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我想添加多个网站,我试图重写代码并更改链接“req.open”并注意完成。任何帮助?我对JavaScript很重要。 感谢如何多个XMLHttpRequest
<html>
<head>
<title>Testing Purposes</title>
<script lang="javascript">
function ready() {
var hell = document.getElementById('hell');
var req = new XMLHttpRequest();
req.open('GET', 'load.php?http://www.140online.com/product/25965/%20%D8%AA%D9%82%D8%B7%D9%8A%D8%B9%20%D9%84%D9%8A%D8%B2%D8%B1', false);
req.send(null);
if(req.status == 200) {
var html_str = req.responseText;
var doc = document.createElement('html');
doc.innerHTML = html_str;
var divs = doc.getElementsByClassName('row-fluid');
for (i = 0; i < divs.length; i++) {
hell.innerHTML = hell.innerHTML +
"<h2>Found:</h2>" +
divs[i].innerText +
"<br />";
}
} else {
alert("Failed to load the page!");
}
}
</script>
</head>
<body onload="ready();">
<div id='hell'>
</div>
</body>
使用''XMLHttpRequest'处理来自服务器 – guest271314
我可以回应的load'事件在这行看到你错了:'req.open('GET','load .PHP HTTP://www.140online.com/product/25965/%20%D8%AA%D9%82%D8%B7%D9%8A%D8%B9%20%D9%84%D9%8A% D8%B2%D8%B1',false);'应该是'req.open('GET','load.php?page = http://www.140online.com/product/25965/%20%D8 %AA%D9%82%D8%B7%D9%8A%D8%B9%20%D9%84%D9%8A%D8%B2%D8%B1',false);' –
参见http:// stackoverflow。 COM /问题/ 14220321 /如何-DO-I-返回的响应,从-的异步调用/ – guest271314