在Javascript中替换正则表达式匹配的第n个实例

Question

我正在尝试编写一个正则表达式函数,该函数将识别并替换字符串中匹配的单个实例,而不会影响其他实例.例如,我有这个字符串:

12||34||56

我想用&符号替换第二组管道来获取此字符串:

12||34&&56

正则表达式的功能需要能够处理管道的X量,让我来代替第n个集管,所以我可以使用相同的功能,使这些替代品:

23||45||45||56||67 -> 23&&45||45||56||67

23||34||98||87 -> 23||34||98&&87

我知道,我可以只拆分/替换/ CONCAT字符串在管,我也知道,我可以匹配/\|\|/,并通过生成的数组迭代,但我想知道如果有可能写一个表达式,可以做这个.请注意,这将是JavaScript,因此有可能产生在运行时使用正则表达式eval(),但它不可能使用任何Perl的正则表达式的具体说明.

Answer 1

更通用的功能

我遇到了这个问题,虽然标题很一般,但是接受的答案只处理问题的具体用例.

我需要一个更通用的解决方案,所以我写了一个并认为我会在这里分享它.

用法

此函数要求您传递以下参数:

• original:您正在搜索的字符串
• pattern:要搜索的字符串,要么是带有捕获组的RegExp .没有捕获组,它将引发错误.这是因为函数调用split原始字符串,并且只有当提供的RegExp包含捕获组时,结果数组才会包含匹配项.
• n:找到的序数发生; 例如,如果你想要第二场比赛,请传球2
• replace:用于替换匹配的字符串,或者用于匹配并返回替换字符串的函数.

例子

// Pipe examples like the OP's
replaceNthMatch("12||34||56", /(\|\|)/, 2, '&&') // "12||34&&56"
replaceNthMatch("23||45||45||56||67", /(\|\|)/, 1, '&&') // "23&&45||45||56||67"

// Replace groups of digits
replaceNthMatch("foo-1-bar-23-stuff-45", /(\d+)/, 3, 'NEW') // "foo-1-bar-23-stuff-NEW"

// Search value can be a string
replaceNthMatch("foo-stuff-foo-stuff-foo", "foo", 2, 'bar') // "foo-stuff-bar-stuff-foo"

// No change if there is no match for the search
replaceNthMatch("hello-world", "goodbye", 2, "adios") // "hello-world"

// No change if there is no Nth match for the search
replaceNthMatch("foo-1-bar-23-stuff-45", /(\d+)/, 6, 'NEW') // "foo-1-bar-23-stuff-45"

// Passing in a function to make the replacement
replaceNthMatch("foo-1-bar-23-stuff-45", /(\d+)/, 2, function(val){
  //increment the given value
  return parseInt(val, 10) + 1;
}); // "foo-1-bar-24-stuff-45"

代码

  var replaceNthMatch = function (original, pattern, n, replace) {
    var parts, tempParts;

    if (pattern.constructor === RegExp) {

      // If there's no match, bail
      if (original.search(pattern) === -1) {
        return original;
      }

      // Every other item should be a matched capture group;
      // between will be non-matching portions of the substring
      parts = original.split(pattern);

      // If there was a capture group, index 1 will be
      // an item that matches the RegExp
      if (parts[1].search(pattern) !== 0) {
        throw {name: "ArgumentError", message: "RegExp must have a capture group"};
      }
    } else if (pattern.constructor === String) {
      parts = original.split(pattern);
      // Need every other item to be the matched string
      tempParts = [];

      for (var i=0; i < parts.length; i++) {
        tempParts.push(parts[i]);

        // Insert between, but don't tack one onto the end
        if (i < parts.length - 1) {
          tempParts.push(pattern);
        }
      }
      parts = tempParts;
    }  else {
      throw {name: "ArgumentError", message: "Must provide either a RegExp or String"};
    }

    // Parens are unnecessary, but explicit. :)
    indexOfNthMatch = (n * 2) - 1;

  if (parts[indexOfNthMatch] === undefined) {
    // There IS no Nth match
    return original;
  }

  if (typeof(replace) === "function") {
    // Call it. After this, we don't need it anymore.
    replace = replace(parts[indexOfNthMatch]);
  }

  // Update our parts array with the new value
  parts[indexOfNthMatch] = replace;

  // Put it back together and return
  return parts.join('');

  }

定义它的替代方法

这个函数最不吸引人的部分是它需要4个参数.通过将它作为方法添加到String原型,可以简化为只需要3个参数,如下所示:

String.prototype.replaceNthMatch = function(pattern, n, replace) {
  // Same code as above, replacing "original" with "this"
};

如果你这样做,你可以在任何字符串上调用方法,如下所示:

"foo-bar-foo".replaceNthMatch("foo", 2, "baz"); // "foo-bar-baz"

通过测试

以下是此功能通过的Jasmine测试.

describe("replaceNthMatch", function() {

  describe("when there is no match", function() {

    it("should return the unmodified original string", function() {
      var str = replaceNthMatch("hello-there", /(\d+)/, 3, 'NEW');
      expect(str).toEqual("hello-there");
    });

  });

  describe("when there is no Nth match", function() {

    it("should return the unmodified original string", function() {
      var str = replaceNthMatch("blah45stuff68hey", /(\d+)/, 3, 'NEW');
      expect(str).toEqual("blah45stuff68hey");
    });

  });

  describe("when the search argument is a RegExp", function() {

    describe("when it has a capture group", function () {

      it("should replace correctly when the match is in the middle", function(){
        var str = replaceNthMatch("this_937_thing_38_has_21_numbers", /(\d+)/, 2, 'NEW');
        expect(str).toEqual("this_937_thing_NEW_has_21_numbers");
      });

      it("should replace correctly when the match is at the beginning", function(){
        var str = replaceNthMatch("123_this_937_thing_38_has_21_numbers", /(\d+)/, 2, 'NEW');
        expect(str).toEqual("123_this_NEW_thing_38_has_21_numbers");
      });

    });

    describe("when it has no capture group", function() {

      it("should throw an error", function(){
        expect(function(){
          replaceNthMatch("one_1_two_2", /\d+/, 2, 'NEW');
        }).toThrow('RegExp must have a capture group');
      });

    });


  });

  describe("when the search argument is a string", function() {

    it("should should match and replace correctly", function(){
      var str = replaceNthMatch("blah45stuff68hey", 'stuff', 1, 'NEW');
      expect(str).toEqual("blah45NEW68hey");
    });

  });

  describe("when the replacement argument is a function", function() {

    it("should call it on the Nth match and replace with the return value", function(){

      // Look for the second number surrounded by brackets
      var str = replaceNthMatch("foo[1][2]", /(\[\d+\])/, 2, function(val) {

        // Get the number without the [ and ]
        var number = val.slice(1,-1);

        // Add 1
        number = parseInt(number,10) + 1;

        // Re-format and return
        return '[' + number + ']';
      });
      expect(str).toEqual("foo[1][3]");

    });

  });

});

可能无法在IE7中运行

该代码可以在IE7中失败,因为该浏览器使用正则表达式正确分割字符串,如讨论在这里.[在IE7上握拳].我相信这是解决方案; 如果你需要支持IE7,祝你好运.:)

Answer 2

这是有用的东西:

"23||45||45||56||67".replace(/^((?:[0-9]+\|\|){n})([0-9]+)\|\|/,"$1$2&&")

其中n是小于第n个管道的那个(当然,如果n = 0,则不需要第一个子表达式)

如果你想要一个函数来做到这一点:

function pipe_replace(str,n) {
   var RE = new RegExp("^((?:[0-9]+\\|\\|){" + (n-1) + "})([0-9]+)\|\|");
   return str.replace(RE,"$1$2&&");
}