我正在使用以下脚本来启动文件下载：

if (file_exists($newfilename)) {
    header('Content-Description: File Transfer');
    header('Content-Type: application/octet-stream');
    header('Content-Disposition: attachment; filename='.basename($newfilename));
    header('Content-Transfer-Encoding: binary');
    header('Expires: 0');
    header('Cache-Control: must-revalidate');
    header('Pragma: public');
    header('Content-Length: ' . filesize($newfilename));
    ob_clean();
    flush();
    readfile($newfilename);
    exit;
}

当我直接打开页面时，它工作正常，但事实是，我需要从另一个页面通过Ajax调用此脚本。当我这样做时，下载就不会开始。脚本的其余部分执行了预期的工作。

我以为问题不在于无法以这种方式使用标头函数，但是肯定有一种方法可以使这项工作正常进行吗？

如果有任何帮助，这是Ajax函数：

<script type="text/javascript">
    // function create GetXmlHttpObject
    function GetXmlHttpObject()
    {
    if (window.XMLHttpRequest)
    {
    // code for IE7+, Firefox, Chrome, Opera, Safari
    return new XMLHttpRequest();
    }
    if (window.ActiveXObject)
    {
    // code for IE6, IE5
    return new ActiveXObject("Microsoft.XMLHTTP");
    }
    return null;
    }

    function submitVideoAjax(){
    var myAjaxPostrequest=new GetXmlHttpObject();

    var t2_title=document.video_form.title.value;

    var parameters="title="+t2_title;

    myAjaxPostrequest.open("POST", "newdownloadmanager.php", true);
    myAjaxPostrequest.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
    myAjaxPostrequest.send(parameters);
    myAjaxPostrequest.onreadystatechange=function(){
    if(myAjaxPostrequest.readyState==4){
    if(myAjaxPostrequest.status==200){
    document.getElementById("result").innerHTML=myAjaxPostrequest.responseText;
    document.getElementById("video_form").style.display = "none";

    }
    else    {
    document.getElementById("video_form").innerHTML="An error has occured making the request";
    }
    }
    }
    }
    </script>

形式如下：

<form name='video_form' id='video_form' method="post">
<input type="hidden" name="title" id="title" value="Madelyn2-01.mp4"/>
<button type="button" name="submit_video" id="submit_video" onclick="submitVideoAjax();">Download</button>
</form>