在 Windows 命令行的 Grep 中使用正则表达式 [英] Using regex in Grep for Windows command line

问题描述

我想捕获包含恰好 3 个字段的所有行，其中一个字段是任何字符串(可能为空)后跟一个 |(并且可能有一些最终的行尾的文本).

我设法构建了一个 regex，它似乎完全符合我的要求

^(?:[^|]*|){3}[^|]*$

当我在

I want to capture all lines which contain exactly 3 fields, where a field is any string (possibly empty) followed by a | (and there may be some final text at the end of the line).

I managed to build a regex which seems to do exactly what I want

^(?:[^|]*|){3}[^|]*$

and when I try it on 101regex it seems to work just fine.

However, I am having problems to run this regex on the Windows command line via grep and I guess it has something to do with the proper escaping.

I tried

grep -E '^^(?:[^^^^|]*^^|){3}[^^^^|]*$' test.txt
grep -E '^^(?:[^^^|]*^|){3}[^^^|]*$' test.txt

but nothing helped. Any ideas?

---

Test Input

0|1|2|3
0|1|2|
|1|2|3
|1|2|
|1|2
|1|
0|1|2
0|1|
|1|2|3|4
|1|2|3|
0|1|2|3|4
0|1|2|3|

---

解决方案

In grep, when you use POSIX ERE regex engine, you need to avoid backslashes in bracket expressions and non-capturing groups:

grep -E "^([^|]*|){3}[^|]*$" test.txt

Here, [^|] is turned into [^|] (since POSIX bracket expressions do not treat escaped chars as regex escapes) and (?: is replaced with (, i.e. the group was made capturing since non-capturing ones are not supported.

See proof it is working:

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