用于标识符的Java正则表达式（字母，数字和下划线）(Java regex for identifiers (letters, digits and underscores))

用于标识符的Java正则表达式（字母，数字和下划线）(Java regex for identifiers (letters, digits and underscores))

假设您输入的内容可能如下所示(identifier1 identifier_2 23 4) 。

我想在每个标识符之后添加一个#符号，其中可以包含字母，数字和下划线。 他们只能以字母，数字和下划线的变体开头。 我的做法是这样的：

input.replaceAll("[A-Za-z0-9_]+", "$0#");

但是，这也会在每个我想排除的数字后加上#符号。 结果应该是(identifier1# identifier_2# 23 4) 。 是否有可能用正则表达式解决这个问题？

Let's say you have given an input which could look like this (identifier1 identifier_2 23 4).

I want to add a # symbol after every identifier, which can contain letters, digits and underscores. They can only start with a letter followed by variations of letters, digits and underscores. My approach was something like this:

input.replaceAll("[A-Za-z0-9_]+", "$0#");

However, this also puts # symbols after every single digit which I wanted to exclude. The result should be (identifier1# identifier_2# 23 4). Is it possible to solve this problem with regex?

最满意答案

你目前的正则表达式说

一个或多个大写或小写字母，数字或下划线，无论顺序如何。

根据那个正则表达式， 54是一个有效的标识符。

你真的想写

一封信，然后是任意数量的字母，数字或下划线，无论顺序如何

这将写在代码中：

input.replaceAll("[A-Za-z][A-Za-z0-9_]*", "$0#");

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Wiktor指出，这个正则表达式仍然会匹配那些不是标识符的东西的“标识符”。 为了解决这个问题，你可以使用下面的变体：

input.replaceAll("\\b([A-Za-z][A-Za-z0-9_]*)\\b", "$1#")

这拒绝123ab123作为有效标识符，但接受123 ab123

Your current regex says

one or more upper or lower-case letters, digits, or underscores, in whatever order.

According to that regex, 54 is a valid identifier.

You actually wanted to write

a letter, followed by any number of letters, digits or underscores, in whatever order

That would be written in code as:

input.replaceAll("[A-Za-z][A-Za-z0-9_]*", "$0#");

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Wiktor notes that this regex will still match "identifiers" that are inside something that is not identifier-ish. To solve this, you could use the following variation:

input.replaceAll("\\b([A-Za-z][A-Za-z0-9_]*)\\b", "$1#")

This rejects 123ab123 as a valid identifier, but accepts ab123 in 123 ab123