是否有允许使用正则表达式的 JavaScript 的 String.indexOf() 版本? [英] Is there a version of JavaScript's String.indexOf() that allows for regular expressions?

问题描述

在 javascript 中，是否有 String.indexOf() 的等价物，它采用正则表达式而不是第一个参数的字符串，同时仍然允许第二个参数?

In javascript, is there an equivalent of String.indexOf() that takes a regular expression instead of a string for the first first parameter while still allowing a second parameter ?

我需要做类似的事情

str.indexOf(/[abc]/ , i);

和

str.lastIndexOf(/[abc]/ , i);

虽然 String.search() 将正则表达式作为参数，但它不允许我指定第二个参数！

While String.search() takes a regexp as a parameter it does not allow me to specify a second argument!

结果证明这比我原先想象的要难，所以我编写了一个小的测试函数来测试所有提供的解决方案......它假设 regexIndexOf 和 regexLastIndexOf 已被添加到 String 对象中.

This turned out to be harder than I originally thought so I wrote a small test function to test all the provided solutions... it assumes regexIndexOf and regexLastIndexOf have been added to the String object.

function test (str) {
    var i = str.length +2;
    while (i--) {
        if (str.indexOf('a',i) != str.regexIndexOf(/a/,i)) 
            alert (['failed regexIndexOf ' , str,i , str.indexOf('a',i) , str.regexIndexOf(/a/,i)]) ;
        if (str.lastIndexOf('a',i) != str.regexLastIndexOf(/a/,i) ) 
            alert (['failed regexLastIndexOf ' , str,i,str.lastIndexOf('a',i) , str.regexLastIndexOf(/a/,i)]) ;
    }
}

我正在测试如下以确保至少对于一个字符的正则表达式，结果与我们使用 indexOf 的结果相同

and I am testing as follow to make sure that at least for one character regexp, the result is the same as if we used indexOf

//在xes中寻找a
测试('xxx');
测试('axx');
测试('xax');
测试('xxa');
测试('axa');
测试('xaa');
测试('aax');
测试('aaa')；

//Look for the a among the xes
test('xxx');
test('axx');
test('xax');
test('xxa');
test('axa');
test('xaa');
test('aax');
test('aaa');

推荐答案

结合一些已经提到的方法(indexOf 显然相当简单)，我认为这些是可以解决问题的函数:

Combining a few of the approaches already mentioned (the indexOf is obviously rather simple), I think these are the functions that will do the trick:

function regexIndexOf(string, regex, startpos) {
    var indexOf = string.substring(startpos || 0).search(regex);
    return (indexOf >= 0) ? (indexOf + (startpos || 0)) : indexOf;
}

function regexLastIndexOf(string, regex, startpos) {
    regex = (regex.global) ? regex : new RegExp(regex.source, "g" + (regex.ignoreCase ? "i" : "") + (regex.multiLine ? "m" : ""));
    if(typeof (startpos) == "undefined") {
        startpos = string.length;
    } else if(startpos < 0) {
        startpos = 0;
    }
    var stringToWorkWith = string.substring(0, startpos + 1);
    var lastIndexOf = -1;
    var nextStop = 0;
    while((result = regex.exec(stringToWorkWith)) != null) {
        lastIndexOf = result.index;
        regex.lastIndex = ++nextStop;
    }
    return lastIndexOf;
}

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更新:编辑了 regexLastIndexOf() 以便现在似乎模仿 lastIndexOf().请让我知道它是否仍然失败以及在什么情况下.

---

UPDATE: Edited regexLastIndexOf() so that is seems to mimic lastIndexOf() now. Please let me know if it still fails and under what circumstances.

更新:通过本页评论中发现的所有测试，以及我自己的测试.当然，这并不意味着它是防弹的.任何反馈表示赞赏.

UPDATE: Passes all tests found on in comments on this page, and my own. Of course, that doesn't mean it's bulletproof. Any feedback appreciated.

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