在 REGEXP_SUBSTR 中使用 connect by by 不会将结果破坏为多行

如何解决在 REGEXP_SUBSTR 中使用 connect by by 不会将结果破坏为多行

SELECT CHR(91)||'a-zA-Z0-9._%-'||CHR(93)||'+'|| listagg(REGEXP_SUBSTR('[e-mail],[e-mail]','@'||CHR(91)||'^,'||CHR(93)||'+',1,LEVEL),',') within group (order by level) as domain
FROM DUAL
CONNECT BY REGEXP_SUBSTR('[e-mail],LEVEL) IS NOT NULL
order by 1;

上面的脚本只有@yahoo.com前面的正则表达式

 [a-zA-Z0-9._%-][e-mail],@hotmail.com

预期结果：

[a-zA-Z0-9._%-][e-mail],[a-zA-Z0-9._%-][e-mail]

解决方法

当然；将它们聚合回来。

SQL> SELECT listagg(REGEXP_SUBSTR('[e-mail],[e-mail]','@'||CHR(91)||'^,'||CHR(93)||'+',1,LEVEL),',') within group (order by level) as domain
  2  FROM DUAL
  3  CONNECT BY REGEXP_SUBSTR('[e-mail],LEVEL) IS NOT NULL
  4  order by 1;

DOMAIN
----------------------------------------------------------------------------------------------------
@yahoo.com,@hotmail.com

SQL>

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如果你想把regexp前缀放在所有域中，那么

SQL>     SELECT LISTAGG (   '[a-zA-Z0-9._%-]+'
  2                      || REGEXP_SUBSTR ('[e-mail],3                                        '@' || CHR (91) || '^,' || CHR (93) || '+',4                                        1,5                                        LEVEL),6                      ',')
  7             WITHIN GROUP (ORDER BY LEVEL) AS domain
  8        FROM DUAL
  9  CONNECT BY REGEXP_SUBSTR ('[e-mail],10                            '@' || CHR (91) || '^,11                            1,12                            LEVEL)
 13                IS NOT NULL;

DOMAIN
-----------------------------------------------------------------------------------------------
[a-zA-Z0-9._%-][e-mail],[a-zA-Z0-9._%-][e-mail]

SQL>