服务器端代码
protected void UploadComplete(object sender, AjaxControlToolkit.AsyncFileUploadEventArgs e)
{
rlativepath =GeneratePrefixFileName() + AsyncFileUpload1.PostedFile.FileName;
databasepath = "~/Image/" + rlativepath;
filePath = Request.PhysicalApplicationPath + "image\\"+rlativepath;
AsyncFileUpload1.SaveAs(filePath);
}
客户端代码
<script type="text/javascript">
function upLoadStarted() {
$get("imgDisplay").style.display = "none";
}
function showConfirmation(sender, args) {
var txt = document.getElementById('<%=statusoutput.ClientID %>');
var img = document.getElementById('<%=statusoutput.ClientID %>');
txt.value = "Upload Successful";
var imgDisplay = $get("imgDisplay");
imgDisplay.src = "ajaxupload.jpg";
imgDisplay.style.cssText = "height:100px;width:100px";
var img = new Image();
img.onload = function () {
imgDisplay.style.cssText = "height:100px;width:100px";
imgDisplay.src = img.src;
};
<%# UploadFolderPath1+=rlativepath %>
img.src = "<%=ResolveUrl(UploadFolderPath1) %>"+ args.get_fileName();
alert(img.src);
var imagedescrip = $get("imagedescrip");
imagedescrip.style.cssText = "visibility:visible;";
}
页面:
<asp:ScriptManager ID="ScriptManager1" runat="server">
</asp:ScriptManager>
content of page
FNever increase, beyond what is necessary, the number of entities required to explain anything." William of Ockham (1285-1349)
<asp:AsyncFileUpload ID="AsyncFileUpload1" runat="server" OnUploadedComplete="UploadComplete" ThrobberID="imgLoader" OnClientUploadStarted="upLoadStarted" UploaderStyle="Modern" OnClientUploadComplete="showConfirmation"
Width="400px" CompleteBackColor="White" UploadingBackColor="#CCFFFF"></asp:AsyncFileUpload>
<input type="text" id="statusoutput" runat="server" readonly="readonly" tabindex="-1000" style="border:0px;background-color:transparent;" />
<asp:Image ID="imgLoader" runat="server" ImageUrl="ajaxupload.jpg" Height="100px" Width="100px" />
<img id = "imgDisplay" alt="" src="" style = "display:none;height:100px;width:100px"/>
我正在使用 AsyncFileUpload 上传文件,在将文件保存在服务器上之前,我重命名了所选文件。如何在客户端获取这个新文件名?现在我的问题是我没有在客户端使用 args.get_filename() 重命名文件名。如何获得它?