将 HTML 内容作为字符串返回给定 URL。Javascript 函数

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时间:2020-08-24 02:17:59  来源:igfitidea点击:

Return HTML content as a string, given URL. Javascript Function

javascripthtml

提问by Jason Kim

I want to write a javascript function that returns HTML content as string given URL to the function. I found a similar answer on Stackoverflow.

我想编写一个 javascript 函数,该函数将 HTML 内容作为给定 URL 的字符串返回给函数。我在 Stackoverflow 上找到了类似的答案。

I am trying to use this answerto solve my problem.

我正在尝试使用此答案来解决我的问题。

However, it seems as though document.write()isn't writing anything. When I load the page, I get a a blank screen.

然而,似乎document.write()没有写任何东西。当我加载页面时,我得到一个空白屏幕。

<html>
<head>
</head>
<body>  
  <script type="text/JavaScript">
  function httpGet(theUrl)
  {
    var xmlHttp = null;

    xmlHttp = new XMLHttpRequest();
    xmlHttp.open( "GET", theUrl, false );
    xmlHttp.send( null );
    return xmlHttp.responseText;
  }
  document.write(httpGet("https://stackoverflow.com/"));
  </script>
</body>
</html>

采纳答案by Satya

you need to return when the readystate==4 e.g.

你需要在 readystate==4 时返回,例如

function httpGet(theUrl)
{
    if (window.XMLHttpRequest)
    {// code for IE7+, Firefox, Chrome, Opera, Safari
        xmlhttp=new XMLHttpRequest();
    }
    else
    {// code for IE6, IE5
        xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
    }
    xmlhttp.onreadystatechange=function()
    {
        if (xmlhttp.readyState==4 && xmlhttp.status==200)
        {
            return xmlhttp.responseText;
        }
    }
    xmlhttp.open("GET", theUrl, false );
    xmlhttp.send();    
}

回答by otaxige_aol

The only one i have found for Cross-site, is this function:

我为跨站点找到的唯一一个是这个功能:

<script type="text/javascript">
var your_url = 'http://www.example.com';

</script>

<script type="text/javascript" src="jquery.min.js" ></script>
<script type="text/javascript">
// jquery.xdomainajax.js  ------ from padolsey

jQuery.ajax = (function(_ajax){

    var protocol = location.protocol,
        hostname = location.hostname,
        exRegex = RegExp(protocol + '//' + hostname),
        YQL = 'http' + (/^https/.test(protocol)?'s':'') + '://query.yahooapis.com/v1/public/yql?callback=?',
        query = 'select * from html where url="{URL}" and xpath="*"';

    function isExternal(url) {
        return !exRegex.test(url) && /:\/\//.test(url);
    }

    return function(o) {

        var url = o.url;

        if ( /get/i.test(o.type) && !/json/i.test(o.dataType) && isExternal(url) ) {

            // Manipulate options so that JSONP-x request is made to YQL

            o.url = YQL;
            o.dataType = 'json';

            o.data = {
                q: query.replace(
                    '{URL}',
                    url + (o.data ?
                        (/\?/.test(url) ? '&' : '?') + jQuery.param(o.data)
                    : '')
                ),
                format: 'xml'
            };

            // Since it's a JSONP request
            // complete === success
            if (!o.success && o.complete) {
                o.success = o.complete;
                delete o.complete;
            }

            o.success = (function(_success){
                return function(data) {

                    if (_success) {
                        // Fake XHR callback.
                        _success.call(this, {
                            responseText: data.results[0]
                                // YQL screws with <script>s
                                // Get rid of them
                                .replace(/<script[^>]+?\/>|<script(.|\s)*?\/script>/gi, '')
                        }, 'success');
                    }

                };
            })(o.success);

        }

        return _ajax.apply(this, arguments);

    };

})(jQuery.ajax);



$.ajax({
    url: your_url,
    type: 'GET',
    success: function(res) {
        var text = res.responseText;
        // then you can manipulate your text as you wish
        alert(text);
    }
});

</script>

回答by Pranay Rana

after you get the response just do call this function to append data to your body element

收到响应后,只需调用此函数将数据附加到您的 body 元素

function createDiv(responsetext)
{
    var _body = document.getElementsByTagName('body')[0];
    var _div = document.createElement('div');
    _div.innerHTML = responsetext;
    _body.appendChild(_div);
}

@satya code modified as below

@satya 代码修改如下

function httpGet(theUrl)
{
    if (window.XMLHttpRequest)
    {// code for IE7+, Firefox, Chrome, Opera, Safari
        xmlhttp=new XMLHttpRequest();
    }
    else
    {// code for IE6, IE5
        xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
    }
    xmlhttp.onreadystatechange=function()
    {
        if (xmlhttp.readyState==4 && xmlhttp.status==200)
        {
            createDiv(xmlhttp.responseText);
        }
    }
    xmlhttp.open("GET", theUrl, false);
    xmlhttp.send();    
}