Question

请考虑以下示例表结构：

+-------------------------------+--------------+------+-----+---------+-------+
| Field                         | Type         | Null | Key | Default | Extra |
+-------------------------------+--------------+------+-----+---------+-------+
| msisdn                        | varchar(20)  | NO   | PRI | NULL    |       |
| job_search_text               | varchar(500) | YES  |     | NULL    |       |
| local_job_region_id           | int(11)      | YES  | MUL | NULL    |       |
|   .......                     | .......      | ...  | ... | ....    | ....  |
|   .......                     | .......      | ...  | ... | ....    | ....  |
|   .......                     | .......      | ...  | ... | ....    | ....  |
|   .......                     | .......      | ...  | ... | ....    | ....  |
|   .......                     | .......      | ...  | ... | ....    | ....  |
|   .......                     | .......      | ...  | ... | ....    | ....  |
+-------------------------------+--------------+------+-----+---------+-------+

我想按如下方式计算行数：

If job_search_text='a' then count row as 'A'
Else If job_search_text='b' then count row as 'B'
Else If job_search_text='c' then count row as 'C'
.
.
.
Else count row as 'Other'

我写的查询是：

SELECT COUNT(CASE WHEN job_search_text='a' THEN 1 END) as 'A',COUNT(CASE WHEN job_search_text='b' THEN 1 END) as 'b' FROM subscriber

但是这个查询比较了每个CASE的job_search_text，并且它也没有处理上面提到的Else情况。请帮帮我。

Answer 1

你在正确的轨道上，但是将计数改为SUM（IF（））......类似于

select
      sum( if( s.job_search_text = 'a', 1, 0 ) ) as 'A',
      sum( if( s.job_search_text = 'b', 1, 0 ) ) as 'B',
      sum( if( s.job_search_text = 'c', 1, 0 ) ) as 'C',
      sum( if( s.job_search_text = 'd', 1, 0 ) ) as 'D',
      sum( if( s.job_search_text = 'e', 1, 0 ) ) as 'E',
      sum( if( s.job_search_text IN ( 'a', 'b', 'c', 'd', 'e' ), 0, 1 ) ) as 'Other'
   from 
      subscriber s

另一个的“IN”测试，如果找到已经考虑的东西，那么它就是零值。如果它找不到其中一个a-e条目，它会为你的“其他”计数加上一个。

Answer 2

select count(*), job_Search_Text
FROM table
group by job_Search_text

每个job_Search_Text返回1行，计数如此...就没有“其他”

10 A
15 B
25 C
30 D
50 E
99 F

根据列的不同值将mysql表中的记录计为不同的列

2 个答案: