Count（）返回0而不是NULL

Question

我已閱讀其他問題，答案似乎無法解決我的情況。我有一張公司名稱表和一份與這些公司有關的活動表。

我需要計算一個指定日期範圍內的活動量，並返回0，如果有沒有。我還需要獲取最後一次活動的日期（即使它超出指定的日期範圍）。我使用LEFT OUTER JOIN，我試過ISNULL()（雖然，如果它不存在，它怎麼檢查它是否爲空？），我不知道該從哪裏去。

這是我有：

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 

這是最終的答案：（謝謝大家）

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：嘗試此操作，首先使用v_rpt_company，因爲您嘗試按公司na進行分組我。

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語句，像這樣：

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

Answer 3

有兩個問題與您查詢：

1. 你有WHERE條款來看，這將是錯誤的時候，那裏已經爲這家公司沒有任何活動（因爲從SO_Activity值將是無效的情況下，那麼條件也將爲空）。
2. 如果上次活動的日期可能超出日期範圍，則需要單獨加入。

這裏是固定的版本：

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