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我想在 MySql version 5.1.9
上运行查询,该查询仅返回所选部门的前两个(按 JoiningDate 排序)。
例如,我的数据如下:
+-------+------------------------------------------+----------+------------+
| empid | title | Dept | JoiningDate|
+-------+------------------------------------------+----------+------------+
| 1 | Research and Development | 1 | 2015-08-06 |
| 2 | Consultant | 2 | 2015-08-06 |
| 3 | Medical Consultant | 3 | 2015-08-06 |
| 4 | Officer | 4 | 2015-08-06 |
| 5 | English Translator | 5 | 2015-08-06 |
| 6 | Teacher | 1 | 2015-08-01 |
| 7 | Physical Education | 2 | 2015-08-01 |
| 8 | Accountant | 3 | 2015-08-01 |
| 9 | Science Teacher | 4 | 2015-08-01 |
| 10 | Home Science | 5 | 2015-08-01 |
| 11 | Research Assistant | 1 | 2015-08-05 |
| 12 | Consultant | 2 | 2015-08-05 |
| 13 | Consultant HR | 3 | 2015-08-05 |
| 14 | Technical Lead | 4 | 2015-08-05 |
| 15 | Hindi Translator | 5 | 2015-08-05 |
| 16 | Urdu Teacher | 1 | 2015-08-02 |
| 17 | Physical Education | 2 | 2015-08-02 |
| 18 | Accountant | 3 | 2015-08-02 |
| 19 | Science | 4 | 2015-08-02 |
| 20 | Home Science | 5 | 2015-08-02 |
+-------+------------------------------------------+----------+------------+
我希望查询输出 Dept (1,2,3) 的最新连接的两个 empid,即:
+-------+------------------------------------------+----------+------------+
| empid | title | Dept | JoiningDate|
+-------+------------------------------------------+----------+------------+
| 1 | Research and Development | 1 | 2015-08-06 |
| 11 | Research Assistant | 1 | 2015-08-05 |
| 2 | Consultant | 2 | 2015-08-06 |
| 12 | Consultant | 2 | 2015-08-05 |
| 3 | Medical Consultant | 3 | 2015-08-06 |
| 13 | Consultant HR | 3 | 2015-08-05 |
+-------+------------------------------------------+----------+------------+
最佳答案
在mysql中你可以使用用户定义的变量来达到你想要的结果
SELECT
t.empid,
t.title,
t.Dept,
t.JoiningDate
FROM
(
SELECT
*,
@r:= CASE WHEN @g = b.Dept THEN @r + 1 ELSE 1 END rounum,
@g:= b.Dept
FROM (
SELECT *
FROM table1
CROSS JOIN (SELECT @r:= NULL,@g:=NULL) a
WHERE Dept IN(1,2,3)
ORDER BY Dept,JoiningDate DESC
) b
) t
WHERE t.rounum <=2
关于mysql - 如何获取所选组的最新N条记录,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/32738654/
我正在尝试用 Swift 编写这段 JavaScript 代码:k_combinations 到目前为止,我在 Swift 中有这个: import Foundation import Cocoa e
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