正则表达式在JavaScript中没有按预期工作 [英] Regex not working as expected in JavaScript

问题描述

我写了以下正则表达式：

I wrote the following regex:

(https?:\/\/)?([da-z\.-]+)\.([a-z]{2,6})(\/(\w|-)*)*\/?

其行为可以在这里看到： http://gskinner.com/RegExr/?34b8m

Its behaviour can be seen here: http://gskinner.com/RegExr/?34b8m

我写了以下JavaScript代码：

I wrote the following JavaScript code:

var urlexp = new RegExp(
    '^(https?:\/\/)?([da-z\.-]+)\.([a-z]{2,6})(\/(\w|-)*)*\/?$', 'gi'
);
document.write(urlexp.test("blaaa"))

它返回 true 即使正则表达式不允许单个单词有效。

And it returns true even though the regex was supposed to not allow single words as valid.

我做错了什么？

推荐答案

您的问题是JavaScript正在查看所有转义序列作为字符串的转义符。所以你的正则表达式会记忆如下：

Your problem is that JavaScript is viewing all your escape sequences as escapes for the string. So your regex goes to memory looking like this:

^(https?://)?([da-z.-]+).([a-z]{2,6})(/(w|-)*)*/?$

当您认为文字句点变为正则表达式通配符时，您可能会注意到中间会出现问题。您可以通过几种方式解决此问题。使用正斜杠正则表达式语法JavaScript提供：

Which you may notice causes a problem in the middle when what you thought was a literal period turns into a regular expressions wildcard. You can solve this in a couple ways. Using the forward slash regular expression syntax JavaScript provides:

var urlexp = /^(https?:\/\/)?([da-z\.-]+)\.([a-z]{2,6})(\/(\w|-)*)*\/?$/gi

或者通过逃避你的反斜杠（而不是你正在做的正斜线 - 这是专属于你的时候）重新使用 / regex / mod 表示法，就像你不必在双引号字符串中转义单引号一样，反之亦然）：

Or by escaping your backslashes (and not your forward slashes, as you had been doing - that's exclusively for when you're using /regex/mod notation, just like you don't have to escape your single quotes in a double quoted string and vice versa):

var urlexp = new RegExp('^(https?://)?([da-z.-]+)\\.([a-z]{2,6})(/(\\w|-)*)*/?$', 'gi')

请注意w之前的双反斜杠 - 也是匹配单词字符所必需的。

Please note the double backslash before the w - also necessary for matching word characters.

关于你的正则表达式本身的几个注释：

A couple notes on your regular expression itself:

[da-z.-]

d 包含在az范围内。除非你的意思是 \d ？在这种情况下，斜杠很重要。

d is contained in the a-z range. Unless you meant \d? In that case, the slash is important.

(/(\w|-)*)*/?

我自己对嵌套的Kleene星的疑虑放在一边，你可以把这个交替变成一个字符类，并完全删除终止 /？，因为尾随的斜杠将由您给出的组匹配。我改写为：

My own misgivings about the nested Kleene stars aside, you can whittle that alternation down into a character class, and drop the terminating /? entirely, as a trailing slash will be match by the group as you've given it. I'd rewrite as:

(/[\w-]*)*

虽然，也许您只想捕捉非空格字符？

Though, maybe you'd just like to catch non space characters?

(/[^/\s]*)*

无论如何，通过这种方式修改你的正则表达式看起来更像：

Anyway, modified this way your regular expression winds up looking more like:

^(https?://)?([\da-z.-]+)\.([a-z]{2,6})(/[\w-]*)*$

请记住，如果你要使用字符串表示法：双重反斜杠。如果您要使用原生 / regex / mod 表示法（我强烈推荐），请转义正斜杠。

Remember, if you're going to use string notation: Double EVERY backslash. If you're going to use native /regex/mod notation (which I highly recommend), escape your forward slashes.

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