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onKeypress Event

Posted on 1998-09-28
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Last Modified: 2010-04-06
I would like to have a editbox in which we can only type numbers and the backspace. I think I have to use the onKeypress event and limit the access to ascii codes but the onkeypress event returns a char.  

Thanks
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Question by:deha2228
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7 Comments
 
LVL 5

Expert Comment

by:JimBob091197
ID: 1340990
What about trapping the WM_CHAR message?

E.g.
In your component's declaration:
procedure WMChar(var Msg: TMessage); message WM_CHAR;

In your code:
procedure TMyEdit.WMChar(var Msg: TMessage);
begin
  // Only allow backspace, ENTER, Escape and 0..9
  if (TWMKey(Msg).CharCode in [8, 13, 27, 48..57]) then
    inherited;
end;

Cheers,
JB
0
 
LVL 2

Accepted Solution

by:
Thaddy earned 100 total points
ID: 1340991
The easy way would be to use a TMaskedEdit. It has all the features you want, Just create an editmask that satisfies your input limitations.
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LVL 10

Expert Comment

by:viktornet
ID: 1340992
Hello all!

I dont like TMaskEdit. It's kinda crappy. I rahter create my own EditBox as deha2228 is going to do. JimBob's solution is prefect for that task and is very easy to implement ;-)

Gotta get back to class ;->

Regards,
Viktor Ivanov
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LVL 27

Expert Comment

by:kretzschmar
ID: 1340993
hi deho2228,

help this codesample

procedure TForm1.Edit1KeyPress(Sender: TObject; var Key: Char);
const validchars = [#8,'0','1','2','3','4','5','6','7','8','9'];
begin
  if not(key in validchars) then key := #0
end;

meikl
0
 
LVL 10

Expert Comment

by:viktornet
ID: 1340994
Here is an easier way...

procedure TForm1.Edit1KeyPress(Sender: TObject; var Key: Char);
begin
  if not key in [#8, #13, #27 '.', '0'..'9'] then key := #0;
end;

Regards,
Viktor Ivanov
0
 
LVL 27

Expert Comment

by:kretzschmar
ID: 1340995
hi viktor,

your comment is just the same, but i ask you for what do you use enter and Escapekey in an editfield?

well my const line could like shown as follows

      const validchars = [#8,'0'..'9'];

and the point if want to use can be added

meikl
0
 
LVL 10

Expert Comment

by:viktornet
ID: 1340996
My point is that you could use the shorter way instead of writing all the chars like so...
const validchars = [#8,'0','1','2','3','4','5','6','7','8','9'];

I added some other characters indeed
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