Hi, probably a silly/easy question but how do I show to a user that the application is busy? I used to do this with a loading form which popsup whenever a heavy task is being done but this doesn’t seem to work.
My old approach was something like this:
Private Sub DoSomethingWhichTakesTime()
LoadingForm.show()
LoadingForm.BringToFront()
‘Do stuff here
LoadingForm.Hide()
End Sub
In a few places in my WiseJ project loading usercontrols take somewhere between 1~2 seconds which is just enough to trigger people thinking it’s not responding.
Any advice would be greatly appreciated!
Vincent
Hi vincent_,
for action like click you can you YouButton.ShowLoader = True (for starting job busy) or YouPanel.ShowLoader = True.
And like that you can set “showLoader” at false in Default.json
Best,
Kevin (ITG)
I’ve managed to fix the issue by setting the “showloader” property to true in Default.json and also setting the timeout to 1000. This works for me and I haven’t tested any other values. (I’ve set the showloader property to false because I do not want users to see it when they first go to the url). This is not ideal for me but they’ll (me) have to live with the loader 🙂
So.. I was a bit fast with accepting this answer. I don’t use async so I used:
Private Sub Button1_Click(sender As Object, e As EventArgs) Handles Button1.Click
Application.ShowLoader = True
Application.Update(Me)
 'Do stuff here
Application.ShowLoader = False
End Sub
This works like a charm on my dev pc but once published I never get to see the loader.
Good I started with “silly/easy”. Thank you!
I will give this a go.
Hi vincent_,
Wisej has already add small tips to show that your applictaion is busy.
Private Async Sub Button1_Click(sender As Object, e As EventArgs) Handles Button1.Click
Application.ShowLoader = True
 'Do stuff here
Application.Update(Me, Sub()
Application.ShowLoader = False
End Sub)
End Sub
Happy coding,
Kevin (ITG)
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