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How do I code for single file manual upload?

Last post 02-15-2012, 11:36 AM by Kenneth. 8 replies.


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02-10-2012, 4:31 PM 72987

dmathews
Joined on 02-10-2012
Posts 10

How do I code for single file manual upload?

Reply Quote

Greetings,

I'm having trouble figuring out how to code the manual upload of a single file (MultipleFilesUpload = False).

------ASPx Page-------

<CuteWebUI:Uploader runat="server" ID="Uploader1" InsertText="Select File" OnFileUploaded="Uploader_FileUploaded" ManualStartUpload="True">
</CuteWebUI:Uploader>

<asp:Button ID="btnUpload" runat="server" Text="Upload" />

------Code Behind---------

Protected Sub Uploader_FileUploaded(ByVal sender As Object, ByVal args As UploaderEventArgs)
        'Copys the uploaded file to a new location.
        args.CopyTo(Server.MapPath("~/uploads/") & args.FileName)
    End Sub

Protected Sub btnUpload_Click(ByVal sender As Object, ByVal e As System.EventArgs) Handles btnUpload.Click

'What goes here? Uploader1.Upload requires file size and file name which I don't know how to retrieve from the control.

End Sub

----------------------------

When I set MultipleFilesUpload = True, everything works well. However, I'd like to validate other information on the form prior to uploading the file. How do I retrieve the file size and file name from the control so that I may then call Uploader1.Upload(filesize, filename, temppath)?

Thanks!

Filed under: manual upload single file size filename

02-12-2012, 9:53 PM 72995 in reply to 72987

Kenneth
Joined on 02-13-2008
Posts 3,886

Re: How do I code for single file manual upload?

Reply Quote

Hi dmathews,

The example below shows you how to get the file info after uploaded.

<%@ Page Language="C#" %>

<%@ Register Namespace="CuteWebUI" Assembly="CuteWebUI.AjaxUploader" TagPrefix="CuteWebUI" %>

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">

<title>Untitled Page</title>

</head>

protected void b1_Click(object sender, EventArgs e)

{

label1.Text += uploader1.Items[0].FileName;

label1.Text += "<br />";

label1.Text += uploader1.Items[0].FileSize;

}

</script>

<body>

<div>

<CuteWebUI:UploadAttachments ID="uploader1" runat="server" MultipleFilesUpload="false">

</CuteWebUI:UploadAttachments>

<asp:Button ID="b1" runat="server" Text="get file info" OnClick="b1_Click" /><br />

<asp:Label ID="label1" runat="server"></asp:Label>

</div>

</form>

</body>

</html>

Regards,

Ken

02-14-2012, 9:33 AM 73024 in reply to 72995

dmathews
Joined on 02-10-2012
Posts 10

Re: How do I code for single file manual upload?

Reply Quote

Thanks Ken.

I was incorrectly using <CuteWebUI:Uploader> instead of <CuteWebUI:UploadAttachments>.

02-14-2012, 10:47 AM 73026 in reply to 72995

dmathews
Joined on 02-10-2012
Posts 10

Re: How do I code for single file manual upload?

Reply Quote

One more issue:

I'm now getting the following error:

"Index out of range Parameter name: index Actual value was 0."

I'm not sure what I'm missing. My code is as follows:

Imports CuteWebUI

Partial Class me_Files
Inherits System.Web.UI.Page

Protected Sub btnUpload_Click(ByVal sender As Object, ByVal e As System.EventArgs) Handles btnUpload.Click
dim filename, filesize as String

filename = uploader1.Items(0).FileName
filesize = uploader1.Items(0).FileSize

End Sub

End Class

----------ASPx Page---------------

<%@ Page Title="" Language="VB" MasterPageFile="~/SSFMaster.master" AutoEventWireup="false" CodeFile="Files.aspx.vb" Inherits="me_Files" %>

<CuteWebUI:UploadAttachments ID="uploader1" runat="server" MultipleFilesUpload="false">
</CuteWebUI:UploadAttachments>

<%@ Register Namespace="CuteWebUI" Assembly="CuteWebUI.AjaxUploader" TagPrefix="CuteWebUI" %>

<%@ MasterType VirtualPath="~/SSFMaster.master" %>

<asp:Content ID="Content1" ContentPlaceHolderID="CHead" Runat="Server">
<link rel="Stylesheet" href="../css/aStyles.css" type="text/css" media="screen" />
</asp:Content>

<asp:Content ID="Content2" ContentPlaceHolderID="CMain" Runat="Server">
<div>

<CuteWebUI:UploadAttachments ID="uploader1" runat="server" MultipleFilesUpload="false">
</CuteWebUI:UploadAttachments>

<asp:Button ID="btnUpload" runat="server" Text="Upload" />

</div>

</asp:Content>

02-15-2012, 6:12 AM 73028 in reply to 73026

Kenneth
Joined on 02-13-2008
Posts 3,886

Re: How do I code for single file manual upload?

Reply Quote

Hi dmathews,

Before you get the item, please ensure that the uploader attachment has items there.

If uploader1.Items.Count <> 0 Then

   Dim name As String = uploader1.Items(0).FileName

End If

Regards,

Ken

02-15-2012, 10:29 AM 73032 in reply to 73028

dmathews
Joined on 02-10-2012
Posts 10

Re: How do I code for single file manual upload?

Reply Quote

Hi Ken,

I'm sorry I didn't clarify that. The problem is that the uploader attachment never has any items. When debugging, after I add an attachment, I click the Submit button and step through my code. The items count is always 0.

Any ideas as to what I'm doing wrong?

Thanks!

02-15-2012, 10:38 AM 73033 in reply to 73032

Kenneth
Joined on 02-13-2008
Posts 3,886

Re: How do I code for single file manual upload?

Reply Quote

Hi dmathews,

Can you try the example I sent above? Does it work for you? If yes, please refer to my code and find out what is the difference with yours.

Regards,

Ken

02-15-2012, 11:17 AM 73034 in reply to 73033

dmathews
Joined on 02-10-2012
Posts 10

Re: How do I code for single file manual upload?

Reply Quote

Ken,

Yes, I was able to get your code to work but when I add 'ManualStartUpload="True"' to the control, I am then unable to get the filename/filesize because the items count is 0.

But if I try to call uploader1.upload, I'm required to supply filename, filesize, and temppath.

So when doing a manual upload, how do I get the filename and filesize required to upload the file since these items aren't in the collection until after upload.

I'm sorry if I'm missing something here.

02-15-2012, 11:36 AM 73035 in reply to 73034

Kenneth
Joined on 02-13-2008
Posts 3,886

Re: How do I code for single file manual upload?

Reply Quote

Hi dmathews,

Try

<%@ Page Language="C#" %>

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">

protected void b1_Click(object sender, EventArgs e)

{

if (Uploader1.Items.Count != 0)

label1.Text = Uploader1.Items[0].FileName;

}

</script>

<title>Start uploading manually</title>

</head>

<body>

<asp:Label ID="label1" runat="server"></asp:Label><br />

<CuteWebUI:UploadAttachments runat="server" ManualStartUpload="true" ID="Uploader1"

InsertText="Browse Files (Max 1M)">

</CuteWebUI:UploadAttachments>

<br />

<asp:Button runat="server" ID="SubmitButton" OnClientClick="return submitbutton_click()"

Text="Submit" /><br />

<p>

after you click on the submit button</p>

<asp:Button ID="b1" runat="server" Text="get info after click the submit button"

OnClick="b1_Click" /><br />

<p>

after you selected the file, before click on the submit button</p>

var fileList;

function submitbutton_click()

{

var submitbutton=document.getElementById('<%=SubmitButton.ClientID %>');

var uploadobj=document.getElementById('<%=Uploader1.ClientID %>');

if(!window.filesuploaded)

{

if(uploadobj.getqueuecount()>0)

{

uploadobj.startupload();

}

return false;

}

window.filesuploaded=false;

return true;

}

function CuteWebUI_AjaxUploader_OnPostback()

{

window.filesuploaded=true;

var submitbutton=document.getElementById('<%=SubmitButton.ClientID %>');

submitbutton.click();

return false;

}

function CuteWebUI_AjaxUploader_OnQueueUI(list)

{

fileList=list;

}

function showFile()

{

alert(fileList[0].FileName);

}

</script>

</form>

</body>

</html>

Regards,

Ken