Question

我试试这个

SELECT regexp_replace(v_gen,'(\b[a-zA-Z]+\b)','$$1%')INTO v_test FROM DUAL;

它应该在'hello world word'

中转换$hello% $world% $word%'

Oracle文档不清楚函数regexp_replace与group ...

Answer 1

regexp_replace('hello world word','([a-zA-Z]+)','$\1%')

\1为您提供与模式匹配的字符串，然后我们将其格式化，附加所需的其他字符。

[a-zA-Z]+查找所有连续的字母序列。

([[:alnum:]]+)将查找字母数字字符序列。如果你想要一个，你可以使用它。

示例：使用多个选项

with my_text as
(
  select 'hello world word' as str from dual
)
SELECT regexp_replace(str,'([[:alnum:]]+)','$\1%') FROM my_text
union all
SELECT regexp_replace(str,'([[:alpha:]]+)','$\1%') FROM my_text
union all
SELECT regexp_replace(str,'([a-zA-Z]+)','$\1%') FROM my_text

REGEXP_REPLACE(STR,'([
----------------------
$hello% $world% $word%
$hello% $world% $word%
$hello% $world% $word%

SQLFiddle Demo

Oracle regexp_replace with group

1 个答案: