转义正则表达式中的变量 [英] Escape a variable within a Regular Expression

问题描述

我试图获取输入文本字段的值并在正则表达式中使用它。这是我必须匹配一行的开头。

I am trying to take the value of an input text field and use it within a regular expression. Here's what I have to match the start of a line.

regex = new RegExp('^' + inputValue, 'i')

对于以字母数字字符开头的常规字符串，它可以正常工作，但是我使用这个字符串来计算金额同样。当输入字段以'$'（美元符号）开头时，我会得到不同的结果。

It works fine for regular strings that start have alphanumeric characters, but I am using this for dollar amounts as well. When the input field starts with a '$' (dollar sign) i get varying results.

如何将上面的变量转义为简单的字符串？我试过这个，但没有运气：

How can I escape the variable above to be viewed as a simple string? I've tried this, but no luck:

regex = new RegExp('^[' + inputValue + ']', 'i')

推荐答案

可能是我更好的实施之一已见过

Probably one of the better implementations I've seen

http://phpjs.org / functions / preg_quote：491

下面为后代复制

function preg_quote (str, delimiter) {
    // Quote regular expression characters plus an optional character  
    // 
    // version: 1109.2015
    // discuss at: http://phpjs.org/functions/preg_quote    // +   original by: booeyOH
    // +   improved by: Ates Goral (http://magnetiq.com)
    // +   improved by: Kevin van Zonneveld (http://kevin.vanzonneveld.net)
    // +   bugfixed by: Onno Marsman
    // +   improved by: Brett Zamir (http://brett-zamir.me)    // *     example 1: preg_quote("$40");
    // *     returns 1: '\$40'
    // *     example 2: preg_quote("*RRRING* Hello?");
    // *     returns 2: '\*RRRING\* Hello\?'
    // *     example 3: preg_quote("\\.+*?[^]$(){}=!<>|:");    // *     returns 3: '\\\.\+\*\?\[\^\]\$\(\)\{\}\=\!\<\>\|\:'
    return (str + '').replace(new RegExp('[.\\\\+*?\\[\\^\\]$(){}=!<>|:\\' + (delimiter || '') + '-]', 'g'), '\\$&');
}

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