此select语句的正确sql脚本是什么？

这是我的剧本

SELECT iBranch_num,
   CASE WHEN iPatient_typ=1 THEN COUNT(iPatient_num) ELSE 0 END AS [New Patient], 
   CASE WHEN iPatient_typ=2 THEN COUNT(iPatient_num) ELSE 0 END AS [Buying Patient],                
   CASE WHEN iPatient_typ=3 THEN COUNT(iPatient_num) ELSE 0 END AS [Active Patient],
   CASE WHEN iPatient_typ=4 THEN COUNT(iPatient_num) ELSE 0 END AS [Inactive Patient]
FROM tblsotransaction
WHERE iStatus_typ=1
   AND iApply_dt BETWEEN 20130401 AND 20130408 
   AND iBranch_num=14
GROUP BY iBranch_num, iPatient_typ

当前结果：

14, 25, 0,  0,  0
14, 0,  8,  0,  0
14, 0,  0,  97, 0
14, 0,  0,  0,  2

我希望结果是这样的

14, 25, 8,  9,  2

---

将整个

CASE

表达式包装在聚合函数

COUNT

中，并从

GROUP BY

子句中删除

iPartient\u typ

：

SELECT 
  iBranch_num,
  COUNT(CASE WHEN iPatient_typ=1 THEN iPatient_num ELSE 0 END) AS [New Patient], 
  COUNT(CASE WHEN iPatient_typ=2 THEN iPatient_num ELSE 0 END) AS [Buying Patient],                
  COUNT(CASE WHEN iPatient_typ=3 THEN iPatient_num ELSE 0 END) AS [Active Patient],
  COUNT(CASE WHEN iPatient_typ=4 THEN iPatient_num ELSE 0 END) AS [Inactive Patient]
FROM tblsotransaction
WHERE iStatus_typ = 1 
  AND iApply_dt BETWEEN 20130401 AND 20130408 
  AND iBranch_num = 14
GROUP BY iBranch_num;

或：

求和

：

SELECT 
  iBranch_num,
  SUM(CASE WHEN iPatient_typ = 1 THEN 1 ELSE 0 END) AS [New Patient], 
  SUM(CASE WHEN iPatient_typ = 2 THEN 1 ELSE 0 END) AS [Buying Patient],
  SUM(CASE WHEN iPatient_typ = 3 THEN 1 ELSE 0 END) AS [Active Patient],
  SUM(CASE WHEN iPatient_typ = 4 THEN 1 ELSE 0 END) AS [Inactive Patient]
FROM tblsotransaction
WHERE iStatus_typ = 1 
  AND iApply_dt BETWEEN 20130401 AND 20130408 
  AND iBranch_num = 14
GROUP BY iBranch_num;

---

您正在使用什么RDBMS？是SQL server吗？我得到了与此相同的结果acript@odlanyer-抱歉，从

GROUP BY

子句中删除

iPartient\u typ

，我更新了我的答案。无法对包含聚合或子查询的表达式执行聚合函数。在您的第二个答案中，结果正确，但计数是三倍！