我的数据库中有一些数据:
id user
1 zhangsan
2 zhangsan
3 zhangsan
4 lisi
5 lisi
6 lisi
7 zhangsan
8 zhangsan
我想保持顺序,并合并附近相同的用户项目,该怎么做? 当我使用shell脚本时,我会(文件test中的数据。):
cat test|cut -d " " -f2|uniq -c
这将得到结果:
3 zhangsan
3 lisi
2 zhangsan
但是如何使用sql来做呢?
最佳答案
如果你尝试:
SET @name:='',@num:=0;
SELECT id,
@num:= if(@name = user, @num, @num + 1) as number,
@name := user as user
FROM foo
ORDER BY id ASC;
这给出:
+------+--------+------+
| id | number | user |
+------+--------+------+
| 1 | 1 | a |
| 2 | 1 | a |
| 3 | 1 | a |
| 4 | 2 | b |
| 5 | 2 | b |
| 6 | 2 | b |
| 7 | 3 | a |
| 8 | 3 | a |
+------+--------+------+
那么你可以尝试:
SET @name:='',@num:=0;
SELECT COUNT(*) as count, user
FROM (
SELECT @num:= if(@name = user, @num, @num + 1) as number,
@name := user as user
FROM foo
ORDER BY id ASC
) x
GROUP BY number;
这给出
+-------+------+
| count | user |
+-------+------+
| 3 | a |
| 3 | b |
| 2 | a |
+-------+------+
(我将我的表命名为foo
,并且只使用了名称a
和b
,因为我懒得写zhangsan
和 lisi
一遍又一遍)。
关于mysql - 如何通过SQL合并相似的项目?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/9172006/