ajax获取服务器响应信息

window.onload = function(){
  document.getElementById('btn').onclick = function(){
    var req = new XMLHttpRequest();
    req.open('get', 'demo.php');
    req.onreadystatechange = function(){
      if(req.readyState == 4 && req.statu == 200){
        var str = req.responseText;
        alert(str)
      }
    }
      req.send();
   }
}

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转载自www.cnblogs.com/wangshengl9263/p/9029578.html