNER JOIN B 
ORDER BY A.i,A.j

如何计算（SQL）中的行？

5 个答案: