关于聚合函数的Where子句条件

Question

我有以下简单的查询,

SELECT US_LOGON_NAME as Username, 
COUNT(I.IS_ISSUE_NO) as Issues
FROM ISSUES I JOIN USERS U ON I.IS_ASSIGNED_USER_ID = U.US_USER_ID
WHERE I.IS_RECEIVED_DATETIME BETWEEN 20110101000000 AND 20110107000000
GROUP BY U.US_LOGON_NAME; 

我想在选择列表中添加额外的COUNT()函数,但在某些条件下强加它们.这是以某种方式用CASE()语句完成的吗？我尝试在选择列表中放置Where子句,但似乎不允许这样做.我不确定这里是否真的需要子查询,但我不这么认为.

例如,我想要一个COUNT()函数,它只计算某个范围内的问题,然后计算另一个范围内的问题或其他各种条件等:

 SELECT US_LOGON_NAME as Username, 
 COUNT(I.IS_ISSUE_NO (condition here)
 COUNT(I.IS_ISSUE_NO (a different condition here)

等等...

仍按登录名称分组.

谢谢.

Answer 1

SELECT
  SUM(CASE WHEN I.IS_ISSUE_NO (condition here) THEN 1 ELSE 0 END) AS COND1
  SUM(CASE WHEN I.IS_ISSUE_NO (condition here) THEN 1 ELSE 0 END) AS COND2

Answer 2

几个解决方案.

您可以利用SQL不COUNT NULL值的事实:

  SELECT US_LOGON_NAME as Username, 
  COUNT(CASE WHEN <cond>       THEN I.IS_ISSUE_NO ELSE NULL END)
  COUNT(CASE WHEN <other cond> THEN I.IS_ISSUE_NO ELSE NULL END)
  . . .

或者您可以使用SUM而不是COUNT:

  SELECT US_LOGON_NAME as Username, 
  SUM(CASE WHEN <cond>       THEN 1 ELSE 0 END)
  SUM(CASE WHEN <other cond> THEN 1 ELSE 0 END)
  . . .

在任何一种情况下,您都可以根据需要重复多次.