如何在同一个正则表达式中使用捕获的组 [英] How can I use the captured group in the same regex

问题描述

mFoo = foo;
mBar = bar;
// convert to
this.foo = foo;
this.bar = bar;

如何使用正则表达式来处理这种替换?请帮忙.这是我在Android Studio(IntelliJ IDEA)Edit -> Find -> Replace in Path

中使用的方法

Text to find: m([A-Z])([A-Za-z0-9]+) = L$1$2
Replace with: this\.L$1$2 = L$1$2

更新

上面的

L是一个错字.根据 JetBrains的文档，它应该为\L

解决方案

您可以使用带有反向引用的模式，并将"=:"后面的最后一个单词分组

要查找的文本:

m([A-Z])([A-Za-z0-9] +)=(L \ 1 \ 2)

替换为:

this.L $ 1 $ 2 = $ 3

在您发表评论后，我了解到您在使用小写/大写字符时也遇到了问题.因此，请尝试这种模式(我也简化了正则表达式):

m(\p{Alpha})(\w+) = (((?i)\1)\2)

和此替换字符串:

this\.L$1$2 = $3

在您的示例中输入文本:

mContext = context

您获得了:

this.LContext = context

我不知道您的文本/替换字符串中指定的"L"是您的错字错误还是其他错误，但是如果是这样，您可以通过以下方式更改替换字符串":

this\.$3 = $3

因此您可以获取此信息:

this.context = context

让我知道这是否对您有帮助！

mFoo = foo;
mBar = bar;
// convert to
this.foo = foo;
this.bar = bar;

How to use a regex to handle this substitution? Please help. Here is the method I used in Android Studio (IntelliJ IDEA) Edit -> Find -> Replace in Path

Text to find: m([A-Z])([A-Za-z0-9]+) = L$1$2
Replace with: this\.L$1$2 = L$1$2

Update

L above is a typo. It should be \L according to JetBrains' document

解决方案

You can use a pattern with back-reference and grouping the last words after "=":

Text to find:

m([A-Z])([A-Za-z0-9]+) = (L\1\2)

Replace with:

this.L$1$2 = $3

After your comments, I understand that you have also problems with lowercase/uppercase characters. So try this pattern (I have also simplified the regex):

m(\p{Alpha})(\w+) = (((?i)\1)\2)

and this replace string:

this\.L$1$2 = $3

So with your example with an input text:

mContext = context

you obtain this:

this.LContext = context

I don't know if "L" specified in your text/replace string is your typo error or other but if it's so you can change the "replace string" in the following way:

this\.$3 = $3

So you can obtain this:

this.context = context

Let me know if this help you!

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