I use ASP.NET and i need an easy way to upload a file asynchronously. So I tried to use asyncfileupload (Ajax control toolkit) but I also need to pass parameters to the server side. How can I do that ? thanks.
我使用ASP.NET,我需要一种简单的方法来异步上传文件。所以我尝试使用asyncfileupload(Ajax控件工具包),但我还需要将参数传递给服务器端。我怎样才能做到这一点 ?谢谢。
Here is my code :
这是我的代码:
on client side :
在客户端:
<asp:ToolkitScriptManager ID="ToolkitScriptManager1" runat="server">
</asp:ToolkitScriptManager>
<asp:AsyncFileUpload ID="afuMedia" runat="server" UploaderStyle="Modern" OnUploadedComplete="afuMedia_UploadedComplete" />
on server side :
在服务器端:
protected void afuMedia_UploadedComplete(object sender, AsyncFileUploadEventArgs e)
{
//int id = int.Parse(Request.QueryString["id"]);
string mediaPath = ConfigurationParameters.MediaPath;
string filePath = CurrentBrand.BrandCode + "\\" + CurrentCulture.CultureCode + "\\" + "highlights-" + id;
string physicalPath = Path.Combine(MapPath("~/" + mediaPath), filePath);
afuMedia.SaveAs(physicalPath);
}
1 个解决方案
#1
2
Add client handler for upload start via the OnClientUploadStarted
property and use it as below:
通过OnClientUploadStarted属性添加上传启动的客户端处理程序,并使用它如下所示:
<asp:AsyncFileUpload ID="afuMedia" runat="server" UploaderStyle="Modern"
OnUploadedComplete="afuMedia_UploadedComplete"
OnClientUploadStarted="afuMedia_OnClientUploadStarted" />
function afuMedia_OnClientUploadStarted(sender, args){
var id = 123;
var url = sender.get_postBackUrl();
url += url.indexOf("?") === -1 ? "?" : "&";
url += ("id=" + id.toString());
sender.set_postBackUrl(url);
}
With this code all that you need to do on you own it's to provide correct id value;
使用此代码,您需要做的就是提供正确的id值;
#1
2
Add client handler for upload start via the OnClientUploadStarted
property and use it as below:
通过OnClientUploadStarted属性添加上传启动的客户端处理程序,并使用它如下所示:
<asp:AsyncFileUpload ID="afuMedia" runat="server" UploaderStyle="Modern"
OnUploadedComplete="afuMedia_UploadedComplete"
OnClientUploadStarted="afuMedia_OnClientUploadStarted" />
function afuMedia_OnClientUploadStarted(sender, args){
var id = 123;
var url = sender.get_postBackUrl();
url += url.indexOf("?") === -1 ? "?" : "&";
url += ("id=" + id.toString());
sender.set_postBackUrl(url);
}
With this code all that you need to do on you own it's to provide correct id value;
使用此代码,您需要做的就是提供正确的id值;