替换 Javascript 中正则表达式匹配的第 n 个实例

Question

I'm trying to write a regex function that will identify and replace a single instance of a match within a string without affecting the other instances. For example, I have this string:

12||34||56

I want to replace the second set of pipes with ampersands to get this string:

12||34&&56

The regex function needs to be able to handle x amount of pipes and allow me to replace the nth set of pipes, so I could use the same function to make these replacements:

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

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

I know that I could just split/replace/concat the string at the pipes, and I also know that I can match on /\|\|/ and iterate through the resulting array, but I'm interested to know if it's possible to write a single expression that can do this. Note that this would be for Javascript, so it's possible to generate a regex at runtime using eval() , but it's not possible to use any Perl-specific regex instructions.

Answer 1

A more general-purpose function

I came across this question and, although the title is very general, the accepted answer handles only the question's specific use case.

I needed a more general-purpose solution, so I wrote one and thought I'd share it here.

Usage

This function requires that you pass it the following arguments:

• original : the string you're searching in
• pattern : either a string to search for, or a RegExp with a capture group . Without a capture group, it will throw an error. This is because the function calls split on the original string, and only if the supplied RegExp contains a capture group will the resulting array contain the matches .
• n : the ordinal occurrence to find; eg, if you want the 2nd match, pass in 2
• replace : Either a string to replace the match with, or a function which will take in the match and return a replacement string.

Examples

// 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"

The Code

  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('');

  }

An Alternate Way To Define It

The least appealing part of this function is that it takes 4 arguments. It could be simplified to need only 3 arguments by adding it as a method to the String prototype, like this:

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

If you do that, you can call the method on any string, like this:

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

Passing Tests

The following are the Jasmine tests that this function passes.

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]");

    });

  });

});

May not work in IE7

This code may fail in IE7 because that browser incorrectly splits strings using a regex, as discussed here . [shakes fist at IE7]. I believe that this is the solution; if you need to support IE7, good luck. :)

Answer 2

here's something that works:

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

where n is the one less than the nth pipe, (of course you don't need that first subexpression if n = 0)

And if you'd like a function to do this:

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

Answer 3

function pipe_replace(str,n) {
    m = 0;
    return str.replace(/\|\|/g, function (x) {
        //was n++ should have been m++
        m++;
        if (n==m) {
            return "&&";
        } else {
            return x;
        }
    });
}

Answer 4

Thanks Binda, I have modified the code for generic uses:

private replaceNthMatch(original, pattern, n, replace) {
    let m = -1;
    return original.replaceAll(pattern, x => {
        m++;
        if ( n == m ) {
            return replace;
        } else {
            return x;
        }
    });
}