如何在javascript中的替换内循环正则表达式的匹配？ [英] How do I loop through a regex's matches inside a replace in javascript?

问题描述

我有以下JavaScript（< P> 中的空格是非破坏的）：

I have the following JavaScript (the spaces in the <P>s are non-breaking):

var html = '...<li>sub 2</li></ol></li></ol>\n\
<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; i.&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; subsub 1</p>\n\
<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; ii.&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; subsub 2</p>\n\
<ol start="2"> \n\
<ol start="3"> \n\
<li>sub 3</li></ol>...';

$(function () {
    var nestedListFixer = /(?:<\/li><\/ol>\s*)+(?:<p>(?:&(?:nbsp|\#0*160|x0*A0);)+(?:\s[ivxlcdm]+\.)(?:&(?:nbsp|\#0*160|x0*A0);)+\s(.*?)<\/p>\s*)+(?:<ol(?:\sstyle=\"[^\"]+\")?\sstart=\"[^\"]+\">\s*)+/i;
    html = html.replace(nestedListFixer, function($0, $1){
        var lis = ""
        $.each($1, function () {
            lis += "<li>" + this + "</li>\n";
        });
        alert("<ol>\n" + lis + "</ol></li>");
        return "<ol>\n" + lis + "</ol></li>";
    });
});

alert（）的输出：

<ol>
<li>s</li>
<li>u</li>
<li>b</li>
<li>s</li>
<li>u</li>
<li>b</li>
<li> </li>
<li>2</li>
</ol></li>

以下是我希望的结果：

<ol>
<li>subsub 1</li>
<li>subsub 2</li>
</ol></li>

如何在 $ 1 ？

更新：简化模式和匹配示例：

Update: simplified pattern and matching example:

var text = '1ab2cb3cd4ab5cb6cd7';

$(function () {
    var textFixer = /(?:a)(?:b(.*?)c)+(?:d)/i;
    text = text.replace(textFixer, function($0, $1){
        var numbers = "";
        $.each($1, function () {
            numbers += this;
        });
        alert(numbers);
        return numbers;
    });
    alert(text);
});

// actual text:
// 13467
// desired text:
// 1234567

推荐答案

以下是我提出的解决方案，但效率似乎不高：

Here's the solution that I came up with but it doesn't seem to be very efficient:

$(function () {
    var nestedListFixer = /(?:<\/li><\/ol>\s*)+(?:<p>(?:&(?:nbsp|\#0*160|x0*A0);)+(?:\s[ivxlcdm]+\.)(?:&(?:nbsp|\#0*160|x0*A0);)+\s(.*?)<\/p>\s*)+(?:<ol(?:\sstyle=\"[^\"]+\")?\sstart=\"[^\"]+\">\s*)+/gi;
    var nestedListItem = /<p>(?:&(?:nbsp|\#0*160|x0*A0);)+(?:\s[ivxlcdm]+\.)(?:&(?:nbsp|\#0*160|x0*A0);)+\s(.*?)<\/p>\s*/gi;

    // fix nested lists
    html = html.replace(nestedListFixer, function($0, $1){
        var lis = ""
        $.each($0.match(nestedListItem), function () {
            lis += "<li>" + this.replace(nestedListItem, "$1") + "</li>\n";
        });
        return "<ol>\n" + lis + "</ol></li>";
    });
});

...或上面的简单示例：

...or for the simpler example above:

$(function () {
    var textFixer = /(?:a)(?:b(.*?)c)+(?:d)/gi;
    var textItem = /b(.*?)c/gi;

    text = text.replace(textFixer, function($0, $1){
        var numbers = "";
        $.each($0.match(textItem), function () {
            numbers += this.replace(textItem, "$1");
        });
        return numbers;
    });
});

执行 .replace（）替换，在一个 .match（）数组的循环内，在自定义 .replace（）函数内部不会看起来很经济。它确实给了我正在寻找的输出。

Doing a .replace() substitution, inside a loop of a .match() array, inside of a custom .replace() function just doesn't seem like it is very economical. It does give me the output that I was looking for though.

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