如何让正则表达式忽略括号之间的所有内容? [英] How to let regex ignore everything between brackets?

问题描述

考虑以下字符串:

I have been driving to {Palm.!.Beach:100} and it . was . great!!

我使用以下正则表达式删除所有标点符号:

I use the following regex to delete all punctuation:

$string preg_replace('/[^a-zA-Z ]+/', '', $string);

输出:

I have been driving to PalmBeach and it  was  great!!

但我需要正则表达式始终忽略 { 和 } 之间的任何内容.所以期望的输出是:

But I need the regex to always ignore whatever is in between { and }. So the desired output would be:

I have been driving to {Palm.!.Beach:100} and it  was  great

如何让正则表达式忽略 { 和 } 之间的内容?

How can I let the regex ignore what is between { and }?

推荐答案

试试这个

[^a-zA-Z {}]+(?![^{]*})

在 Regexr 上查看

表示匹配任何不包含在否定字符类中的东西，但前提是前面没有右括号而前面没有开括号，这是通过否定前瞻(?![^{]*}).

Means match anything that is not included in the negated character class, but only if there is no closing bracket ahead without a opening before, this is done by the negative lookahead (?![^{]*}).

$string preg_replace('/[^a-zA-Z {}]+(?![^{]*})/', '', $string);

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