如何将输入文件名作为变量传递给xmlhttp.open

Question

我想传递所选文件的名称。它可以工作，如果我把“文件名”而不是网址。这是我的测试代码。请帮我搞定。

<input id="upload" type="file" />

<script>
document.getElementById('upload').onchange = uploadOnChange;

function uploadOnChange() {
var filename = this.value;
var lastIndex = filename.lastIndexOf("\\");
if (lastIndex >= 0) {
    filename = filename.substring(lastIndex + 1);
}

var url='"'+filename+'"'; 

if (window.XMLHttpRequest)
{// code for IE7+, Firefox, Chrome, Opera, Safari
xmlhttp=new XMLHttpRequest();
}
else
{// code for IE6, IE5
xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
}

xmlhttp.open("GET",url,false);
xmlhttp.send();
xmlDoc=xmlhttp.responseXML; 

var tag=xmlDoc.getElementsByTagName("tag"); 

}
</script>