如何使用Listag和多列组合SQL（oracle）中的多行？

我有一个SQL语句，它返回如下所示的数据：

SOURCEDID   ORGSOURCEDIDS   ROLE    USERNAME    EMAIL   GIVENNAME   FAMILYNAME
123456  0   teacher test.teacher@   test.teacher@   Test    Teacher
123456  1802    teacher test.teacher@test.teacher@  Test    Teacher
123456  1803    teacher test.teacher@   test.teacher@   Test    Teacher
123456  1804    teacher test.teacher@   test.teacher@   Test    Teacher

SOURCEDID   ORGSOURCEDIDS
123456  0,1802,1803,1804

SELECT TO_CHAR(TEACHERNUMBER) as sourcedId, 
   LISTAGG(schoolid, ',') WITHIN GROUP (ORDER BY schoolid) orgSourcedIds,
   LISTAGG(EMAIL_ADDR, ',') WITHIN GROUP (ORDER BY EMAIL_ADDR) EMAIL_ADDR
FROM teachers 
group by TEACHERNUMBER
order by TEACHERNUMBER;

SOURCEDID   ORGSOURCEDIDS   EMAIL_ADDR
123456  0,1802,1803,1804    test.teacher@,test.teacher@,test.teacher@,test.teacher@

我正在尝试将具有相同“SOURCEDID”的行与逗号分隔的列表中的“ORGSOURCEDIDS”合并为一行，例如：“0180218031804”。除ORGSOURCEDIDS之外的所有其他数据始终相同（当您编辑任何一条记录时，系统会在每个组织中创建和更新单独的记录）

我正在尝试使用以下SQL：

SELECT TO_CHAR(TEACHERNUMBER) as sourcedId, 
   LISTAGG(schoolid, ',') WITHIN GROUP (ORDER BY schoolid) orgSourcedIds
FROM teachers 
group by TEACHERNUMBER

这给了我类似的东西：

SOURCEDID   ORGSOURCEDIDS   ROLE    USERNAME    EMAIL   GIVENNAME   FAMILYNAME
123456  0   teacher test.teacher@   test.teacher@   Test    Teacher
123456  1802    teacher test.teacher@test.teacher@  Test    Teacher
123456  1803    teacher test.teacher@   test.teacher@   Test    Teacher
123456  1804    teacher test.teacher@   test.teacher@   Test    Teacher

SOURCEDID   ORGSOURCEDIDS
123456  0,1802,1803,1804

SELECT TO_CHAR(TEACHERNUMBER) as sourcedId, 
   LISTAGG(schoolid, ',') WITHIN GROUP (ORDER BY schoolid) orgSourcedIds,
   LISTAGG(EMAIL_ADDR, ',') WITHIN GROUP (ORDER BY EMAIL_ADDR) EMAIL_ADDR
FROM teachers 
group by TEACHERNUMBER
order by TEACHERNUMBER;

SOURCEDID   ORGSOURCEDIDS   EMAIL_ADDR
123456  0,1802,1803,1804    test.teacher@,test.teacher@,test.teacher@,test.teacher@

但我不知道如何在上面的示例中添加其他不变的列（角色、用户名、电子邮件、givenname、familyname）。如果我尝试添加其他列，我会得到“非表达式分组”

我也试过这样的方法：

SOURCEDID   ORGSOURCEDIDS   ROLE    USERNAME    EMAIL   GIVENNAME   FAMILYNAME
123456  0   teacher test.teacher@   test.teacher@   Test    Teacher
123456  1802    teacher test.teacher@test.teacher@  Test    Teacher
123456  1803    teacher test.teacher@   test.teacher@   Test    Teacher
123456  1804    teacher test.teacher@   test.teacher@   Test    Teacher

SOURCEDID   ORGSOURCEDIDS
123456  0,1802,1803,1804

SELECT TO_CHAR(TEACHERNUMBER) as sourcedId, 
   LISTAGG(schoolid, ',') WITHIN GROUP (ORDER BY schoolid) orgSourcedIds,
   LISTAGG(EMAIL_ADDR, ',') WITHIN GROUP (ORDER BY EMAIL_ADDR) EMAIL_ADDR
FROM teachers 
group by TEACHERNUMBER
order by TEACHERNUMBER;

SOURCEDID   ORGSOURCEDIDS   EMAIL_ADDR
123456  0,1802,1803,1804    test.teacher@,test.teacher@,test.teacher@,test.teacher@

返回如下内容：

SOURCEDID   ORGSOURCEDIDS   ROLE    USERNAME    EMAIL   GIVENNAME   FAMILYNAME
123456  0   teacher test.teacher@   test.teacher@   Test    Teacher
123456  1802    teacher test.teacher@test.teacher@  Test    Teacher
123456  1803    teacher test.teacher@   test.teacher@   Test    Teacher
123456  1804    teacher test.teacher@   test.teacher@   Test    Teacher

SOURCEDID   ORGSOURCEDIDS
123456  0,1802,1803,1804

SELECT TO_CHAR(TEACHERNUMBER) as sourcedId, 
   LISTAGG(schoolid, ',') WITHIN GROUP (ORDER BY schoolid) orgSourcedIds,
   LISTAGG(EMAIL_ADDR, ',') WITHIN GROUP (ORDER BY EMAIL_ADDR) EMAIL_ADDR
FROM teachers 
group by TEACHERNUMBER
order by TEACHERNUMBER;

SOURCEDID   ORGSOURCEDIDS   EMAIL_ADDR
123456  0,1802,1803,1804    test.teacher@,test.teacher@,test.teacher@,test.teacher@

---

但在我的情况下，电子邮件地址对于具有相同源DID的每一行始终是相同的，因此在该字段上使用ListAg没有帮助。

只需将它们添加到

选择
和
分组依据中即可：

SELECT TO_CHAR(TEACHERNUMBER) as sourcedId, 
       LISTAGG(schoolid, ',') WITHIN GROUP (ORDER BY schoolid) as orgSourcedIds,
       ROLE, USERNAME, EMAIL, GIVENNAME, FAMILYNAME
FROM teachers 
GROUP BY TEACHERNUMBER, ROLE, USERNAME, EMAIL, GIVENNAME, FAMILYNAME

Listagg
是一个聚合函数，因此对于其余的列，您必须1）将它们移动到
group by
子句中，如@GordonLinoff或2所建议的那样）使用任何聚合函数，例如
max（）
或3）检查它们是否重复，如果是，则使用
listag，如果否-使用
max（）
：

结果:

 SOURCEDID ORGS                 ROLES                                   EMAILS
---------- -------------------- --------------------------------------- ------------------
    123456 0, 1802, 1803, 1804  teacher1, teacher2, teacher1, teacher1  test@teachers.org

角色不同，所以它们以列表的形式呈现，电子邮件很常见，所以只有一个。
很简单，我知道我必须错过一些愚蠢的东西，谢谢！