Oracle to select 通过正则表达式将所有匹配的事件匹配到一行中的一列，并带有分隔符

Question

我知道 select 通过正则表达式将所有匹配项匹配到多行中的一列，如下所示：

SELECT REGEXP_SUBSTR(str, '\+?\d{3,11}', 1, LEVEL) AS substr
FROM (
    SELECT 'Txa233141b Ta233141 Ta233142 Ta233147 Ta233148' AS str FROM DUAL
)
CONNECT BY LEVEL <= LENGTH(REGEXP_REPLACE(str, '[^ ]+')) + 1;

结果是

但是我想要如下结果，我该怎么办？

233141,233141,233142,233147,233148

Answer 1

您还可以在查询之上使用 listagg，如下所示：

SELECT listagg(REGEXP_SUBSTR(str, '\+?\d{3,11}', 1, LEVEL),',') within group (order by 1) AS substr
FROM (
    SELECT 'Txa233141b Ta233141 Ta233142 Ta233147 Ta233148' AS str FROM DUAL
)
CONNECT BY LEVEL <= LENGTH(REGEXP_REPLACE(str, '[^ ]+')) + 1;

干杯！！

Answer 2

使用简单的字符串函数，而不是尝试拆分字符串和聚合它：

查询:

WITH test_data ( str ) AS (
  SELECT 'Txa233141b Ta233141 Ta233142 Ta233147 Ta233148' AS str FROM DUAL
)
SELECT TRIM(
         TRAILING ',' FROM
         REGEXP_REPLACE(
           str,
           '.*?(\+?\d{3,11}|$)',
           ','
         )
       ) AS replaced_str
FROM   test_data

输出:

| REPLACED_STR                       |
| :--------------------------------- |
| 233141,233141,233142,233147,233148 |

db<>fiddle here

Oracle to select 通过正则表达式将所有匹配的事件匹配到一行中的一列，并带有分隔符

Oracle to select all matched occurrences by a regular expression to one column in one row with separator

regex

oracle

select

connect-by

regexp-substr