如何在子查询中使用 group by / having 子句？答案

【问题标题】：How do you use a group by / having clause in a sub-query for?如何在子查询中使用 group by / having 子句？
【发布时间】：2013-09-22 03:02:21
【问题描述】：

我正在尝试使用 group by 和 having 子句从以下子查询中提取 ClientID，但出现以下错误：

消息 116，第 16 级，状态 1，第 1 行
不引入子查询时，选择列表中只能指定一个表达式存在。

查询：

select 
    ClientID 
from 
    SurveyResponses 
where 
     ClientID in (select ClientID, count (surveyresponseid) 
                  from SurveyResponses
                  where SurveyID in (1988,1989,2750,3206,15561) 
group by 
     ClientID
having count (SurveyResponseID) > 1) and SurveyID = 1989

【问题讨论】：

标签： sql sql-server

【解决方案1】：

你可以这样尝试：-

select ClientID from SurveyResponses where ClientID in
(select ClientID from SurveyResponses
where SurveyID in (1988,1989,2750,3206,15561) 
group by ClientID
having count (SurveyResponseID) > 1) and SurveyID = 1989

【讨论】：

【解决方案2】：

您在子查询中拉出两列，并且只能拉出一列，因为您告诉 sql，检查客户 ID 存在于调查响应计数中的位置以及 SurveyResponses 表中的客户 ID。

试试这个，它是未经测试的

select ClientID from SurveyResponses 
where ClientID in (select ClientID  from SurveyResponses
where SurveyID in (1988,1989,2750,3206,15561) 
group by ClientID
having count (SurveyResponseID) > 1) and SurveyID = 1989

【讨论】：

你应该不在你的代码 sn-ps 之前使用> char - 使它更难阅读，禁用语法突出显示......
啊，我不知道。谢谢。

【解决方案3】：

select 
    ClientID 
from 
    SurveyResponses 
where 
     ClientID in (select ClientID,  
                  from SurveyResponses
                  where SurveyID in (1988,1989,2750,3206,15561) 
group by 
     ClientID
having count (SurveyResponseID) > 1) and SurveyID = 1989

此子查询无效，因为您选择了多个字段 ClientID, count (surveyresponseid) 。如果您想处理subquery 中的多个字段以处理Where In 条件，试试这个。Handling multiple columns

【讨论】：

【解决方案4】：

所有其他答案都可以正常工作，您也可以使用EXISTS 语法：

SELECT clientID
FROM SurveyResponses
WHERE EXISTS (
         SELECT * 
         FROM SurveyResponses SR 
         WHERE SR.SurveyId IN (1988, 1989, 2759, 3206, 15561) 
               AND SR.ClientId = ClientID)
GROUP BY ClientID
HAVING COUNT(SurveyResponseID) > 1 AND SurveyID = 1989

【讨论】：

【解决方案5】：

从 select 子句中删除 count()

   select ClientID from SurveyResponses where ClientID in
    (select ClientID from SurveyResponses
    where SurveyID in (1988,1989,2750,3206,15561) 
    group by ClientID
    having count (SurveyResponseID) > 1) and SurveyID = 1989

【讨论】：