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社区首页 >问答首页 >当与内部连接一起使用时,填充函数返回重复的记录

当与内部连接一起使用时,填充函数返回重复的记录
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Stack Overflow用户
提问于 2019-11-11 15:18:21
回答 2查看 110关注 0票数 1

你好,我想从责任表中获取用户责任列表,并从用户表中匹配user_id,但是当我在这些表上应用内部连接时,我得到了重复的记录。

用户表

责任表

下面是查询。

代码语言:javascript
复制
SELECT DISTINCT u.[user_id], 
                u.first_name, 
                u.mobile, 
                responsibility = Stuff((SELECT ',' + responsibility_nm 
                                        FROM   dms_responsibility AS dr 
                                        WHERE  responsibility_id = 
                                               dur.responsibility_id 
                                        FOR xml path('')), 1, 1, '') 
FROM   dms_user_responsibilities AS dur 
       INNER JOIN dms_user AS u 
               ON u.[user_id] = dur.[user_id] 
                  AND u.territory_cd = dur.territory_cd 
GROUP  BY u.[user_id], 
          u.first_name, 
          dur.responsibility_id, 
          u.mobile 

此查询返回重复。但是,如果我尝试在没有内部联接的情况下获取责任,那么我会得到正确的结果。

代码语言:javascript
复制
select STUFF((select ',' + responsibility_nm from DMS_Responsibility 
where responsibility_id in (1,2) FOR XML PATH('')), 1, 1, '') as res

我搞不懂是什么导致了这个问题。

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回答 2

Stack Overflow用户

回答已采纳

发布于 2019-11-18 01:45:33

最后在我的查询中找到了解决方案和解决方案。我所做的是从DMS_User_Responsibilities中获取单个特定记录的所有记录,这就是为什么它从DMS_Responsibilities表中返回所有职责的原因。

我的查询修复方法是。

代码语言:javascript
复制
select distinct u.[user_id], u.first_name + ' ' + u.Name + ' (' + 
convert(varchar(10), u.[user_id]) + ')' as Name, u.mobile,
u.email, u.start_dt, u.end_dt,
responsibility = STUFF((select distinct ',' + responsibility_nm from 
DMS_Responsibility as dr
where responsibility_id in (
select dr.responsibility_id from DMS_Responsibility as dr
inner join DMS_user_Responsibilities as dur
on dur.Responsibility_id = dr.Responsibility_id
where dur.[User_id] = u.[user_id] and dur.territory_cd = u.territory_cd -- This is the fix which i did, now its getting only the particular userid for the record.
) FOR XML PATH('')), 1, 1, '')
from dms_user_responsibilities as dur 
inner join dms_user as u
on u.[user_id] = dur.[user_id] and u.territory_cd = dur.territory_cd

票数 0
EN

Stack Overflow用户

发布于 2019-11-11 15:30:07

当您为XML生成时,您的usersresposibilities之间没有关系。尝试在构建XML期间添加关系。

代码语言:javascript
复制
SELECT DISTINCT u.[user_id], 
                u.first_name, 
                u.mobile, 
                responsibility = Stuff ((SELECT ',' + responsibility_nm 
                                         FROM   dms_responsibility AS dr 
                                 INNER JOIN dms_user_responsibilities t1 
                                         ON dr.responsibility_id = 
                                            t1.responsibility_id 
                                 INNER JOIN dms_user t2 
                                         ON t2.[user_id] = t1.[user_id] 
                                            AND t2.territory_cd = 
                                                t1.territory_cd 
                                         WHERE  t1.responsibility_id = 
                                                dur.responsibility_id 
                                         FOR xml path('')), 1, 1, '') 
FROM   dms_user_responsibilities AS dur 
       INNER JOIN dms_user AS u 
               ON u.[user_id] = dur.[user_id] 
                  AND u.territory_cd = dur.territory_cd 
票数 0
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页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接:

https://stackoverflow.com/questions/58797026

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