如何使用正则表达式仅替换括号内的内容? [英] How to replace only the contents within brackets using regular expressions?

问题描述

如何使用正则表达式只替换括号内的内容?

String = "这是我的字符串 [123]"

我想用 456 替换 123

Desired_String = "这是我的字符串 [456]"

我的尝试:

regex = '.*\[(.*?)\].*'re.sub(正则表达式，'456'，字符串)>>456

我不确定要使用什么正则表达式，以便将 123 替换为 456.

解决方案

如果你稍微修改一下正则表达式和替换字符串，你就会明白:

regex = '\[.*?\]'re.sub(regex,'[456]',String)

您无需将整个字符串与 .* 匹配即可进行替换.

此外，这里并不完全需要，但原始正则表达式字符串是一种很好的做法:

regex = r'\[.*?\]'^

<小时>

另一种选择是使用否定类，这也会更快！

regex = r'[^\[\]]+(?=\])'re.sub(正则表达式，'456'，字符串)

可能有点混乱，但是[^\[\]]+ 匹配除[ 和] 之外的任何字符.>

regex101 演示

How to replace only the contents within brackets using regular expressions?

String = "This is my string [123]"

I want to replace 123 with 456

Desired_String = "This is my string [456]"

My attempt:

regex = '.*\[(.*?)\].*'
re.sub(regex,'456',String)
>> 456

I'm not sure what regex to use so that it replaces the 123 with 456.

解决方案

If you modify your regex and your replacement string a little, you get it:

regex = '\[.*?\]'
re.sub(regex,'[456]',String)

You don't need to match the entire string with .* for the substitution to work.

Also, it's not entirely required here, but it's good practice to raw your regex strings:

regex = r'\[.*?\]'
        ^

---

Another option would be to use negated classes, which will be faster as well!

regex = r'[^\[\]]+(?=\])'
re.sub(regex,'456',String)

It might be a little confusing, but [^\[\]]+ matches any character except [ and ].

regex101 demo

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