计数大于1的MySQL group by with where子句

我知道这应该很简单，但就我的一生而言，我无法让我的查询满足我的需要

我正在查找按

ref

分组的具有特定状态（

paid

）的所有行，但前提是找到的行数超过1行

这是我的示例表：

+-----+----------+----------+-------+
| id  | deleted  | status   |  ref  |
+-----+----------+----------+-------+
|   1 |        0 | pending  | 10001 |
|   2 |        0 | paid     | 10001 |
|   3 |        0 | paid     | 10001 |
|   4 |        0 | paid     | 10002 |
|   5 |        1 | pending  | 10002 |
|   6 |        1 | paid     | 10002 |
|   7 |        0 | pending  | 10003 |
|   8 |        0 | paid     | 10003 |
|   9 |        0 | paid     | 10003 |
|  10 |        0 | paid     | 10003 |
|  11 |        0 | pending  | 10004 |
|  12 |        0 | paid     | 10004 |
|  13 |        1 | pending  | 10005 |
|  14 |        1 | paid     | 10005 |
|  15 |        1 | paid     | 10005 |
|  16 |        0 | paid     | 10005 |
|  17 |        0 | pending  | 10006 |
|  18 |        0 | paid     | 10006 |
|  19 |        0 | paid     | 10006 |
+-----+----------+----------+-------+

这是我的SQL：

SELECT * FROM `orders`
WHERE `deleted` = 0 AND `status` = 'paid'
GROUP BY SUBSTR(`ref`,0,5)
HAVING COUNT(*) > 1
ORDER BY `id` DESC

我需要通过SUBSTR进行匹配，因为

ref

有时包含附加的数字

问题是我的查询返回以下内容：

+-----+----------+---------+-------+
| id  | deleted  | status  |  ref  |
+-----+----------+---------+-------+
|   2 |        0 | paid    | 10001 |
+-----+----------+---------+-------+

当我希望它返回

ref

s

10001

，

10003

和

10006

时

有人能帮我找出我做错了什么吗

谢谢

试试看

SELECT * FROM `orders`
WHERE `deleted` = 0 AND `status` = 'paid'
GROUP BY SUBSTR(`ref`,1,5)
HAVING COUNT(*) > 1
ORDER BY `id` DESC

SUBSTR的位置参数以1开头，而不是0。

来自SUBSTR文档：

对于所有形式的子字符串（），要从中提取子字符串的字符串中的第一个字符的位置被计算为1

所以试试这个：

SELECT * FROM `orders`
WHERE `deleted` = 0 AND `status` = 'paid'
GROUP BY SUBSTR(`ref`,1,5)
HAVING COUNT(*) > 1
ORDER BY `id` DESC

查询应该是

SELECT * from order
WHERE status="paid"
GROUP BY SUBSTRING('ref',1,5)
HAVING COUNT(*) > 1;

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我被8秒打败了……）塔克斯！感激。。我确实引用了医生的话。。这就是我花了8秒钟的代价：）