Count（）返回0而不是NULL

Question

我已阅读其他问题，答案似乎无法解决我的情况。我有一张公司名称表和一份与这些公司有关的活动表。

我需要计算一个指定日期范围内的活动量，并返回0，如果有没有。我还需要获取最后一次活动的日期（即使它超出指定的日期范围）。我使用LEFT OUTER JOIN，我试过ISNULL()（虽然，如果它不存在，它怎么检查它是否为空？），我不知道该从哪里去。

这是我有：

SELECT  v_rpt_Company.Company_Name, COUNT(DISTINCT SO_Activity.SO_Activity_Recid) as touches, MAX(SO_Activity.Date_Entered) As lasttouch 
FROM   v_rpt_Member LEFT OUTER JOIN 
         Company_Team ON v_rpt_Member.Member_RecID = Company_Team.Member_RecID LEFT OUTER JOIN 
         v_rpt_Company ON Company_Team.Company_RecID = v_rpt_Company.Company_RecID LEFT OUTER JOIN 
         SO_Activity ON v_rpt_Company.Company_RecID = SO_Activity.Company_RecID 
WHERE  (Company_Team.AcctMgr_Flag = 1) AND (v_rpt_Member.Member_ID = @member) AND ((SO_Activity.Last_Update >= CONVERT(datetime, @date_start, 101)) AND (SO_Activity.Last_Update <= CONVERT(datetime, @date_end, 101))) 
GROUP BY v_rpt_Company.Company_Name 
ORDER BY v_rpt_Company.Company_Name ASC,lasttouch DESC,touches DESC 

这是最终的答案：（谢谢大家）

SELECT a.touches, a.name,b.lasttouch 

FROM (

SELECT  v_rpt_Company.Company_Name as name, 
COUNT(DISTINCT SO_Activity.SO_Activity_Recid) as touches 
FROM v_rpt_Company 
LEFT OUTER JOIN Company_Team ON Company_Team.Company_RecID = v_rpt_Company.Company_RecID AND Company_Team.AcctMgr_Flag = 1 
LEFT OUTER JOIN v_rpt_Member ON v_rpt_Member.Member_RecID = Company_Team.Member_RecID 
LEFT OUTER JOIN SO_Activity ON v_rpt_Company.Company_RecID = SO_Activity.Company_RecID AND ((SO_Activity.Last_Update >= CONVERT(datetime, @date_start, 101)) AND (SO_Activity.Last_Update <= CONVERT(datetime, @date_end, 101))) 
WHERE v_rpt_Member.Member_ID = @member 
Group By v_rpt_Company.Company_Name) As a 

LEFT OUTER JOIN 

(SELECT MAX(SO_Activity.Date_Entered) As lasttouch, v_rpt_Company.Company_Name as name 

FROM v_rpt_Company 
LEFT OUTER JOIN Company_Team ON Company_Team.Company_RecID = v_rpt_Company.Company_RecID AND Company_Team.AcctMgr_Flag = 1 
LEFT OUTER JOIN v_rpt_Member ON v_rpt_Member.Member_RecID = Company_Team.Member_RecID 
LEFT OUTER JOIN SO_Activity ON v_rpt_Company.Company_RecID = SO_Activity.Company_RecID 

WHERE v_rpt_Member.Member_ID = @member 

GROUP BY v_rpt_Company.Company_Name) as b 

ON a.name = b.name 

Answer 1

更新：尝试此操作，首先使用v_rpt_company，因为您尝试按公司na进行分组我。

SELECT  v_rpt_Company.Company_Name, 
COUNT(DISTINCT SO_Activity.SO_Activity_Recid) as touches, 
MAX(SO_Activity.Date_Entered) As lasttouch 
FROM v_rpt_Company 
LEFT OUTER JOIN Company_Team ON Company_Team.Company_RecID = v_rpt_Company.Company_RecID AND Company_Team.AcctMgr_Flag = 1 
LEFT OUTER JOIN v_rpt_Member ON v_rpt_Member.Member_RecID = Company_Team.Member_RecID 
LEFT OUTER JOIN SO_Activity ON v_rpt_Company.Company_RecID = SO_Activity.Company_RecID AND ((SO_Activity.Last_Update >= CONVERT(datetime, @date_start, 101)) AND (SO_Activity.Last_Update <= CONVERT(datetime, @date_end, 101))) 
WHERE v_rpt_Member.Member_ID = @member 
GROUP BY v_rpt_Company.Company_Name 
ORDER BY v_rpt_Company.Company_Name ASC,lasttouch DESC,touches DESC 

Answer 2

修改您的count语句，像这样：

COALESCE(COUNT(DISTINCT SO_Activity.SO_Activity_Recid),0) 

Answer 3

有两个问题与您查询：

1. 你有WHERE条款来看，这将是错误的时候，那里已经为这家公司没有任何活动（因为从SO_Activity值将是无效的情况下，那么条件也将为空）。
2. 如果上次活动的日期可能超出日期范围，则需要单独加入。

这里是固定的版本：

SELECT 
v_rpt_Company.Company_Name, 
ISNULL(COUNT(DISTINCT SO_Activity.SO_Activity_Recid), 0) as touches, 
MAX(SO2.Date_Entered) As lasttouch 
FROM v_rpt_Member 
LEFT OUTER JOIN Company_Team ON v_rpt_Member.Member_RecID = Company_Team.Member_RecID 
LEFT OUTER JOIN v_rpt_Company ON Company_Team.Company_RecID = v_rpt_Company.Company_RecID 
LEFT OUTER JOIN SO_Activity ON v_rpt_Company.Company_RecID = SO_Activity.Company_RecID 
         AND SO_Activity.Last_Update >= CONVERT(datetime, @date_start, 101) 
         AND SO_Activity.Last_Update <= CONVERT(datetime, @date_end, 101) 
LEFT OUTER JOIN SO_Activity SO2 ON v_rpt_Company.Company_RecID = SO_Activity.Company_RecID 
WHERE Company_Team.AcctMgr_Flag = 1 AND v_rpt_Member.Member_ID = @member 
GROUP BY v_rpt_Company.Company_Name 
ORDER BY v_rpt_Company.Company_Name ASC,lasttouch DESC,touches DESC