捕获组与量词Regexp匹配 [英] Capture groups match with quantifier Regexp

问题描述

我是regex世界中的新手，我需要捕获一些不同类型的字符串.

I am newbie in regex world, I need to capture some different types of strings.

通过这种方式，请提出更愚蠢的方式来捕获此类字符串. n =任何正数(不相同)

By the way please suggest more elagant way to capture such strings. n = any positive number(not the same)

|n||0||0||0||0|
|n||n||0||0||0|
|n||n||n||0||0|
|n||n||n||n||0|
|n||n||n||n||n|

我试图使用这样的正则表达式来捕获字符串的第一和第二类型

I have tried to use such regular expression for capturing first and secodn types of strings

^\|([1-9]+)\|(?:([1-9]+)\|){4}|(?:(0)\|){4}$

零应视为单独的字符， 我需要捕获每个数字或零

Zero should be treated as separate char, I need to capture each number or zero

现在的问题是，它仅捕获第一个匹配的字符和最后一个匹配的字符

The problem now that it only captures first matched character and last one

但不能捕获其他数字

请帮助使用此正则表达式，如果有人提供更多卑​​鄙的方式(最后，我必须编写4个ors语句来捕获我的字符串类型)，那就太好了

Please help with this regular expression and it would be great if someone provides more elagant way ( at the end, I have to write 4 ors statements to capture my string types)

谢谢

推荐答案

我不确定是否足以满足您需求:

I'm not sure wheter it is sufficient for you or not:

\|(?:(0)|([0-9]+))\|

https://regex101.com/r/fX5xI4/2

现在，您必须将您的匹配项分成x个元素组，其中x是列数.我想那应该没事.

Now u have to split your matches into groups of x elements where x is number of colums. I suppose that should be just fine.

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