Question

SELECT field_name,
SUM(CASE WHEN field_name is not null  THEN 1 ELSE 0 END) as count from table_name group by department


RESULT IS:

field_name    count 

AAA              9
BBB              0
CCC              7

但是我怎么才能得到第二列中多于0的行。

Answer 1

使用HAVING子句：

  SELECT field_name,
     SUM(CASE WHEN field_name IS NOT NULL 
              THEN 1 ELSE 0 END) AS `count` 
    FROM table_name
GROUP BY department
  HAVING `count` > 0

_{我认为您选择的是department而不是field_name，或者您需要按field_name分组}

Answer 2

  SELECT field_name,
         SUM(CASE WHEN field_name is not null
                  THEN 1
                  ELSE 0
             END) as count
    from table_name
group by department
  HAVING `count` > 0

HAVING子句在 WHERE和GROUP BY之后执行，因此您可以解决聚合函数

Answer 3

尝试以下。

SELECT field_name, SUM(CASE WHEN field_name is not null THEN 1 ELSE 0 END) as count 
from table_name
where field_name is not null

实际上，第二个查询很好理解，但第一个查询更好：）

SELECT field_name, SUM(CASE WHEN field_name is not null THEN 1 ELSE 0 END) as count 
from table_name
where SUM(CASE WHEN field_name is not null THEN 1 ELSE 0 END) > 0

如何从MySQL中的CASE语句中获得该结果？

3 个答案: