|
he he he I'm sorry, when I was looking for "Vertical ProgressBar" I'd write "Vertical ProgresssBar". Sorry.
I'd need the vertical progressbar because my form it's little limited of space, but I find space.
Anyway thanks.
SINCERELY.
ANTHONY ACUÑA
PREFERED PHRASE:
SOMEBODY TELL ME WHY IS MORE REAL WHEN I DREAM THAT I AM WAKE?
|
|
|
|
|
Hi,
I have an unmanged exe and I would somehow, like to have it run inside my form. I don't care how, but is it possible?
I just want my app to be a container, or to contain the other unmanaged app.
Any ideas?
/\ |_ E X E GG
|
|
|
|
|
You can probably hack this together by making your unmanaged app a child window in your managed app. I would not do this for production code ,tho.
Christian Graus - Microsoft MVP - C++
Metal Musings - Rex and my new metal blog
|
|
|
|
|
Is it possible to render an event from a USB card reader is the card swiped is not magnetic?
Thanks
I am trying to display MsgBox.Tex = "Card is not Authorized"
|
|
|
|
|
oskardiazdeleon wrote: the card swiped is not magnetic?
Is there input data from the device when a non-magnetic card is swiped? If not then No.
led mike
|
|
|
|
|
|
Ok, in my text editor, when a user clicks new when the document hasn't been saved, it prompts the message:
MessageBox.Show("This document has not been saved, would you like to save this document?", "Save", MessageBoxButtons.YesNoCancel, MessageBoxIcon.Exclamation, MessageBoxDefaultButton.Button1);
But how do I now get returns from this function, so I can use an if statement such as:
if(WHATERVER GOES HERE == OK){Messagebox.Show("User clicked OK");
Thanks,
Any reply is appreciated.
|
|
|
|
|
MEssageBox.Show returns a DialogResult value. That's the return value you want.
--
Rules of thumb should not be taken for the whole hand.
|
|
|
|
|
So if I used:
if(MessageBox.Show == DialogResult.OK)<br />
{<br />
MessageBox.Show("User selected OK");<br />
}<br />
that should work?
Hmm, how does the compiler know that MessageBox.Show in the if statement is the same as the one with the buttons?
Do I have to do something like:
MessageBox myBox = new MessageBox();<br />
myBox.Show(ALL THE CODE HERE);<br />
if(myBox == DialogResult.OK)<br />
{<br />
CODE HERE<br />
}
Should that work?
|
|
|
|
|
You can do something like this.
<br />
if (MessageBox.Show("Press Ok or Cancel", "Test", MessageBoxButtons.OKCancel) == DialogResult.OK)<br />
{<br />
MessageBox.Show("You pressed Ok");<br />
}<br />
else<br />
{<br />
MessageBox.Show("You pressed Cancel");<br />
}<br />
You cannot use new on MessageBox because its a static class (all functions are static and the constructor is private). Also since the MessageBox.Show function does not return unless the user press a button, there is no confusion for the compiler.
---
"Drawing on my superior command of language I said nothing."
|
|
|
|
|
Thanks, I'll try that after I finish my ICT Essay :P
And yeah, I realised you cannot use MessageBox as a normal class, using new etc.
Out of interest, why is this the case?
|
|
|
|
|
The way I think about it is like this:
We create normal classes (with constructor) because we might want to have multiple instances of that class at the same time. There is no need to create multiple instances of the MessageBox class because the only thing you use a MessageBox is to show it to the user and wait for a input. The MessageBox is a modal dialog and hence the application will have to sit and wait till user clicks a button. Once the user clicks the button the only information you will need is which button the user clicked (if there is more than one button). So the entire purpose for which MessageBox class exists is finished when we call the Show method and check its return value. Hence the natural way is to make it a static class so that its very easy for use (means write less code). User never needs to create an instance.
---
"Drawing on my superior command of language I said nothing."
|
|
|
|
|
Blekk wrote: And yeah, I realised you cannot use MessageBox as a normal class, using new etc.
That's because MessageBox
a) doesn't have any constructors
b) the Show method is declared static, which means that you don't need an instance of the class and call it directly like Class.Method
hope this helps,
regards
modified 12-Sep-18 21:01pm.
|
|
|
|
|
Thanks Pradeep C and Greeeg, I now understand why you don't have to create an instance and you have both also made the whole concept of the C# language a little more clear.
There are still many things I don't really know at this stage, but probably should know, meh, I would rather find it out this way than reading through endless chapters of a book and then forgetting what it was at the end of the book. 
Thanks.
|
|
|
|
|
|
DialogResult result = MessageBox.Show("do something?", "title here", MessageBoxButtons.YesNo);
if(result == DialogResult.Yes)
{
}
else
{
}
|
|
|
|
|
Hi Blekk.
Exists two way for using.
1)
if( MessageBox.Show(Handle, msg, caption, MessageBoxButtons.RetryCancel, MessageBoxIcon.Question) == DialogResult.Retry)<br />
{<br />
}
2)
DialogResult p = MessageBox.Show(Handle, msg, caption, MessageBoxButtons.RetryCancel, MessageBoxIcon.Question);<br />
if( p == DialogResult.Retry )<br />
{<br />
}
SINCERELY.
ANTHONY ACUÑA
PREFERED PHRASE:
SOMEBODY TELL ME WHY IS MORE REAL WHEN I DREAM THAT I AM WAKE?
|
|
|
|
|
Thankyou also, I will use both methods and see which I find the easiest to understand etc.
Thanks for all the help.
|
|
|
|
|
Thanks all for your help, I've never known anybody respond so quickly on any forum ever, with so much useful information too.
Thanks all for your help.
|
|
|
|
|
These two programs are giving errors while compiling.
WAP that counts the no. of occurence of particular character in a line of text.
class demo
{
private int Occurance(string Input, string Find)
{
System.Text.RegularExpressions.Regex rx =
new System.Text.RegularExpressions.Regex(Find);
// to display them do this
System.Text.RegularExpressions.Match mt = rx.Match(Input);
while (mt.Success)
{
System.Console.WriteLine("Found >>" + Find + "<< at position " + mt.Index.ToString());
mt = mt.NextMatch();
}
// end display
return rx.Matches(Input).Count;
}
}
This program is giving error that entry point is not defined.
-------------------------------
WAP to take 2 dates using objects(day,month,year)& display both dates,their comparison also that which one is greater(take the dates with the help of three parameter constructors)
class demo
{
private DateTime getDate(int day, int month, int year)
{
return new DateTime(year, month, day);
}
private void compareDates(DateTime a, DateTime b)
{
if (a > b)
Console.WriteLine(a.ToShortDateString() + " > " + b.ToShortDateString());
else if (a == b)
Console.WriteLine(a.ToShortDateString() + " = " + b.ToShortDateString());
else
Console.WriteLine(a.ToShortDateString() + " < " + b.ToShortDateString());
}
}
this program is giving error "type or namespace couldnot be found(are u missing a using directive or an assembly reference).
|
|
|
|
|
1. You haven't defined an entry point. Where's your Main()?
2. Are you creating a dll?
BTW: Make your title more descriptive in the future.
|
|
|
|
|
using System;<br />
<br />
namespace demo<br />
{<br />
public class demo<br />
{<br />
public demo()<br />
{<br />
}<br />
<br />
public Main()<br />
{<br />
this.demo();<br />
}<br />
<br />
private int Occurance(string Input, string Find) <br />
{ <br />
System.Text.RegularExpressions.Regex rx = new System.Text.RegularExpressions.Regex(Find);
System.Text.RegularExpressions.Match mt = rx.Match(Input); <br />
while (mt.Success) <br />
{ <br />
System.Console.WriteLine("Found >>" + Find + "<< at position " + mt.Index.ToString()); <br />
mt = mt.NextMatch(); <br />
}<br />
<br />
return rx.Matches(Input).Count; <br />
} <br />
<br />
private DateTime getDate(int day, int month, int year) <br />
{ <br />
return new DateTime(year, month, day); <br />
} <br />
<br />
private void compareDates(DateTime a, DateTime b) <br />
{ <br />
if (a > b) <br />
{<br />
Console.WriteLine(a.ToShortDateString() + " > " + b.ToShortDateString()); <br />
}<br />
else if (a == b) <br />
{<br />
Console.WriteLine(a.ToShortDateString() + " = " + b.ToShortDateString()); <br />
}<br />
else <br />
{<br />
Console.WriteLine(a.ToShortDateString() + " < " + b.ToShortDateString()); <br />
}<br />
} <br />
<br />
~demo()<br />
{<br />
}<br />
<br />
}<br />
}
SINCERELY.
ANTHONY ACUÑA
PREFERED PHRASE:
SOMEBODY TELL ME WHY IS MORE REAL WHEN I DREAM THAT I AM WAKE?
|
|
|
|
|
I hope I'm just missing something basic.
Is it possible to have a custom background (say, gradient) for a System.Windows.Forms.Button without having to do *ALL* the drawing yourself.
OnPaintBackground does not appear to be called; the paint code for Button appears to be integrated with it's background logic.
|
|
|
|
|
I think you should be able to do this without much problem by creating a csutom control which derives from Button class.
You can override the OnPaint method and call base.OnPaint() so that it would do all the normal painting required for creating a Button and the you could draw whatever you want, like gradient or ...
---
"Drawing on my superior command of language I said nothing."
|
|
|
|
|
That solution will not let you paint the background and ignore the foreground. The need is to have a custom background, but let .NET draw the foreground image and text. The calculations of *where* the image and text appear is fairly convoluted due to the ImageAlign, TextAlign, and TextImageRelation properties.
I have pretty much confirmed this is not possible as .NET draws the button background and image/text in one operation, giving derived classes no opportunity to inject in the process.
Thank you for the answer.
|
|
|
|
General 
News 
Suggestion 
Question 
Bug 
Answer 
Joke 
Praise 
Rant 
Admin 
Submit an article or tip
