helloxielan 2014-11-10 12:21 采纳率: 0%
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更新ajax send()内容

I'm trying to receive data from ajax call and then send this data back using ajax.
javascript code:
setInterval (function () { var xmlhttp = new XMLHttpRequest(); var content = "data=1"; xmlhttp.onreadystatechange = function() { if (xmlhttp.readyState == 4 && xmlhttp.status == 200) { content = "data=" + xmlhttp.responseText; alert (xmlhttp.responseText); } } xmlhttp.open("POST" , "execute-page.php" , true); xmlhttp.setRequestHeader("Content-type","application/x-www-form-urlencoded"); xmlhttp.send(content); },5000);
the problem now is that it keeps sending the old content. how can I update content variable with ajax response text?

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  • weixin_33713350 2014-11-10 12:25
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    Move the line var content = "data=1"; out of the function:

    var content = "data=1";
    setInterval (function ()
        {
          var xmlhttp = new XMLHttpRequest();
          xmlhttp.onreadystatechange = function()
            {
                if (xmlhttp.readyState == 4 && xmlhttp.status == 200)
                  {
                    content = "data=" + xmlhttp.responseText;
                    alert (xmlhttp.responseText);
                  }
            }
          xmlhttp.open("POST" , "execute-page.php" , true);
          xmlhttp.setRequestHeader("Content-type","application/x-www-form-urlencoded");
          xmlhttp.send(content);
        },5000);
    
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