Question

目前，此查询仅返回“SFDC Contact ID”列中存在值的情况。好奇我如何更改查询以返回所有结果？我认为这需要一个外连接？

SELECT cmsuser.userid as "User ID", cmsuser.username as "Username", cmsuserprofile.value as "SFDC Contact ID"
FROM cmsuser, cmsuserprofile 
WHERE cmsuser.userid = cmsuserprofile.userid and fieldid = '5004';

Answer 1

您需要以下内容：

         SELECT cmsuser.userid as "User ID", 
                cmsuser.username as "Username", 
                cmsuserprofile.value as "SFDC Contact ID"
           FROM cmsuser
LEFT OUTER JOIN cmsuserprofile 
             ON cmsuser.userid = cmsuserprofile.userid 
           WHERE fieldid = '5004';

即使LEFT JOIN上没有匹配项，cmsuser也会返回cmsuserprofile（联接左侧的表格）中的所有记录。

将JOIN和WHERE条件分开以使您的代码更易于理解和维护是一种很好的做法，这样您就可以更轻松地获得加入条件和什么只是where子句。

Answer 2

据推测，alist *= 0位于个人资料表中。并且，“所有结果”表示“所有用户”。如果是这样，那么你确实需要这样的fieldid：

left join

注意：您应该使用正确的显式SELECT u.userid as "User ID", u.username as "Username", p.value as "SFDC Contact ID" FROM cmsuser u LEFT JOIN cmsuserprofile p ON u.userid = p.userid and p.fieldid = '5004';语法。一个简单的规则：永远不要在join子句中使用逗号。

PostgreSQL - 需要一种方法来返回所有结果

2 个答案: