How to change label control data while opening a file.

Question

1.00/5 (2 votes)

See more:

VB

ASP.NET

Hi,

I want to update the label control data(text) when I open any file like Excel file.

Please reply.

Posted 15-Apr-13 1:14am

Archana Parate

Add a Solution

Comments

Prasad Khandekar 15-Apr-13 7:21am

Can you explain your problem in a bit more detail?

Archana Parate 15-Apr-13 7:46am

I have a imagebutton.

When I click on Imagebutton, it will open a file.
While opening a file, side by side a text will appear on label that on which date the file get opend. It means " File Read on 15-April-2013" text will appear on label control when file is opening.

the code is like below:

Dim fl As New FileInfo("C:\inetpub\ftproot\NewFolder1\External\abc.doc")
If (fl.Exists()) Then
lbl_readon.Text = "File Read on 15-April-2013 "
FileOpen(fl.FullName)
end if

Archana Parate 15-Apr-13 8:04am

I have a imagebutton.

When I click on Imagebutton, it will open a file.
While opening a file, side by side a text will appear on label that on which date the file get opend. It means " File Read on 15-April-2013" text will appear on label control when file is opening.

the code is like below:

Dim fl As New FileInfo("C:\inetpub\ftproot\NewFolder1\External\abc.doc")
If (fl.Exists()) Then
lbl_readon.Text = "File Read on 15-April-2013 "
FileOpen(fl.FullName)
end if

ZurdoDev 15-Apr-13 7:43am

OK. Where are you stuck?

Archana Parate 15-Apr-13 8:06am

The code is like Below :

Dim fl As New FileInfo("C:\inetpub\ftproot\NewFolder1\External\abc.doc")

If (fl.Exists()) Then
lbl_readon.Text = "File Read on 15-April-2013 " FileOpen(fl.FullName)
end if

Actually.

There is no error in the given code.

Actually it opens a file but it does not show the label text while opening a file.

Please tell me how to do it..

ZurdoDev 15-Apr-13 8:09am

You need to post more code. Is this part of an ajax postback? That would be why your label is not updated. If it is a full postback, it should work.

1 solution

Add a Solution

Add your solution here

Treat my content as plain text, not as HTML

Preview 0

…

Existing Members

Sign in to your account

...or Join us

Download, Vote, Comment, Publish.

Your Email
Password
Forgot your password?

Your Email
This email is in use. Do you need your password?
Optional Password

I have read and agree to the Terms of Service and Privacy Policy
Please subscribe me to the CodeProject newsletters

When answering a question please:

Read the question carefully.
Understand that English isn't everyone's first language so be lenient of bad spelling and grammar.
If a question is poorly phrased then either ask for clarification, ignore it, or edit the question and fix the problem. Insults are not welcome.
Don't tell someone to read the manual. Chances are they have and don't get it. Provide an answer or move on to the next question.

Let's work to help developers, not make them feel stupid.

This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

Dave Kreskowiak · Answer 1 · 2013-04-15T02:15:00

It doesn't work because you're opening the file on the UI (startup) thread. You're setting the text of the label, but since the thread is busy opening and, possibly, processing the file, the thread cannot get around to processing the WM_PAINT messages your app got to repaint the label with the new text.

Put the code for opening and processing the file in a background thread and it'll work. You can use explicit Threads, a BackgroundWorker component, Tasks from the Task Parallel library, or what have you.

Oh! And don't think you can just move the code for setting the label text to the background thread. It won't work, reliably anyway. You cannot touch UI controls from anything other than the UI thread.