在javascript中,是否有一个string . indexof()的等效,为第一个参数接受正则表达式而不是字符串,同时仍然允许第二个参数?

我需要做点什么

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

and

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

虽然String.search()接受regexp作为参数,但它不允许我指定第二个参数!

编辑: 这比我最初想象的要难,所以我写了一个小测试函数来测试所有提供的解决方案……它假设regexIndexOf和regexLastIndexOf已经添加到String对象中。

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)]) ;
    }
}

我正在进行如下测试,以确保至少对于一个字符regexp,如果我们使用indexOf,结果是相同的

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”);


你可以使用substr。

str.substr(i).match(/[abc]/);

String构造函数的实例有一个.search()方法,该方法接受RegExp并返回第一个匹配项的索引。

要从特定位置开始搜索(伪造.indexOf()的第二个参数),可以切掉前i个字符:

str.slice(i).search(/re/)

但这将获得较短字符串的索引(在第一部分被切掉之后),因此如果返回的索引不是-1,则需要将切掉的部分(i)的长度添加到返回的索引中。这将给你原始字符串的索引:

function regexIndexOf(text, re, i) {
    var indexInSuffix = text.slice(i).search(re);
    return indexInSuffix < 0 ? indexInSuffix : indexInSuffix + i;
}

它不是原生的,但您当然可以添加此功能

<script type="text/javascript">

String.prototype.regexIndexOf = function( pattern, startIndex )
{
    startIndex = startIndex || 0;
    var searchResult = this.substr( startIndex ).search( pattern );
    return ( -1 === searchResult ) ? -1 : searchResult + startIndex;
}

String.prototype.regexLastIndexOf = function( pattern, startIndex )
{
    startIndex = startIndex === undefined ? this.length : startIndex;
    var searchResult = this.substr( 0, startIndex ).reverse().regexIndexOf( pattern, 0 );
    return ( -1 === searchResult ) ? -1 : this.length - ++searchResult;
}

String.prototype.reverse = function()
{
    return this.split('').reverse().join('');
}

// Indexes 0123456789
var str = 'caabbccdda';

alert( [
        str.regexIndexOf( /[cd]/, 4 )
    ,   str.regexLastIndexOf( /[cd]/, 4 )
    ,   str.regexIndexOf( /[yz]/, 4 )
    ,   str.regexLastIndexOf( /[yz]/, 4 )
    ,   str.lastIndexOf( 'd', 4 )
    ,   str.regexLastIndexOf( /d/, 4 )
    ,   str.lastIndexOf( 'd' )
    ,   str.regexLastIndexOf( /d/ )
    ]
);

</script>

我没有完全测试这些方法,但到目前为止,它们似乎是有效的。


根据BaileyP的回答。主要的区别是,如果模式不能匹配,这些方法将返回-1。

编辑:感谢Jason Bunting的回答,我有了一个想法。为什么不修改正则表达式的.lastIndex属性?尽管这只适用于带有全局标志(/g)的模式。

编辑:更新以通过测试用例。

String.prototype.regexIndexOf = function(re, startPos) {
    startPos = startPos || 0;

    if (!re.global) {
        var flags = "g" + (re.multiline?"m":"") + (re.ignoreCase?"i":"");
        re = new RegExp(re.source, flags);
    }

    re.lastIndex = startPos;
    var match = re.exec(this);

    if (match) return match.index;
    else return -1;
}

String.prototype.regexLastIndexOf = function(re, startPos) {
    startPos = startPos === undefined ? this.length : startPos;

    if (!re.global) {
        var flags = "g" + (re.multiline?"m":"") + (re.ignoreCase?"i":"");
        re = new RegExp(re.source, flags);
    }

    var lastSuccess = -1;
    for (var pos = 0; pos <= startPos; pos++) {
        re.lastIndex = pos;

        var match = re.exec(this);
        if (!match) break;

        pos = match.index;
        if (pos <= startPos) lastSuccess = pos;
    }

    return lastSuccess;
}

好吧,因为你只是想匹配字符的位置,regex可能是多余的。

我假设你想要的不是,找到这些字符中的第一个,而是找到这些字符中的第一个。

这当然是一个简单的答案,但做到了你的问题所要做的事情,尽管没有正则表达式部分(因为你没有明确说明为什么它必须是一个正则表达式)

function mIndexOf( str , chars, offset )
{
   var first  = -1; 
   for( var i = 0; i < chars.length;  i++ )
   {
      var p = str.indexOf( chars[i] , offset ); 
      if( p < first || first === -1 )
      {
           first = p;
      }
   }
   return first; 
}
String.prototype.mIndexOf = function( chars, offset )
{
   return mIndexOf( this, chars, offset ); # I'm really averse to monkey patching.  
};
mIndexOf( "hello world", ['a','o','w'], 0 );
>> 4 
mIndexOf( "hello world", ['a'], 0 );
>> -1 
mIndexOf( "hello world", ['a','o','w'], 4 );
>> 4
mIndexOf( "hello world", ['a','o','w'], 5 );
>> 6
mIndexOf( "hello world", ['a','o','w'], 7 );
>> -1 
mIndexOf( "hello world", ['a','o','w','d'], 7 );
>> 10
mIndexOf( "hello world", ['a','o','w','d'], 10 );
>> 10
mIndexOf( "hello world", ['a','o','w','d'], 11 );
>> -1

结合已经提到的一些方法(indexOf显然相当简单),我认为这些函数将达到目的:

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;
}

更新:编辑regexLastIndexOf(),所以它似乎是模仿lastIndexOf()现在。请让我知道它是否仍然失败,在什么情况下。


更新:通过本页评论和我自己的所有测试。当然,这并不意味着它是防弹的。感谢任何反馈。


在所有建议的解决方案都以这样或那样的方式失败了我的测试之后(编辑:在我写这篇文章之后,一些解决方案被更新以通过测试),我找到了Array的mozilla实现。indexOf和Array.lastIndexOf

我使用这些来实现我的版本的String.prototype.regexIndexOf和String.prototype.regexLastIndexOf如下:

String.prototype.regexIndexOf = function(elt /*, from*/)
  {
    var arr = this.split('');
    var len = arr.length;

    var from = Number(arguments[1]) || 0;
    from = (from < 0) ? Math.ceil(from) : Math.floor(from);
    if (from < 0)
      from += len;

    for (; from < len; from++) {
      if (from in arr && elt.exec(arr[from]) ) 
        return from;
    }
    return -1;
};

String.prototype.regexLastIndexOf = function(elt /*, from*/)
  {
    var arr = this.split('');
    var len = arr.length;

    var from = Number(arguments[1]);
    if (isNaN(from)) {
      from = len - 1;
    } else {
      from = (from < 0) ? Math.ceil(from) : Math.floor(from);
      if (from < 0)
        from += len;
      else if (from >= len)
        from = len - 1;
    }

    for (; from > -1; from--) {
      if (from in arr && elt.exec(arr[from]) )
        return from;
    }
    return -1;
  };

它们似乎通过了我在问题中提供的测试函数。

显然,它们只在正则表达式匹配一个字符时才有效,但这对于我的目的来说已经足够了,因为我将使用它来处理([abc], \s, \W, \D)之类的事情。

我将继续关注这个问题,以防有人提供更好/更快/更干净/更通用的实现,适用于任何正则表达式。


RexExp实例已经有一个lastIndex属性(如果它们是全局的),所以我所做的是复制正则表达式,稍微修改它以适应我们的目的,在字符串上执行它并查看lastIndex。这将不可避免地比在字符串上循环更快。(你有足够的例子,如何把它放在字符串原型,对吧?)

function reIndexOf(reIn, str, startIndex) {
    var re = new RegExp(reIn.source, 'g' + (reIn.ignoreCase ? 'i' : '') + (reIn.multiLine ? 'm' : ''));
    re.lastIndex = startIndex || 0;
    var res = re.exec(str);
    if(!res) return -1;
    return re.lastIndex - res[0].length;
};

function reLastIndexOf(reIn, str, startIndex) {
    var src = /\$$/.test(reIn.source) && !/\\\$$/.test(reIn.source) ? reIn.source : reIn.source + '(?![\\S\\s]*' + reIn.source + ')';
    var re = new RegExp(src, 'g' + (reIn.ignoreCase ? 'i' : '') + (reIn.multiLine ? 'm' : ''));
    re.lastIndex = startIndex || 0;
    var res = re.exec(str);
    if(!res) return -1;
    return re.lastIndex - res[0].length;
};

reIndexOf(/[abc]/, "tommy can eat");  // Returns 6
reIndexOf(/[abc]/, "tommy can eat", 8);  // Returns 11
reLastIndexOf(/[abc]/, "tommy can eat"); // Returns 11

你也可以在RegExp对象上创建函数原型:

RegExp.prototype.indexOf = function(str, startIndex) {
    var re = new RegExp(this.source, 'g' + (this.ignoreCase ? 'i' : '') + (this.multiLine ? 'm' : ''));
    re.lastIndex = startIndex || 0;
    var res = re.exec(str);
    if(!res) return -1;
    return re.lastIndex - res[0].length;
};

RegExp.prototype.lastIndexOf = function(str, startIndex) {
    var src = /\$$/.test(this.source) && !/\\\$$/.test(this.source) ? this.source : this.source + '(?![\\S\\s]*' + this.source + ')';
    var re = new RegExp(src, 'g' + (this.ignoreCase ? 'i' : '') + (this.multiLine ? 'm' : ''));
    re.lastIndex = startIndex || 0;
    var res = re.exec(str);
    if(!res) return -1;
    return re.lastIndex - res[0].length;
};


/[abc]/.indexOf("tommy can eat");  // Returns 6
/[abc]/.indexOf("tommy can eat", 8);  // Returns 11
/[abc]/.lastIndexOf("tommy can eat"); // Returns 11

快速解释一下如何修改RegExp:对于indexOf,我只需要确保设置了全局标志。对于lastIndexOf,除非RegExp已经在字符串的末尾匹配,否则我将使用负向前查找最后一次出现。


我还需要一个regexIndexOf函数用于数组,所以我自己编写了一个。然而,我怀疑,这是优化,但我猜它应该工作正常。

Array.prototype.regexIndexOf = function (regex, startpos = 0) {
    len = this.length;
    for(x = startpos; x < len; x++){
        if(typeof this[x] != 'undefined' && (''+this[x]).match(regex)){
            return x;
        }
    }
    return -1;
}

arr = [];
arr.push(null);
arr.push(NaN);
arr[3] = 7;
arr.push('asdf');
arr.push('qwer');
arr.push(9);
arr.push('...');
console.log(arr);
arr.regexIndexOf(/\d/, 4);

在某些简单的情况下,您可以通过使用split简化向后搜索。

function regexlast(string,re){
  var tokens=string.split(re);
  return (tokens.length>1)?(string.length-tokens[tokens.length-1].length):null;
}

这有一些严重的问题:

重叠的匹配不会显示出来 返回的索引是匹配的结束,而不是开始(如果你的regex是一个常量,没问题)

但从好的方面来看,它的代码更少。对于一个不能重叠的定长正则表达式(比如/\s\w/用于查找单词边界),这已经足够好了。


我有一个简短的版本给你。这对我来说很有效!

var match      = str.match(/[abc]/gi);
var firstIndex = str.indexOf(match[0]);
var lastIndex  = str.lastIndexOf(match[match.length-1]);

如果你想要一个原型版本:

String.prototype.indexOfRegex = function(regex){
  var match = this.match(regex);
  return match ? this.indexOf(match[0]) : -1;
}

String.prototype.lastIndexOfRegex = function(regex){
  var match = this.match(regex);
  return match ? this.lastIndexOf(match[match.length-1]) : -1;
}

编辑:如果你想添加对fromIndex的支持

String.prototype.indexOfRegex = function(regex, fromIndex){
  var str = fromIndex ? this.substring(fromIndex) : this;
  var match = str.match(regex);
  return match ? str.indexOf(match[0]) + fromIndex : -1;
}

String.prototype.lastIndexOfRegex = function(regex, fromIndex){
  var str = fromIndex ? this.substring(0, fromIndex) : this;
  var match = str.match(regex);
  return match ? str.lastIndexOf(match[match.length-1]) : -1;
}

要使用它,就像这样简单:

var firstIndex = str.indexOfRegex(/[abc]/gi);
var lastIndex  = str.lastIndexOfRegex(/[abc]/gi);

对于具有稀疏匹配的数据,使用字符串。跨浏览器搜索速度最快。它每次迭代都会重新切片字符串:

function lastIndexOfSearch(string, regex, index) {
  if(index === 0 || index)
     string = string.slice(0, Math.max(0,index));
  var idx;
  var offset = -1;
  while ((idx = string.search(regex)) !== -1) {
    offset += idx + 1;
    string = string.slice(idx + 1);
  }
  return offset;
}

对于密集的数据,我做了这个。与执行方法相比,它比较复杂,但对于密集数据,它比我尝试过的其他方法快2-10倍,比公认的解决方案快100倍左右。要点如下:

It calls exec on the regex passed in once to verify there is a match or quit early. I do this using (?= in a similar method, but on IE checking with exec is dramatically faster. It constructs and caches a modified regex in the format '(r).(?!.?r)' The new regex is executed and the results from either that exec, or the first exec, are returned; function lastIndexOfGroupSimple(string, regex, index) { if (index === 0 || index) string = string.slice(0, Math.max(0, index + 1)); regex.lastIndex = 0; var lastRegex, index flags = 'g' + (regex.multiline ? 'm' : '') + (regex.ignoreCase ? 'i' : ''), key = regex.source + '$' + flags, match = regex.exec(string); if (!match) return -1; if (lastIndexOfGroupSimple.cache === undefined) lastIndexOfGroupSimple.cache = {}; lastRegex = lastIndexOfGroupSimple.cache[key]; if (!lastRegex) lastIndexOfGroupSimple.cache[key] = lastRegex = new RegExp('.*(' + regex.source + ')(?!.*?' + regex.source + ')', flags); index = match.index; lastRegex.lastIndex = match.index; return (match = lastRegex.exec(string)) ? lastRegex.lastIndex - match[1].length : index; };

方法的jsPerf

我不明白上面这些测试的目的。需要正则表达式的情况是不可能与调用indexOf进行比较的,我认为这是首先创建该方法的目的。为了让测试通过,使用'xxx+(?!x)'比调整regex迭代的方式更有意义。


杰森·邦廷的最后一个指数不成立。我的方法不是最优的,但有效。

//Jason Bunting's
String.prototype.regexIndexOf = function(regex, startpos) {
var indexOf = this.substring(startpos || 0).search(regex);
return (indexOf >= 0) ? (indexOf + (startpos || 0)) : indexOf;
}

String.prototype.regexLastIndexOf = function(regex, startpos) {
var lastIndex = -1;
var index = this.regexIndexOf( regex );
startpos = startpos === undefined ? this.length : startpos;

while ( index >= 0 && index < startpos )
{
    lastIndex = index;
    index = this.regexIndexOf( regex, index + 1 );
}
return lastIndex;
}

Use:

str.search(regex)

请在这里查看文档。


仍然没有执行请求任务的本机方法。

这是我正在使用的代码。它模仿了string .prototype. indexof和string .prototype. lastindexof方法的行为,但是除了表示要搜索的值的字符串之外,它们还接受RegExp作为搜索参数。

是的,这是一个相当长的答案,因为它试图遵循当前的标准尽可能接近,当然包含了合理数量的JSDOC评论。然而,一旦缩小,代码只有2.27k,一旦gzip传输,它只有1023字节。

这个添加到String中的2个方法。prototype(使用Object.defineProperty如果可用)是:

搜索的 搜索尾页

它通过了OP发布的所有测试,此外,我在日常使用中已经相当彻底地测试了这些例程,并试图确保它们在多个环境中工作,但反馈/问题总是受欢迎的。

/*jslint maxlen:80, browser:true */ /* * Properties used by searchOf and searchLastOf implementation. */ /*property MAX_SAFE_INTEGER, abs, add, apply, call, configurable, defineProperty, enumerable, exec, floor, global, hasOwnProperty, ignoreCase, index, lastIndex, lastIndexOf, length, max, min, multiline, pow, prototype, remove, replace, searchLastOf, searchOf, source, toString, value, writable */ /* * Properties used in the testing of searchOf and searchLastOf implimentation. */ /*property appendChild, createTextNode, getElementById, indexOf, lastIndexOf, length, searchLastOf, searchOf, unshift */ (function () { 'use strict'; var MAX_SAFE_INTEGER = Number.MAX_SAFE_INTEGER || Math.pow(2, 53) - 1, getNativeFlags = new RegExp('\\/([a-z]*)$', 'i'), clipDups = new RegExp('([\\s\\S])(?=[\\s\\S]*\\1)', 'g'), pToString = Object.prototype.toString, pHasOwn = Object.prototype.hasOwnProperty, stringTagRegExp; /** * Defines a new property directly on an object, or modifies an existing * property on an object, and returns the object. * * @private * @function * @param {Object} object * @param {string} property * @param {Object} descriptor * @returns {Object} * @see https://goo.gl/CZnEqg */ function $defineProperty(object, property, descriptor) { if (Object.defineProperty) { Object.defineProperty(object, property, descriptor); } else { object[property] = descriptor.value; } return object; } /** * Returns true if the operands are strictly equal with no type conversion. * * @private * @function * @param {*} a * @param {*} b * @returns {boolean} * @see http://www.ecma-international.org/ecma-262/5.1/#sec-11.9.4 */ function $strictEqual(a, b) { return a === b; } /** * Returns true if the operand inputArg is undefined. * * @private * @function * @param {*} inputArg * @returns {boolean} */ function $isUndefined(inputArg) { return $strictEqual(typeof inputArg, 'undefined'); } /** * Provides a string representation of the supplied object in the form * "[object type]", where type is the object type. * * @private * @function * @param {*} inputArg The object for which a class string represntation * is required. * @returns {string} A string value of the form "[object type]". * @see http://www.ecma-international.org/ecma-262/5.1/#sec-15.2.4.2 */ function $toStringTag(inputArg) { var val; if (inputArg === null) { val = '[object Null]'; } else if ($isUndefined(inputArg)) { val = '[object Undefined]'; } else { val = pToString.call(inputArg); } return val; } /** * The string tag representation of a RegExp object. * * @private * @type {string} */ stringTagRegExp = $toStringTag(getNativeFlags); /** * Returns true if the operand inputArg is a RegExp. * * @private * @function * @param {*} inputArg * @returns {boolean} */ function $isRegExp(inputArg) { return $toStringTag(inputArg) === stringTagRegExp && pHasOwn.call(inputArg, 'ignoreCase') && typeof inputArg.ignoreCase === 'boolean' && pHasOwn.call(inputArg, 'global') && typeof inputArg.global === 'boolean' && pHasOwn.call(inputArg, 'multiline') && typeof inputArg.multiline === 'boolean' && pHasOwn.call(inputArg, 'source') && typeof inputArg.source === 'string'; } /** * The abstract operation throws an error if its argument is a value that * cannot be converted to an Object, otherwise returns the argument. * * @private * @function * @param {*} inputArg The object to be tested. * @throws {TypeError} If inputArg is null or undefined. * @returns {*} The inputArg if coercible. * @see https://goo.gl/5GcmVq */ function $requireObjectCoercible(inputArg) { var errStr; if (inputArg === null || $isUndefined(inputArg)) { errStr = 'Cannot convert argument to object: ' + inputArg; throw new TypeError(errStr); } return inputArg; } /** * The abstract operation converts its argument to a value of type string * * @private * @function * @param {*} inputArg * @returns {string} * @see https://people.mozilla.org/~jorendorff/es6-draft.html#sec-tostring */ function $toString(inputArg) { var type, val; if (inputArg === null) { val = 'null'; } else { type = typeof inputArg; if (type === 'string') { val = inputArg; } else if (type === 'undefined') { val = type; } else { if (type === 'symbol') { throw new TypeError('Cannot convert symbol to string'); } val = String(inputArg); } } return val; } /** * Returns a string only if the arguments is coercible otherwise throws an * error. * * @private * @function * @param {*} inputArg * @throws {TypeError} If inputArg is null or undefined. * @returns {string} */ function $onlyCoercibleToString(inputArg) { return $toString($requireObjectCoercible(inputArg)); } /** * The function evaluates the passed value and converts it to an integer. * * @private * @function * @param {*} inputArg The object to be converted to an integer. * @returns {number} If the target value is NaN, null or undefined, 0 is * returned. If the target value is false, 0 is returned * and if true, 1 is returned. * @see http://www.ecma-international.org/ecma-262/5.1/#sec-9.4 */ function $toInteger(inputArg) { var number = +inputArg, val = 0; if ($strictEqual(number, number)) { if (!number || number === Infinity || number === -Infinity) { val = number; } else { val = (number > 0 || -1) * Math.floor(Math.abs(number)); } } return val; } /** * Copies a regex object. Allows adding and removing native flags while * copying the regex. * * @private * @function * @param {RegExp} regex Regex to copy. * @param {Object} [options] Allows specifying native flags to add or * remove while copying the regex. * @returns {RegExp} Copy of the provided regex, possibly with modified * flags. */ function $copyRegExp(regex, options) { var flags, opts, rx; if (options !== null && typeof options === 'object') { opts = options; } else { opts = {}; } // Get native flags in use flags = getNativeFlags.exec($toString(regex))[1]; flags = $onlyCoercibleToString(flags); if (opts.add) { flags += opts.add; flags = flags.replace(clipDups, ''); } if (opts.remove) { // Would need to escape `options.remove` if this was public rx = new RegExp('[' + opts.remove + ']+', 'g'); flags = flags.replace(rx, ''); } return new RegExp(regex.source, flags); } /** * The abstract operation ToLength converts its argument to an integer * suitable for use as the length of an array-like object. * * @private * @function * @param {*} inputArg The object to be converted to a length. * @returns {number} If len <= +0 then +0 else if len is +INFINITY then * 2^53-1 else min(len, 2^53-1). * @see https://people.mozilla.org/~jorendorff/es6-draft.html#sec-tolength */ function $toLength(inputArg) { return Math.min(Math.max($toInteger(inputArg), 0), MAX_SAFE_INTEGER); } /** * Copies a regex object so that it is suitable for use with searchOf and * searchLastOf methods. * * @private * @function * @param {RegExp} regex Regex to copy. * @returns {RegExp} */ function $toSearchRegExp(regex) { return $copyRegExp(regex, { add: 'g', remove: 'y' }); } /** * Returns true if the operand inputArg is a member of one of the types * Undefined, Null, Boolean, Number, Symbol, or String. * * @private * @function * @param {*} inputArg * @returns {boolean} * @see https://goo.gl/W68ywJ * @see https://goo.gl/ev7881 */ function $isPrimitive(inputArg) { var type = typeof inputArg; return type === 'undefined' || inputArg === null || type === 'boolean' || type === 'string' || type === 'number' || type === 'symbol'; } /** * The abstract operation converts its argument to a value of type Object * but fixes some environment bugs. * * @private * @function * @param {*} inputArg The argument to be converted to an object. * @throws {TypeError} If inputArg is not coercible to an object. * @returns {Object} Value of inputArg as type Object. * @see http://www.ecma-international.org/ecma-262/5.1/#sec-9.9 */ function $toObject(inputArg) { var object; if ($isPrimitive($requireObjectCoercible(inputArg))) { object = Object(inputArg); } else { object = inputArg; } return object; } /** * Converts a single argument that is an array-like object or list (eg. * arguments, NodeList, DOMTokenList (used by classList), NamedNodeMap * (used by attributes property)) into a new Array() and returns it. * This is a partial implementation of the ES6 Array.from * * @private * @function * @param {Object} arrayLike * @returns {Array} */ function $toArray(arrayLike) { var object = $toObject(arrayLike), length = $toLength(object.length), array = [], index = 0; array.length = length; while (index < length) { array[index] = object[index]; index += 1; } return array; } if (!String.prototype.searchOf) { /** * This method returns the index within the calling String object of * the first occurrence of the specified value, starting the search at * fromIndex. Returns -1 if the value is not found. * * @function * @this {string} * @param {RegExp|string} regex A regular expression object or a String. * Anything else is implicitly converted to * a String. * @param {Number} [fromIndex] The location within the calling string * to start the search from. It can be any * integer. The default value is 0. If * fromIndex < 0 the entire string is * searched (same as passing 0). If * fromIndex >= str.length, the method will * return -1 unless searchValue is an empty * string in which case str.length is * returned. * @returns {Number} If successful, returns the index of the first * match of the regular expression inside the * string. Otherwise, it returns -1. */ $defineProperty(String.prototype, 'searchOf', { enumerable: false, configurable: true, writable: true, value: function (regex) { var str = $onlyCoercibleToString(this), args = $toArray(arguments), result = -1, fromIndex, match, rx; if (!$isRegExp(regex)) { return String.prototype.indexOf.apply(str, args); } if ($toLength(args.length) > 1) { fromIndex = +args[1]; if (fromIndex < 0) { fromIndex = 0; } } else { fromIndex = 0; } if (fromIndex >= $toLength(str.length)) { return result; } rx = $toSearchRegExp(regex); rx.lastIndex = fromIndex; match = rx.exec(str); if (match) { result = +match.index; } return result; } }); } if (!String.prototype.searchLastOf) { /** * This method returns the index within the calling String object of * the last occurrence of the specified value, or -1 if not found. * The calling string is searched backward, starting at fromIndex. * * @function * @this {string} * @param {RegExp|string} regex A regular expression object or a String. * Anything else is implicitly converted to * a String. * @param {Number} [fromIndex] Optional. The location within the * calling string to start the search at, * indexed from left to right. It can be * any integer. The default value is * str.length. If it is negative, it is * treated as 0. If fromIndex > str.length, * fromIndex is treated as str.length. * @returns {Number} If successful, returns the index of the first * match of the regular expression inside the * string. Otherwise, it returns -1. */ $defineProperty(String.prototype, 'searchLastOf', { enumerable: false, configurable: true, writable: true, value: function (regex) { var str = $onlyCoercibleToString(this), args = $toArray(arguments), result = -1, fromIndex, length, match, pos, rx; if (!$isRegExp(regex)) { return String.prototype.lastIndexOf.apply(str, args); } length = $toLength(str.length); if (!$strictEqual(args[1], args[1])) { fromIndex = length; } else { if ($toLength(args.length) > 1) { fromIndex = $toInteger(args[1]); } else { fromIndex = length - 1; } } if (fromIndex >= 0) { fromIndex = Math.min(fromIndex, length - 1); } else { fromIndex = length - Math.abs(fromIndex); } pos = 0; rx = $toSearchRegExp(regex); while (pos <= fromIndex) { rx.lastIndex = pos; match = rx.exec(str); if (!match) { break; } pos = +match.index; if (pos <= fromIndex) { result = pos; } pos += 1; } return result; } }); } }()); (function () { 'use strict'; /* * testing as follow to make sure that at least for one character regexp, * the result is the same as if we used indexOf */ var pre = document.getElementById('out'); function log(result) { pre.appendChild(document.createTextNode(result + '\n')); } function test(str) { var i = str.length + 2, r, a, b; while (i) { a = str.indexOf('a', i); b = str.searchOf(/a/, i); r = ['Failed', 'searchOf', str, i, a, b]; if (a === b) { r[0] = 'Passed'; } log(r); a = str.lastIndexOf('a', i); b = str.searchLastOf(/a/, i); r = ['Failed', 'searchLastOf', str, i, a, b]; if (a === b) { r[0] = 'Passed'; } log(r); i -= 1; } } /* * Look for the a among the xes */ test('xxx'); test('axx'); test('xax'); test('xxa'); test('axa'); test('xaa'); test('aax'); test('aaa'); }()); <pre id="out"></pre>


如果您正在使用RegExp寻找一个非常简单的lastIndex查找,并且不关心它是否完全模仿了lastIndexOf,那么这可能会引起您的注意。

我只是将字符串反向,并从length - 1中减去第一个出现索引。它碰巧通过了我的测试,但我认为长字符串可能会出现性能问题。

interface String {
  reverse(): string;
  lastIndex(regex: RegExp): number;
}

String.prototype.reverse = function(this: string) {
  return this.split("")
    .reverse()
    .join("");
};

String.prototype.lastIndex = function(this: string, regex: RegExp) {
  const exec = regex.exec(this.reverse());
  return exec === null ? -1 : this.length - 1 - exec.index;
};

我使用string .prototype.match(regex),它返回一个字符串数组,所有找到的匹配给定的正则表达式在字符串(更多信息见这里):

function getLastIndex(text, regex, limit = text.length) { const matches = text.match(regex); // no matches found if (!matches) { return -1; } // matches found but first index greater than limit if (text.indexOf(matches[0] + matches[0].length) > limit) { return -1; } // reduce index until smaller than limit let i = matches.length - 1; let index = text.lastIndexOf(matches[i]); while (index > limit && i >= 0) { i--; index = text.lastIndexOf(matches[i]); } return index > limit ? -1 : index; } // expect -1 as first index === 14 console.log(getLastIndex('First Sentence. Last Sentence. Unfinished', /\. /g, 10)); // expect 29 console.log(getLastIndex('First Sentence. Last Sentence. Unfinished', /\. /g));


var mystring = "abc ab a";
var re  = new RegExp("ab"); // any regex here

if ( re.exec(mystring) != null ){ 
   alert("matches"); // true in this case
}

使用标准正则表达式:

var re  = new RegExp("^ab");  // At front
var re  = new RegExp("ab$");  // At end
var re  = new RegExp("ab(c|d)");  // abc or abd

对于一个比大多数其他答案更简洁的解决方案,你可能想要使用String.prototype.replace函数,它将在每个检测到的模式上运行一个函数。例如:

let firstIndex = -1;
"the 1st numb3r".replace(/\d/,(p,i) => { firstIndex = i; });
// firstIndex === 4

这对于“last index”的情况特别有用:

let lastIndex = -1;
"the l4st numb3r".replace(/\d/g,(p,i) => { lastIndex = i; });
// lastIndex === 13

在这里,重要的是要包括“g”修饰符,以便对所有发生的情况进行计算。如果没有找到正则表达式,这些版本也会导致-1。

最后,下面是包含起始索引的更通用的函数:

function indexOfRegex(str,regex,start = 0) {
    regex = regex.global ? regex : new RegExp(regex.source,regex.flags + "g");
    let index = -1;
    str.replace(regex,function() {
        const pos = arguments[arguments.length - 2];
        if(index < 0 && pos >= start)
            index = pos;
    });
    return index;
}

function lastIndexOfRegex(str,regex,start = str.length - 1) {
    regex = regex.global ? regex : new RegExp(regex.source,regex.flags + "g");
    let index = -1;
    str.replace(regex,function() {
        const pos = arguments[arguments.length - 2];
        if(pos <= start)
            index = pos;
    });
    return index;
}

这些函数特别避免在开始索引处分割字符串,我认为这在Unicode时代是有风险的。它们不会修改常见Javascript类的原型(尽管您可以自己这么做)。它们接受更多的RegExp标志,例如“u”或“s”以及将来可能添加的任何标志。我发现回调函数比for/while循环更容易解释。


让正则表达式;//你的RegExp 加勒比海盗。map(x => !!x. tostring ().match(regExp)).indexOf(true)


来自Jason Bunting的regexIndexOf可以更简单地反向,并且仍然支持UTF8字符:

function regexLastIndexOf(string, regex, startpos=0) {
    return text.length - regexIndexOf([...text].reverse().join(""), regex, startpos) - 1;
}

你可以使用String.prototype.matchAll()和方便的Array.prototype.at():

const str = "foo a foo B"; const matches =[…str.matchAll(/[abc]/gi)]; If (matches.length) { const indexOfFirstMatch = matches.at(0).index; const indexOfLastMatch = matches.at(-1).index; console.log (indexOfFirstMatch indexOfLastMatch) }