CODEWITHSUNDEEP
asp.netajaxajaxcontroltoolkit
DotnetSparrow
DotnetSparrow

Reputation: 27996

getting asynfileupload controls file name on button click

I am using ajaxfileupload control on my asp.net page. After image is uploaded, I call uploadcomplete method to save the image on disk and show in image control using following javascript:

  string fileName = Guid.NewGuid() + Path.GetExtension(PhotoAFU.FileName.Trim()); // encrypt filename 

                    string filePath = Path.Combine(storagePath, fileName);
                    string fileVirtPath = GetImageUrl(fileName); 

       int rnd = new Random((int)DateTime.Now.Ticks).Next(1, 1000);
                    ScriptManager.RegisterClientScriptBlock(PhotoAFU, PhotoAFU.GetType(), "img",
                        String.Format(
                            @"top.document.getElementById('{0}').src='{1}?x={2}';
                            top.document.getElementById('{3}').value = '{4}'",
                            EditPhotoImage.ClientID,
                            fileVirtPath,
                            rnd,
                            UploadedImageFileNameHF.ClientID,
                            fileName),
                        true
                );

Now I click on a save button and try to get the image using following code:

Path.GetFileName(EditPhotoImage.ImageUrl) // shows old image


or


Path.GetFileName(PhotoAFU.FileName) // it shows actual image name not encrypted one

but they both show old image not the current image or actual image name not encrypted name. How can I get the filename from above method in this method ? I tried using viewstate but it is not working properly.

Upvotes: 2

Views: 773

Answers (1)

IUnknown
IUnknown

Reputation: 22448

You can pass data from server to client in PostedUrl property of AjaxFileUploadEventArgs parameter in UploadComplete event handler as JSON obeject and get this data in OnClientUploadComplete handler on client side:

protected void AjaxFileUpload1_OnUploadComplete(object sender, AjaxFileUploadEventArgs e)
{
    string fileName = Guid.NewGuid().ToString();
    string fileVirtPath = "foobar";

    e.PostedUrl = string.Format("{{ fileName: '{0}', imageSrc: '{1}?x={2}' }}", 
        fileName, fileVirtPath, new Random((int)DateTime.Now.Ticks).Next(1, 1000));
}

 function AjaxFileUpload1_OnClientUploadComplete(sender, args) {
      var fileInfo = Sys.Serialization.JavaScriptSerializer.deserialize(args.get_postedUrl());
      $get("<%= EditPhotoImage.ClientID %>").src = fileInfo.imageSrc;
      $get("<%= UploadedImageFileNameHF.ClientID %>").value = fileInfo.fileName;
 }

Upvotes: 2

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