I'm developed an XMLHTTPRequest Formdata Uploader using upload api. My Javascript runs without error and upload status ends correct with upload complete. I'm using windows Server 2008 ans ASP Classic. But i think, ASP Classic isn't the problem. I thinking, i have a data transfer problem. Below is my code.
我使用upload api开发了一个XMLHTTPRequest Formdata Uploader。我的Javascript运行没有错误,上传完成后上传状态结束。我正在使用Windows Server 2008和ASP Classic。但我认为,ASP Classic不是问题所在。我想,我有数据传输问题。以下是我的代码。
function sendForm(form, btn) {
for (x=0; x < document.getElementById("auswahl").files.length; x++){
uploadFiles(document.getElementById("auswahl"));
}
}
var url= "http://www.centil-europe.ch/db/Images/test/";
//$(document).ready(function(){
// document.getElementById('upload').addEventListener('change', function(e) {
function uploadFiles(filecntrl){
var file = filecntrl.files[0];
var xhr = new XMLHttpRequest();
xhr.file = file; // not necessary if you create scopes like this
xhr.addEventListener('progress', function(e) {
var done = e.position || e.loaded, total = e.totalSize || e.total;
console.log('xhr progress: ' + (Math.floor(done/total*1000)/10) + '%');
}, false);
if ( xhr.upload ) {
xhr.upload.onprogress = function(e) {
var done = e.position || e.loaded, total = e.totalSize || e.total;
console.log('xhr.upload progress: ' + done + ' / ' + total + ' = ' + (Math.floor(done/total*1000)/10) + '%');
};
}
xhr.onreadystatechange = function(e) {
if ( 4 == this.readyState ) {
console.log(['xhr upload complete', e]);
}
};
xhr.open('post', url, true);
xhr.setRequestHeader("Content-Type","multipart/form-data");
var formData = new FormData();
formData.append("thefile", file);
xhr.send(formData);
}
// }, false);
//});
And the HTML-Part
和HTML部分
<BODY bgcolor="#FFFFFF" link="#999900" vlink="#CCCC33" alink="#999966">
<table cellSpacing="0" cellPadding="4" width="767" align="center" height="100" border="0" bgcolor="#FFFFFF">
<tbody>
<tr>
<td vAlign="middle" width="100%" height="100" bgcolor="#F5F5F5">
<FORM METHOD="POST" ENCTYPE="multipart/form-data" id="iduploadfrm">
<Input Type="hidden" Name="ID" value="<%=ID%>">
<TABLE BORDER=0 class="Tabelle">
<tr>
<td> </td>
</tr>
<tr>
<td><b>File zum uploaden auswählen:</b><br>
<input type=FILE size=50 name="FILE1" class="Textfield" id="auswahl" multiple>
</td>
</tr>
<tr><td><!--//Database <INPUT TYPE=RADIO NAME="saveto" value="database">//-->
</td>
</tr>
<tr align="right">
<td>
<INPUT TYPE=SUBMIT VALUE="Upload!" class="Button">
</td>
</tr>
</TABLE>
<!-- The table listing the files available for upload/download -->
<table role="presentation" class="table table-striped" id="tblpresentation">
<tbody class="files">
<tr class="template-upload fade in" id="filetbl" style="visibility:hidden">
<td width="17%">
<span id="idimage" class="preview"></span>
</td>
<td width="16%">
<p class="name"> </p>
<strong class="error text-danger"></strong>
</td>
<td width="26%">
<p class="size" id="size"> </p></td>
<td width="41%">
<button class="btn btn-primary start" id="idstart" onClick="sendForm(document.getElementById('iduploadfrm'),this);">
<span>Start</span>
</button>
<button class="btn btn-warning cancel">
<i class="glyphicon glyphicon-ban-circle"></i>
<span>Cancel</span>
</button>
</td>
</tr>
</tbody>
</table>
</FORM>
</table>
I have a preview image table with a button. When clicking this button the event calls the function sendForm with button as parameter. Later i will check, wich button id is clicked to upload images seperately for each previewed image. So, this function calls uploadFiles function with file-control as parameter. The code running without any problems, but the image isn't storing on servers directory. Please can anywhere tell me, wich problem i having?
我有一个带按钮的预览图像表。单击此按钮时,事件将使用按钮作为参数调用函数sendForm。稍后我将检查,点击按钮ID为每个预览图像单独上传图像。因此,此函数使用file-control作为参数调用uploadFiles函数。代码运行没有任何问题,但图像不存储在服务器目录中。请随便告诉我,我有问题吗?
On Server-Side the Folder has IIS ans Network_Service has read and write rights.
在服务器端,文件夹具有IIS和Network_Service具有读写权限。
thanks a lot for any solution René
非常感谢René的任何解决方案
1 个解决方案
#1
0
You've said:
xhr.setRequestHeader("Content-Type","multipart/form-data");
… which will overwrite the default header and destroy the separator information (so the server won't know where one part ends and the next begins).
...将覆盖默认标头并销毁分隔符信息(因此服务器将不知道一个部件的结束位置和下一个部件的开始位置)。
Don't do that.
不要那样做。
#1
0
You've said:
xhr.setRequestHeader("Content-Type","multipart/form-data");
… which will overwrite the default header and destroy the separator information (so the server won't know where one part ends and the next begins).
...将覆盖默认标头并销毁分隔符信息(因此服务器将不知道一个部件的结束位置和下一个部件的开始位置)。
Don't do that.
不要那样做。