如何使用正则表达式使用字符串作为分隔符来拆分字符串? [英] How can I use regex to split a string, using a string as a delimiter?
问题描述
我正在尝试在Oracle存储过程中使用字符串作为分隔符来拆分字符串.我可以轻松地使用instr
,但是我正在尝试学习如何使用正则表达式,因为我了解它功能强大且高效.
I'm trying to split a string using a string as a delimiter, in an Oracle store procedure. I can use instr
easily, but I'm trying to learn how to do this with regex, as I understand that it is powerful and efficient.
在阅读了一些文章之后,我认为我可以做到这一点(预期结果是"Hello
"):
After reading some articles, I thought I could do this (expected result was "Hello
"):
select regexp_substr('Hello My Delimiter World', '( My Delimiter )+', 1, 1)
from dual
结果:
我的分隔符
My Delimiter
和(预期结果为"World
"):
and (expected result was "World
"):
select regexp_substr('Hello My Delimiter World', '( My Delimiter )+', 1, 2)
from dual
结果:
空
此要求的正确regex_substr是什么?
What is the correct regex_substr for this requirement?
我正在寻找类似下面的内容.通过一次操作,它将选择字符串中的子字符串:
I'm looking for something like the below. In a single pass, it selects the sub-string within the string:
例如选择regexp_substr('Hello World', '[^ ]+', 1, 2) from dual
,但是此示例仅适用于单个字符.
E.g. select regexp_substr('Hello World', '[^ ]+', 1, 2) from dual
But this sample only works with a single character.
推荐答案
尝试这些方法.
这将获得您最初要求的第一个元素:
This gets the first element as you originally asked for:
SQL> with tbl(str) as (
select 'Hello My Delimiter World' from dual
)
SELECT REGEXP_SUBSTR( str ,'(.*?)( My Delimiter |$)', 1, 1, NULL, 1 ) AS element
FROM tbl;
ELEME
-----
Hello
此版本解析整个字符串.添加了NULL元素以显示它可用于缺少的元素:
This version parses the whole string. NULL elements added to show it works with missing elements:
SQL> with tbl(str) as (
select ' My Delimiter Hello My Delimiter World My Delimiter My Delimiter test My Delimiter ' from dual
)
SELECT LEVEL AS element,
REGEXP_SUBSTR( str ,'(.*?)( My Delimiter |$)', 1, LEVEL, NULL, 1 ) AS element_value
FROM tbl
CONNECT BY LEVEL <= regexp_count(str, ' My Delimiter ')+1;
ELEMENT ELEMENT_VALUE
---------- --------------------
1
2 Hello
3 World
4
5 test
6
6 rows selected.
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