Visual Basic

Re: How to draw a spectral plot to a picturebox

12-Nov-06 10:18

Why do so many people try to draw onto picture boxes ? You draw onto an Image or Bitmap, THEN you can worry about displaying it ( use a picturebox if you must ).

It's easier to handle your paint event and draw onto that directly.

buzzztah14-Nov-06 1:22

buzzztah

14-Nov-06 1:22

Hi Christian,

Thanks for your reply.. it did put thins in another perspective.

Since i am a newbie in the .Net environment, graphics is completely new to me. Could someone explain me in a few words how I can draw a line onto a graphical image and then put it in a Picturebox on the screen?

Thanks a lot for helping!!

Cheers
BaZ

Re: How to draw a spectral plot to a picturebox

Christian Graus14-Nov-06 9:41

14-Nov-06 9:41

Bitmap bm = new Bitmap (200, 200); // I think you need to set a pixel depth as well )
using (Graphics gr = Graphics.FromBitmap(bm))
{
gr.DrawLine(20, 20, Pens.White); // Something like that
}

You don't need a picturebox, but you can pass the bitmap to a picturebox here if you like
I'd prefer to make the bitmap a member, or to draw directly onto the graphics object in the paint event.

Reading Registry Value

Puggsley12-Nov-06 9:24

12-Nov-06 9:24

I have two option boxes, "optEnabled" and "optDisabled". I need to read a REG_BINARY Value from the key "HKEY_CURRENT_USER\Software\MyApp\DSBkgnd". If the value is "00000000" I need to set optEnabled to Checked an if the value is "01000000" set optDisabled to Checked. I have followed examples on msdn but they wouldn't work. Cry | :((

If anyone can provide the code or suggestions i would be very grateful Big Grin | :-D

Christian Graus12-Nov-06 10:15

12-Nov-06 10:15

Puggsley wrote:
I have followed examples on msdn but they wouldn't work.

Then you didn't follow them properly. Try posting your code. That would also tell us if you're using VB6 or VB.NET.

Puggsley12-Nov-06 10:30

12-Nov-06 10:30

Dim readValue As String
readValue = My.Computer.Registry.GetValue_(HKEY_CURRENT_USER\Software\Microsoft\MessengerService", "DSBkgndMode", 0)
If readValue = "00000000" Then
OptEnabled.Checked = True
OptDisabled.Checked = False
Else
OptEnabled.Checked = False
OptDisabled.Checked = True
End If

I have tried using a message box to display the value of readValue but it was always blank

Christian Graus12-Nov-06 11:03

12-Nov-06 11:03

Well, this won't compile.

readValue = My.Computer.Registry.GetValue("HKEY_CURRENT_USER\Software\Microsoft\MessengerService", "DSBkgndMode", "0")

You probably then want to convert the value to a number, and compare that.

Why have an enabled checkbox and a disabled one ? Isn't that redundant ?

Christian Graus12-Nov-06 11:05

12-Nov-06 11:05

Also - have you checked if that key exists at all ? DSBkgndMode doesn't exist on my machine.

Puggsley12-Nov-06 11:13

12-Nov-06 11:13

I checked that they key exists, and it does. The program compiles with out any errors or warnings, and i have checked again to make sure.

Christian Graus12-Nov-06 11:21

12-Nov-06 11:21

It can't, the code you posted is missing a quote.

Did you try the code I posted ?

Puggsley12-Nov-06 11:24

12-Nov-06 11:24

Soz, it's missing a quaote because i access the internet on a different pc. Where is the code you posted?

Puggsley12-Nov-06 11:37

12-Nov-06 11:37

Soz, Its missing a quote cos i'm on the internet on a different pc. Where is the code you posted?

Christian Graus12-Nov-06 11:40

12-Nov-06 11:40

Please don't double post. Here is my response ( which is visible above, but anyhow.... )

readValue = My.Computer.Registry.GetValue("HKEY_CURRENT_USER\Software\Microsoft\MessengerService", "DSBkgndMode", "0")

You probably then want to convert the value to a number, and compare that.

Why have an enabled checkbox and a disabled one ? Isn't that redundant ?

Puggsley12-Nov-06 11:52

Sorting a one-dimensional array [modified]

12-Nov-06 11:52

I've had to restart the pc, and now im getting build errors in "Private Sub InitializeComponent()" There is 12 differnet errors, all for code generated by the form designer. Im gonna redo this form from scratch then get back tomorrow.

Thanks for the help

maddogjbm12-Nov-06 8:31

maddogjbm

12-Nov-06 8:31

Hi everyone,
I've been working on this for 2 weeks and am at my wits end.
I have to read in 20 random numbers between 10 and 100, inclusive, and then display them with no duplicate numbers. The random array is no problem, but I can't seem to be able to loose the duplicate numbers. I started with the array unsorted, had not luck, so I sorted it first, but still no luck. Here's some snippets to show the direction I've been trying. (I hope I'm posting this right)

<Private Sub cmdSort_Click(ByVal sender As System.Object, _
ByVal e As System.EventArgs) Handles cmdSort.Click

Dim array1 As Integer() = New Integer() {randomObject.Next(10, 101), _
randomObject.Next(10, 101), randomObject.Next(10, 101), randomObject.Next(10, 101), _
randomObject.Next(10, 101), randomObject.Next(10, 101), randomObject.Next(10, 101), _
randomObject.Next(10, 101), randomObject.Next(10, 101), randomObject.Next(10, 101), _
randomObject.Next(10, 101), randomObject.Next(10, 101), randomObject.Next(10, 101), _
randomObject.Next(10, 101), randomObject.Next(10, 101), randomObject.Next(10, 101), _
randomObject.Next(10, 101), randomObject.Next(10, 101), randomObject.Next(10, 101), _
randomObject.Next(10, 101)}
Dim i, j As Integer
txtOutput.Text = ""
' omitted code for bubblesort
BubbleSort(array1)

For i = 0 To array1.GetUpperBound(0)
BinarySearch(array1, array1(i))
If array1(i) > 0 Then
txtOutput.Text &= array1(i) & " "
End If
Next
End Sub
Function BinarySearch(ByVal array As Integer(), _
ByVal key As Integer) As Integer

Dim low As Integer = 0 ' low index
Dim high As Integer = array.GetUpperBound(0) ' high index
Dim middle As Integer ' middle index
Dim number As Integer
key = number
While low <= high
middle = (low + high) \ 2

If key = array(middle) Then ' match
Return -1
ElseIf key < array(middle) Then ' search low end
high = middle - 1 ' of array
Else
low = middle + 1
End If

End While

Return number ' search key not found

End Function>

maddogjbm

-- modified at 14:39 Sunday 12th November, 2006

Guffa12-Nov-06 9:07

Guffa

12-Nov-06 9:07

Hm... your code raises more questions than it answers...

Why do you call the BinarySearch function without taking care of the return value?
As the random numbers are always between 10 and 100, when do you expect the if statement that checks if it is larger than zero to exclude any value?
Why do you write your own BinarySearch, when there already exists one in the framework?
Why do you have the parameter key in the BinarySearch method, when you don't use it?
Why don't assign any value to the variable number?

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b { font-weight: normal; }

maddogjbm12-Nov-06 11:29

maddogjbm

12-Nov-06 11:29

OK, to start this is my first semester in VB.NET, I didn't grow up doing this, in fact I hadn't even touched a computer until 2001.
I'm trying to learn, if I knew the answers I wouldn't need to ask you.
In my mind the value return would be the number returned if it is not a duplicate? The reason for the statement that checks if it is larger than zero is because at one point the BinarySearch function was returning (and cmdSort was displaying) -1, that statement stopped it from displaying. I used my own BinarySearch because it was from my course text, I didn't realise framework had one, I'm still trying to make sense of the framework library, to me it's hard to understand. The book is a little easier, but still complicated.
As for the rest, my code started out much simpler but wasn't producing the results I desired so I've been altering it to try and get it to do what I need it to do. I believe I mentioned I've been working on this for 2 weeks, it's due by midnight tonight and I'm no closer to solving it now than I was before I asked for help here.

Syed Shahid Hussain12-Nov-06 9:26

Syed Shahid Hussain

12-Nov-06 9:26

Hi Friend,

First of all try to make your post concise (dont mind).

now why not you simply declare an array then loop to 20 and generate random no.s then place an inner check that searches the random no. is already placed or not if yes loop variable's ending condition must be one incremented (or what ever u like).

I tried to answer. If not correct tell me your requirments then i'll try to solve.

Bye

Syed Shahid Hussain

Christian Graus12-Nov-06 10:17

12-Nov-06 10:17

Wow -what a mess.

List<int> list = new List<int>();

while(list.Count < 20)
{
int n = randomObject.Next(10,101);
if (!list.Contains(n)) list.Add(n); // ! is Not in VB.NET
}

list.Sort();

That should do it in C#, hopefully you can follow my logic and impliment it in VB.NET.

Guffa12-Nov-06 11:20

Guffa

12-Nov-06 11:20

Why not make the numbers sorted and without duplicates from the start? Smile | :)

List<int> list = new List<int>();
int count = 20, n = 10;
Random rnd = new Random();
while (count > 0) {
	if (rnd.Next(101 - n) < count) {
		list.Add(n);
		count--;
	}
	n++;
}

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b { font-weight: normal; }

Christian Graus12-Nov-06 11:24

12-Nov-06 11:24

Because that's harder to read ? :P Also, because I think it's worth pointing out that the container classes have methods like Sort, there's no need to be writing them.