将 LISTAGG 转换为 XMLAGG Oracle

Question

我有一个下面的 SQL，我正在尝试将 LISTAGG 转换为 XMLAGG。我的数据库版本是 12.1，因此 LISTAGG 不支持超过 4K 的字符。

原始查询：--

SELECT LISTAGG (se1.EMAIL, ', ')
WITHIN GROUP (ORDER BY se1.EMAIL)
AS listofsgs
FROM abc se1, xyz team , mno sne
                WHERE     team.employee_id = se1.employee_id
                      AND sne.non_employee_id = team.non_employee_id

转换后的查询：--

SELECT XMLAGG(XMLELEMENT(se1.EMAIL||',')).EXTRACT('//text()')  
WITHIN GROUP (ORDER BY se1.EMAIL)
AS listofsgs
FROM abc se1, xyz team , mno sne
   WHERE     team.employee_id = se1.employee_id
   AND sne.non_employee_id = team.non_employee_id

但是每当我运行它时，它总是会因为缺少逗号的语法错误而失败。我转换后的查询是正确的还是我做错了？

Answer 1

试试下面的查询替换LISTAGG

SELECT RTRIM (
          XMLAGG (XMLELEMENT (e, se1.email || ',') ORDER BY se1.email).EXTRACT (
             '//text()'),
          ',')
          listofsgs
  FROM abc se1, xyz team, mno sne
 WHERE     team.employee_id = se1.employee_id
       AND sne.non_employee_id = team.non_employee_id

Answer 2

我遇到了同样的问题，下面为我工作：

Select RTRIM((XMLAGG(XMLELEMENT(e, se1.EMAIL || ',') ORDER BY se1.EMAIL).EXTRACT ('//text()')).getclobval(),',') listofsgs
FROM abc se1, xyz team , mno sne
   WHERE     team.employee_id = se1.employee_id
   AND sne.non_employee_id = team.non_employee_id

请注意：我添加 getclobval() 是因为我收到“ORA-19011：字符串缓冲区太小错误”。如果不需要，可以将其删除。

Answer 3

试试这个查询：

SELECT XMLAGG(XMLELEMENT(e,se1.EMAIL||',')).EXTRACT('//text()', ',')
    WITHIN GROUP (ORDER BY se1.EMAIL)
    AS listofsgs
    FROM abc se1, xyz team , mno sne
       WHERE     team.employee_id = se1.employee_id
       AND sne.non_employee_id = team.non_employee_id;