从通过 XMLHTTPRequest 加载的网页中提取 URL 的最佳方法? [英] Optimal way to extract a URL from web page loaded via XMLHTTPRequest?

问题描述

问题概览

• 我有一个动态生成的网页，X，它由链接到网页的搜索结果组成，Y1、Y2、Y3 等
• Y1 包含资源 URL R1，Y2 包含资源 URL R2，依此类推.
• 我想动态地增强页面 X 链接到资源 R1、R2 等

• I have a dynamically produced web page, X, which consists of search results linking to web pages, Y1, Y2, Y3 etc.
• Y1 contains a resource URL R1, Y2 contains resource URL R2, and so on.
• I would like to dynamically enhance page X with links to resources R1, R2 etc.

可能的解决方案

我目前正在考虑使用 JavaScript 和 XMLHTTPRequest 从网页 Y1、Y2 等中检索 HTML，并然后使用正则表达式提取网址.

I'm currently thinking of using JavaScript and XMLHTTPRequest to retrieve the HTML from web pages Y1, Y2, etc., and to then use a regular expression to extract the URL.

页面Y1、Y2等在每个30-100KB HTML区域内.

Pages Y1, Y2, etc. are in the region of 30-100KB HTML each.

这听起来是个好计划吗?或者我最好以 JSON 格式检索每个网页并从中提取资源 URL?如果 HTML 是可行的方法，您是否有任何建议的优化/捷径来搜索 30-100 KB 的文本?

Does this sound like a good plan? Or would I be better retrieving each web page in JSON format and extracting the resource URL from there? If HTML is the way to go, do you have any suggested optimisations/short cuts for searching 30-100 KB of text?

推荐答案

您不想使用正则表达式来提取 URL.我建议使用jQuery来执行AJAX请求，然后使用jQuery从服务器返回的HTML中解析过滤出URL.

You don't want to use regex to extract the URL. I suggest using jQuery to perform the AJAX request, and then use jQuery to parse and filter out the URLs from the HTML that is returned from the server.

jQuery.ajax({
    url: "http://my.url.here",
    dataType: "html";
    ...
    success: function(data) {
        jQuery("a", data).each(function() {
            var $link = jQuery(this);
            ...
            ...
        });
    }
    ...
});

如果 jQuery 不是一个选项，您可以在收到回复时执行以下操作:

If jQuery is not an option, you can do something like this when you get your response back:

var html = XHR.responseText;
var div = document.createElement("div");
div.innerHTML = html;

//you can now search for nodes inside your div.
//The following gives you all the anchor tags
div.getElementsByTagName('a'); 
...

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