提取方括号"[]"之间的数据使用Perl [英] Extract data between square brackets "[]" using Perl

问题描述

我正在使用正则表达式从弧形括号(或括号")中提取数据，就像从(a,b)中提取a,b一样，如下所示.我有一个文件，其中每一行都会像

I was using a regex for extracting data from curved brackets (or "parentheses") like extracting a,b from (a,b) as shown below. I have a file in which every line will be like

this is the range of values (a1,b1) and [b1|a1]
this is the range of values (a2,b2) and [b2|a2]
this is the range of values (a3,b3) and [b3|a3]

我正在使用以下字符串提取a1,b1，a2,b2等...

I'm using the following string to extract a1,b1, a2,b2, etc...

@numbers = $_ =~ /\((.*),(.*)\)/

但是，如果我想从方括号[]中提取数据，该怎么办?例如

However, if I want to extract the data from square brackets [], how can I do it? For example

this is the range of values (a1,b1) and [b1|a1]
this is the range of values (a1,b1) and [b2|a2]

我只需要提取/匹配方括号中的数据，而不提取/匹配括号中的数据.

I need to extract/match only the data in square brackets and not the curved brackets.

推荐答案

[更新] 同时，我写了一篇有关.*特定问题的博客文章，如下所述: 为什么在正则表达式中几乎不使用.*

[Update] In the meantime, I've written a blog post about the specific issue with .* I describe below: Why Using .* in Regular Expressions Is Almost Never What You Actually Want

如果标识符a1，b1等本身不包含逗号或方括号，则应按照以下方式使用模式，以避免回溯地狱:

If your identifiers a1, b1 etc. never contain commas or square brackets themselves, you should use a pattern along the lines of the following to avoid backtracking hell:

/\[([^,\]]+),([^,\]]+)\]/

这是Regex101上的工作示例.

Here's a working example on Regex101.

像.*这样的贪婪量词的问题在于，一开始您很可能会消耗过多，因此正则表达式引擎必须进行大量的回溯.即使您使用非贪婪的量词，引擎也会进行比实际需要更多的匹配尝试，因为它一次只消耗一个字符，然后尝试提升模式中的位置.

The issue with greedy quantifiers like .* is that you'll very likely consume too much in the beginning so that the regex engine has to do extensive backtracking. Even if you use non-greedy quantifiers, the engine will do more attempts to match than necessary because it'll only consume one character at a time and then try to advance the position in the pattern.

(您甚至可以使用原子组使匹配更加有效.)

(You could even use atomic groups to make the matching even more performant.)

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