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＆符替换第二组管道来获取此字符串：
12||34&&56
正则表达式函数需要能够处理x个管道并允许我替换第n组管道，所以我可以使用相同的函数来进行这些替换：
23||45||45||56||67 -> 23&&45||45||56||67
23||34||98||87 -> 23||34||98&&87
我知道我可以在管道上拆分/替换/连接字符串，我也知道我可以匹配/\|\|/并迭代生成的数组，但我很想知道它是否是可以写一个可以做到这一点的表达式。请注意，这适用于Javascript，因此可以使用eval()在运行时生成正则表达式，但是不可能使用任何特定于Perl的正则表达式指令。
答案 0 :(得分：31)
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。
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。这是因为函数调用原始字符串split和only if the supplied RegExp contains a capture group will the resulting array contain the matches。n：要查找的序数事件;例如，如果您想要第二场比赛，请传入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。// 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"
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('');
}
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原型中，可以简化为仅需要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该方法，如下所示：
"foo-bar-foo".replaceNthMatch("foo", 2, "baz"); // "foo-bar-baz"
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]");
});
});
});
此代码可能在IE7中失败，因为浏览器使用正则表达式错误地拆分字符串，如here所述。 [在IE7上握拳]。我相信this是解决方案;如果你需要支持IE7，祝你好运。 ：）
答案 1 :(得分：20)
这里有用的东西：
"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）
如果您想要一个功能来执行此操作：
function pipe_replace(str,n) {
var RE = new RegExp("^((?:[0-9]+\\|\\|){" + (n-1) + "})([0-9]+)\|\|");
return str.replace(RE,"$1$2&&");
}
答案 2 :(得分：0)
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;
}
});
}