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javascript Ajax获取远程url的返回判断

答案:
复制代码 代码如下:

  <SCRIPT LANGUAGE="JavaScript">
  <!--
  function ajaxByJyking(){
  var xmlhttp_request = "";
  try{
  if( window.ActiveXObject ){
  for( var i = 5; i; i-- ){
  try{
  if( i == 2 ){
  xmlhttp_request = new ActiveXObject( "Microsoft.XMLHTTP" ); }
  else{
  xmlhttp_request = new ActiveXObject( "Msxml2.XMLHTTP." + i + ".0" );
  xmlhttp_request.setRequestHeader("Content-Type","text/xml");
  xmlhttp_request.setRequestHeader("Charset","gb2312"); }
  break;}
  catch(e){
  xmlhttp_request = false; } } }
  else if( window.XMLHttpRequest )
  { xmlhttp_request = new XMLHttpRequest();
  if (xmlhttp_request.overrideMimeType)
  { xmlhttp_request.override.MimeType('text/xml'); } } }
  catch(e){ xmlhttp_request = false; }
  xmlhttp_request.open('GET', 'http://www.zhaoxi.net', true);
  xmlhttp_request.send(null);
  xmlhttp_request.onreadystatechange = function(){
  if (xmlhttp_request.readyState == 4) {
  // 收到完整的服务器响应
  document.write("yes")
  } else{
  alert(1)
  }
  }
  }
  ajaxByJyking();
  //-->
  </SCRIPT>

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