带有正则表达式的.match（）返回null [英] .match() with a regular expression returns null

问题描述

我正在尝试做一些我认为很容易做的事情，即通过匹配正则表达式将字符串限制为某些字符。

I am trying to do something I thought would be pretty easy to do, which is to restrict a string to certain characters by matching a regular expression.

var value = 'FailureStr1ng';
var type = 'ALPHA';
var regex = null;

switch(type) {
    case 'ALPHA':
        regex = '^[a-zA-Z]+$';
        break;
    case 'NUMERIC':
        regex = '^[0-9]+$';
        break;
    case 'ALPHANUMERIC':
        regex = '^[a-zA-Z0-9]+$';
        break;
}

return value.match(regex);

出于某种原因，当使用匹配时，它总是返回 null 。有没有办法解决这个问题，或者有更好的方法来解决这个问题？

For some reason, when using the match it always returns null. Is there a way to fix this, or a better method to do this?

注意：这里的代码是更大代码的代码片段，反过来，值和类型变量通常由另一个方法定义。

Note: The code here is a snippet of much larger code, and in turn the value and type variable are usually defined by another method.

推荐答案

你想要 RegExp .test ，它测试匹配的值而不是检索匹配。使用现有代码，这意味着：

You want RegExp.test, which tests a value for a match instead of retrieving the match. With your existing code, that would mean:

if(!new RegExp(regex).test(value)){
    alert('Your string was invalid.');
}

然而，最好使用RegExp文字而不是字符串，因为它们'效率更高，更清晰，更不容易出错：

However, it would be preferable to use RegExp literals instead of strings, as they're much more efficient and clear, and less prone to error:

var value = 'FailureStr1ng';
var type = 'ALPHA';
var regex = null;

switch(type) {
    case 'ALPHA':
        regex = /^[a-zA-Z]+$/;
        break;
    case 'NUMERIC':
        regex = /^[0-9]+$/;
        break;
    case 'ALPHANUMERIC':
        regex = /^[a-zA-Z0-9]+$/;
        break;
}

if(!regex.test(value)) {
    alert('Your string was invalid.');
}

更好的是，使用字典：

var expressions = {
    ALPHA: /^[a-zA-Z]+$/,
    NUMERIC: /^[0-9]+$/,
    ALPHANUMERIC: /^[a-zA-Z0-9]+$/
};

if(!expressions[type].test(value)) {
    alert('Your string was invalid.');
}

这篇关于带有正则表达式的.match（）返回null的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！