Sql 如何在Oracle10g中按顺序连接多行

如果我有这样的数据：

GROUP  |  SEQUENCE  |  COMMAND
------------------------------
ONE    |     3      |  <message2>MESSAGE</message2>
ONE    |     1      |  <?xml version="1.0" encoding="UTF-8"?>
ONE    |     2      |  <message1>MESSAGE</message1>
TWO    |     2      |  <message2>MESSAGE</message2>
TWO    |     1      |  <?xml version="1.0" encoding="UTF-8"?>
........
TWO    |    10      |  <message9>MESSAGE</message9>

GROUP  |  COMMAND
-----------------
ONE    |  <?xml version="1.0" encoding="UTF-8"?>,<message1>MESSAGE</message1>,<message2>MESSAGE</message2>
TWO    |  <?xml version="1.0" encoding="UTF-8"?>,<message1>MESSAGE</message1>, .. ,<message9>MESSAGE</message9>

---

如有任何意见和建议，将不胜感激^_^

切勿使用

WM\u CONCAT

。阅读

参见本主题

它没有文档记录，任何依赖于

WM_CONCAT

的应用程序在升级到

12c

后将无法工作，因为它已从最新的12c版本中删除

根据数据库版本的不同，进行字符串聚合的方法有很多。见以下几个例子：

11gR2

使用

LIASTAGG

：

SQL> SELECT grp,
  2    listagg(command, ',') WITHIN GROUP(
  3  ORDER BY seq) command
  4  FROM t
  5  GROUP BY grp;

GRP COMMAND
--- --------------------------------------------------------------------------------------------
ONE <?xml version=1.0 encoding=UTF-8?>,<message1>MESSAGE</message1>,<message2>MESSAGE</message2>
TWO <?xml version=1.0 encoding=UTF-8?>,<message2>MESSAGE</message2>,<message9>MESSAGE</message9>

SQL>

---

您好，先生，我在执行过程中遇到了这个错误“ORA-00923:FROM关键字未在预期位置找到”。Oracle10g是否支持Listag？Oracle10g中是否有其他连接解决方案@LalitKumarBOk，请参阅更新的答案。使用

ROW\u NUMBER（）

和

SYS\u CONNECT\u BY\u PATH

。再次感谢您的快速响应。：）@不客气！很高兴我能在第二个主题上帮助您；-）

SQL> SELECT grp,
  2    listagg(command, ',') WITHIN GROUP(
  3  ORDER BY seq) command
  4  FROM t
  5  GROUP BY grp;

GRP COMMAND
--- --------------------------------------------------------------------------------------------
ONE <?xml version=1.0 encoding=UTF-8?>,<message1>MESSAGE</message1>,<message2>MESSAGE</message2>
TWO <?xml version=1.0 encoding=UTF-8?>,<message2>MESSAGE</message2>,<message9>MESSAGE</message9>

SQL>

SQL> SELECT grp,
  2         LTRIM(MAX(SYS_CONNECT_BY_PATH(command,','))
  3         KEEP (DENSE_RANK LAST ORDER BY seq),',') command
  4  FROM   (SELECT grp,
  5                 command,
  6                 seq,
  7                 ROW_NUMBER() OVER (PARTITION BY grp ORDER BY seq) AS curr,
  8                 ROW_NUMBER() OVER (PARTITION BY grp ORDER BY seq) -1 AS prev
  9          FROM   t)
 10  GROUP BY grp
 11  CONNECT BY prev = PRIOR curr AND grp = PRIOR grp
 12  START WITH curr = 1;

GRP COMMAND
--- --------------------------------------------------------------------------------------------
ONE <?xml version=1.0 encoding=UTF-8?>,<message1>MESSAGE</message1>,<message2>MESSAGE</message2>
TWO <?xml version=1.0 encoding=UTF-8?>,<message2>MESSAGE</message2>,<message9>MESSAGE</message9>

SQL>