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var request = new XMLHttpRequest();
request.open("GET","get.php",ture);
request.send();
request.onreadystatechange = function(){
if(request.readyState ===4&& request.status===200)
//执行一些事情request.responseText
}
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原文地址：http://www.cnblogs.com/family-626-77/p/5560949.html
