Visual Basic

Christian Graus1-Nov-09 22:38

1-Nov-09 22:38

Luc Baetsle wrote:
But if a letter is entered, the error I created is given, but after the letter is visible in textbox also.

What error ?

You should also use char.IsControl to check if it's a backspace, etc. And if it's neither, set the Handled property of the eventargs to true. I believe you need to do this in the KeyPress event.

Christian Graus

Driven to the arms of OSX by Vista.

Read my blog to find out how I've worked around bugs in Microsoft tools and frameworks.

dan!sh 1-Nov-09 22:41

dan!sh

1-Nov-09 22:41

That wont fire in case user pastes text in the textbox through mouse.

It's not necessary to be so stupid, either, but people manage it. - Christian Graus, 2009 AD

Christian Graus1-Nov-09 22:49

1-Nov-09 22:49

I didn't know that, but, either way, any code that turns text into a number should use tryparse, and if that's the case, some extra code should be written to deal with it, but the KeyPress event is the best way to handle the user typing into the control.

Christian Graus

Driven to the arms of OSX by Vista.

Read my blog to find out how I've worked around bugs in Microsoft tools and frameworks.

dan!sh 1-Nov-09 23:01

dan!sh

1-Nov-09 23:01

Christian Graus wrote:
some extra code should be written to deal with it

And where would you write that? In the TextChanged or the MouseDoown event, hence what I feel is, it is better to keep the check in one place instead of two.

It's not necessary to be so stupid, either, but people manage it. - Christian Graus, 2009 AD

Christian Graus1-Nov-09 23:03

1-Nov-09 23:03

Well, you could do it all in TextChanged, but at that point, you have ALREADY accepted an unacceptable character, therefore, this solution is unacceptable if you want a clean implimentation, where pressing a letter or other non number/non control character, just does nothing.

Every implimentation I've ever seen, does it the way I'm suggesting. I don't even bother handling mouse pastes, if people want to do that, that's their fault. My code will still detect a non number and show a message accordingly.

Christian Graus

Driven to the arms of OSX by Vista.

Read my blog to find out how I've worked around bugs in Microsoft tools and frameworks.

Luc Baetsle1-Nov-09 23:05

Luc Baetsle

1-Nov-09 23:05

Why not in keypress event? Or will a copy / paste not be recognized then?

Member 37888845-Nov-09 1:00

Member 3788884

5-Nov-09 1:00

Text box named "origens"

Use this code

Private Sub origens_KeyPress(KeyAscii As Integer)
If KeyAscii < 48 Then ' Rejects all ascii < 0.
KeyAscii = 0
Else
If KeyAscii > 57 Then
KeyAscii = 0
End If
End If

End Sub

Create array of Images.

Anubhava Dimri1-Nov-09 21:12

1-Nov-09 21:12

Private imageArray() As Image

For i = 0 To List1.Items.Count - 1
                    imageArray(i) = Image    ' Image is a Image which format is metafile
Next

The above code work properly but imagearray() is not storts images data.

So the error is generated on imagearray(i).width.
error is "Parameter is not valid"

Thanks

If you can think then I Can.

Christian Graus1-Nov-09 21:21

1-Nov-09 21:21

eg_Anubhava wrote:
The above code work properly but imagearray() is not storts images data.

What makes you think that ?

eg_Anubhava wrote:
imageArray(i) = Image

This means that all the items point to the same image, if you change one, they all change. I suggest you read a basic book on VB, esp read about value types and reference types. I'd guess you set Image to null or otherwise dispose of it, thus breaking all the many identical references to it that are in this useless array.

Christian Graus

Driven to the arms of OSX by Vista.

Read my blog to find out how I've worked around bugs in Microsoft tools and frameworks.

Anubhava Dimri1-Nov-09 22:02

1-Nov-09 22:02

i want to be Array of Images.

If you can think then I Can.

Christian Graus1-Nov-09 22:07

1-Nov-09 22:07

I know that. But, you're making an array where all the images are the SAME object. Read my post again. Why do you want all the images in the array to be the same anyhow ?

Christian Graus

Driven to the arms of OSX by Vista.

Read my blog to find out how I've worked around bugs in Microsoft tools and frameworks.

Anubhava Dimri1-Nov-09 22:29

1-Nov-09 22:29

No sir its only a example. I am creating diffrent images array.

If you can think then I Can.

Christian Graus1-Nov-09 22:37

1-Nov-09 22:37

So you want us to help fix your code, but the code you posted is NOT the code you're using ?

Christian Graus

Driven to the arms of OSX by Vista.

Read my blog to find out how I've worked around bugs in Microsoft tools and frameworks.

Anubhava Dimri1-Nov-09 23:06

Read from XML file in word template for Office addin

1-Nov-09 23:06

My code is same but the reason of hiding is Long Code to define.
There Createmage is a function which returns the Image of passing parameter.

Private imageArray() As Image
For i = 0 To List1.Items.Count - 1
imageArray(i) = CreateImage(i)
Next

If you can think then I Can.

nitin_ion1-Nov-09 20:20

nitin_ion

1-Nov-09 20:20

I m new to office addin in VS 2008.
I am creating a word template in VS 2008 now i also created a ribbon button. what i want is when i click the ribbon button it should read xml file and fill the data in the template in which there is a table.

i am reading the file on the click of a button but i am not able to access the word template table in which i want to add the data since both are seperate projects (addin project and template project)

i also used the Microsoft.Office.Interop library to access the word object but i am not able to access the table object through which i want to insert the data in the xml file into the table.

any one has an idea on how to do it?

At the time of Generated Preview ......... [modified]

Anubhava Dimri1-Nov-09 19:27

1-Nov-09 19:27

Hello Friends,

I have generated problem while generating Print priview (When i m printing 4000-50000 Pages).

Error is Given Below

Screen-compatible bitmap cannot be created. The screen bitmap format cannot be determined.

Actually the problem is generated by this line

Me.PrintDocument1.DefaultPageSettings.HardMarginX

In 4900 times it's working normally. But at 4913 it generates error Attempted to divide by zero.

so plese give me the solution for that.

Thanks

If you can think then I Can.

modified on Monday, November 2, 2009 1:43 AM

Christian Graus1-Nov-09 20:03

1-Nov-09 20:03

eg_Anubhava wrote:
Me.PrintDocument1.DefaultPageSettings.HardMarginX

That can't be the whole line, it doesn't do anything. What's the rest of it ?

Christian Graus

Driven to the arms of OSX by Vista.

Read my blog to find out how I've worked around bugs in Microsoft tools and frameworks.

Anubhava Dimri1-Nov-09 20:29

1-Nov-09 20:29

The Proper Line is
If (leftMargin - Me.PrintDocument1.DefaultPageSettings.HardMarginX) >= 0.0F Then

'Code here

end if

If you can think then I Can.

Christian Graus1-Nov-09 20:50

1-Nov-09 20:50

OK, then I guess HardMarginX must internally do some sort of division. Perhaps you need to work out what that is, with reflection, and replace it with your own code which first checks for zero ?

Christian Graus

Driven to the arms of OSX by Vista.

Read my blog to find out how I've worked around bugs in Microsoft tools and frameworks.

Anubhava Dimri1-Nov-09 21:19

1-Nov-09 21:19

Can u explain it.

If you can think then I Can.

Christian Graus1-Nov-09 21:20

Code Changes not executing

1-Nov-09 21:20

Not sure how to make it simpler. If you get a divide by zero exception, that property must, inside .NET, do a divide by a value that's getting set to zero. Use reflektor or similar to find out what the source code for that property is, and instead of accessing that property, write your own that checks the values it uses, but first makes sure the value being divided by is not zero, and if it is, find a way to deal with that.

Christian Graus

Driven to the arms of OSX by Vista.

Read my blog to find out how I've worked around bugs in Microsoft tools and frameworks.

Richard Adler1-Nov-09 18:36

Richard Adler

1-Nov-09 18:36

I am making codoing changes to my Visual Basic.Net project but when I execute the code using the debugger, I see the old code. I added a new button to a form but when I execute the code, the button is not there. It seems like the code is executing from a different library.
Any suggestions?

Re: Code Changes not executing

Christian Graus1-Nov-09 18:40