slow speed - C# Discussion Boards

Re: Setting CollectionClass to Property - Custom Controls

30-Jul-07 2:05

Hello,

N a v a n e e t h wrote:
I succeeded in creating collectionclass

I guess like that:

public class YourItemsCollection : CollectionBase, IDisposable
{
 //...
}

For this class you have to write an converter, which inherits from "System.ComponentModel.ExpandableObjectConverter".

public class YourItemsCollectionConverter: ExpandableObjectConverter
{
 //...
}

In this class you have to override following methods:

public override object ConvertTo(ITypeDescriptorContext context, CultureInfo culture, object value, Type destinationType)

public override PropertyDescriptorCollection GetProperties(ITypeDescriptorContext context, object value, Attribute[] filter)

The "YourItemsCollectionConverter" has to be set as an Attribute to "YourItemsCollection":

[TypeConverter(typeof(YourItemsCollectionConverter))]

This article[^] shows it very good!

-- modified at 8:18 Monday 30th July, 2007

-- modified at 8:34 Monday 30th July, 2007

All the best,

Martin

N a v a n e e t h30-Jul-07 2:21

N a v a n e e t h

30-Jul-07 2:21

Thank you martin. It was excellent. Thank you very much

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Re: Setting CollectionClass to Property - Custom Controls

Martin#30-Jul-07 2:27

Plugin Frmaework in VS 2005 .NET

30-Jul-07 2:27

You are allways wellcome!

All the best,

Martin

dotman130-Jul-07 0:06

dotman1

30-Jul-07 0:06

Hi,

I am a new to Visual Studio.NET but not to development and require assistance; after downloading Plug-in samples from codeproject and the Microsoft sample,

and finding issues with deprecation etc using Visual Studio 2005 .NET.

The MS sample performs some form of conversion and I suspect this is in relation to the PluginCollection class, which now causes an issue with

ConfigurationSettings being replaced with ConfigurationManager, and naturally the following type cast is no longer supported:

m_plugins = (PluginCollection)ConfigurationSettings.GetConfig("plugins");

I am attempting to develop a plug-in host capable of displaying usercontrols plugins on a panel in the same manner as the codeproject sample

(ExtensibleUI_demo) but using the Microsoft framework by adding the IPlugin Interfaces to the welcome class. E.g.:
public partial class Welcome : UserControl, IPlugin

I have managed to develop an application based on the ExtensibleUI_demo to work, but the plug-in name details are not added to the listbox, and I am not sure

how much of this framework is no longer applicable or appropriate best practice.

Can anyone assist? it is my intention to use plug-ins to replace database forms, displaying them on a parent container such as a panel.

Regards

dotman1

Re: Plugin Frmaework in VS 2005 .NET

Joseph Guadagno30-Jul-07 4:23

Joseph Guadagno

30-Jul-07 4:23

Did you see my article on creating Plug-ins with .NET? It even has a sample application to test the plug-ins.

Here are some of my articles.
Hope this helps!
Joseph Guadagno
http://www.josephguadagno.net

Re: Plugin Frmaework in VS 2005 .NET

dotman130-Jul-07 23:14

dotman1

30-Jul-07 23:14

Thanks Joseph,
No I had not noticed your article; the article is well documented and clear, the demo code what can I say this is exactly what I was trying to achieve, its good to see someone else also appreciate class diagrams. I will revisit my prototype as I did manage to get a prototype working with the support of an article “Plugins in C# by Redth”, obviously there is nothing wrong with this approach yours lends itself to my personal style and approach.

Greatly appreciated.

dotman1

slow speed

Xmen Real 29-Jul-07 23:01

29-Jul-07 23:01

hi friends,
i m making a program where it open a text file in rtb1 and when button1 pressed it take one part from rtb1 and put it into rtb2
the codes are working good but problem is that its taking so much time

rtb1 have 34609 lines and putting 123 lines in rtb2 from rtb1 in 31 secs

i ve pentium D 2.8Ghz and 512mb ram, i think its enough for it but 31 secs are too

and here is the coding

richTextBox1.Find("[ID] = 1234");
     int selectionlength = 0;
     int c = richTextBox1.GetLineFromCharIndex(richTextBox1.SelectionStart);
     int d = richTextBox1.GetFirstCharIndexFromLine(c);

     for (int a = c; a < richTextBox1.Lines.Length; a++)

     {

         selectionlength += richTextBox1.Lines[a].Length + 1;

         if (richTextBox1.Lines[a].Trim() == "[End]")
         {
             break;
         }

     }

     richTextBox1.SelectionStart = d;
     richTextBox1.SelectionLength = selectionlength - 2;
     richTextBox2.Text = richTextBox1.SelectedText;

and the text file is look like :

[ID] = 0000
lines
[End]
[ID] = 0001
lines
[End]
.................

so somebody know any solution to increase its speed... Smile | :)

???

Becoming Programmer...

Luc Pattyn29-Jul-07 23:23

Luc Pattyn

29-Jul-07 23:23

Hi
two comments

1.
take richTextBox1.Lines out of the for line and the for block,
so do:

string[] lines=richTextBox1.Lines;
int linesLength=lines.Lenght;
for (int a = c; a < linesLength; a++) {
    selectionlength += lines[a].Length + 1;
    ...
}

That will give a dramatic difference because now the string array has to
be generated once instead of 100,000 times.

2.
If you don't need the extra capabilities of RichTextBox, don't use it at all.
Your app seems line-orineted; I can't tell whether you use colors, bold, etc.
If not, consider a ListBox, it is faster at anything it does (and it only
selects entire lines).

Smile | :)

Luc Pattyn

try { [Search CP Articles] [Search CP Forums] [Forum Guidelines] [My Articles] }
catch { [Google] }

Xmen Real 30-Jul-07 2:31

30-Jul-07 2:31

Luc Pattyn wrote:
1.
take richTextBox1.Lines out of the for line and the for block,

this must be good idea and i will try it.

Luc Pattyn wrote:
I can't tell whether you use colors, bold, etc.
If not, consider a ListBox, it is faster at anything it does (and it only
selects entire lines).

yep

Luc Pattyn wrote:
2.
If not, consider a ListBox, it is faster at anything it does (and it only
selects entire lines).

but how can i find "[ID]=1234" in listbox

thanks buddy

Becoming Programmer...

Re: slow speed [modified]

Luc Pattyn30-Jul-07 2:46

Luc Pattyn

30-Jul-07 2:46

Hi,

ListBox (and ComboBox) have two find methods: FindString and FindStringExact;
and these have an overload that takes a starting index.

If interested I suggest you look it up in MSDN...

But before changing all your code, give it a try with my first suggestion
(and tell us the new speed).

Smile | :)

Luc Pattyn

try { [Search CP Articles] [Search CP Forums] [Forum Guidelines] [My Articles] }
catch { [Google] }

Xmen Real 30-Jul-07 3:00

30-Jul-07 3:00

heee

its working about 2 secs to do

thank you alot

-- modified at 9:14 Monday 30th July, 2007

Becoming Programmer...

girm29-Jul-07 23:37

girm

29-Jul-07 23:37

Hi,

I may suggest that (considering you are just manipulating text) you should rely on the Find() method of the rtf :

int positionFound=Find(text,startPosition,endPosition,options)

I think you should use 'Find' to get the position of the '[END]' tag and not use a loop , as in your exemple.

Hope it helps

Xmen Real 30-Jul-07 2:33

30-Jul-07 2:33

i m not getting you, can you explain little bit more??

"I think you should use 'Find' to get the position of the '[END]' tag and not use a loop , as in your exemple"

but how can i know the lenght of between [ID] and [End]???

Becoming Programmer...

girm30-Jul-07 21:51

girm

30-Jul-07 21:51

Sure...

* assuming you have a form with 2 rtf , and a button to copy from richTextBox1 to richTextBox2
* assuming you only deal with text

the code below should do the trick in no time

using System;
using System.Collections.Generic;
using System.ComponentModel;
using System.Data;
using System.Drawing;
using System.Text;
using System.Windows.Forms;

namespace rtfTest
{
    public partial class Form1 : Form
    {
        public Form1()
        {
            InitializeComponent();
            richTextBox1.Text = "[ID] = 0000\nlines\n[END]\n[ID] = 1234\nlines for 1234\n[END]";
        }

        private void button1_Click(object sender, EventArgs e)
        {
            string beginTag = "[ID] = 1234";
            string endTag = "[END]";
            int startPosition = richTextBox1.Find(beginTag);
            if (startPosition >= 0)
            {
                startPosition = startPosition + beginTag.Length+1;// we start at the first line after beginTag
                int endPosition = richTextBox1.Find(endTag,startPosition,RichTextBoxFinds.MatchCase);
                if (endPosition > startPosition)
                    richTextBox2.Text = richTextBox1.Text.Substring(startPosition , endPosition - startPosition );
            }
                
        }
    }
}

Xmen Real 31-Jul-07 19:08

31-Jul-07 19:08

yep thats also good way, i like, thanks

Becoming Programmer...

C# Project corrupted?

Vikas K.29-Jul-07 21:21

29-Jul-07 21:21

Hi,

Sorry, If I have posted this query to a wrong group!

I'm facing an issue with C# project. I use VS 2005 express edition.

I was developing a C# project and had couple of UI controls that I had placed
using UI editor. Everything was fine but now I can't view the "UI designer" now.

If I run the project I do get the UI that I had designed but cant edit it now?

How can I see the disigner again?

Thanks
Vikas

T.EDY29-Jul-07 21:53

T.EDY

29-Jul-07 21:53

Hi,

Try to close the form and Rebuild Solution..and if the rebuild has succeeded..try to open your form again,sometimes i get the same problem but when i do that my UI designer can be view again

Kind Regards,

Tomi

Vikas K.29-Jul-07 23:29

29-Jul-07 23:29

I tried but still I can't see the form design! I guess, I will have to port the entire code into a new project now Cry | :((

Re: C# Project corrupted? [modified]

Developer61130-Jul-07 0:05

Developer611

30-Jul-07 0:05

Hi there .

I think , it's better to check your reference files.
Maybe you missed one of your reference . if it is , remove you reference from reference list and add it to your project reference again , and then Rebuild your solution or project , what ever as you want . Smile | :)

DMASTER

Martin#30-Jul-07 0:08

30-Jul-07 0:08

Hello,

What you can try:
In your project explorer, rightclick the Form or usercontrol you can not edit and change the "buildaction" to an other setting and then back to "compile" (I have not the english version and only 2003 so I don't know the exact names).
Then make a rebuild.

Hope it helps!

All the best,

Martin

Vikas K.30-Jul-07 19:54

30-Jul-07 19:54

Hi, Ok I know why this is happening but pls tell me whats wrong in doing this?

1. My form is like this: Everything is proper and I can view the designer

<br />
//file Form1.cs<br />
<br />
public partial class Form1 : Form<br />
{<br />
 // many functions/events related to UI controls are here<br />
}<br />

2. I need to have a special control class say MyPictureBox derived from PictureBox so added the code as follows in file Form1.cs:

<br />
//file Form1.cs<br />
<br />
public class MyPictureBox : System.Windows.Forms.PictureBox<br />
{<br />
  // some special code will be put here<br />
}<br />
<br />
public partial class Form1 : Form<br />
{<br />
 // many functions/events related to UI controls are here<br />
}<br />

Thats it: Now If I try to open the designer I get the error:

<br />
One or more errors encountered while loading the designer. The errors are listed below. Some errors can be fixed by rebuilding your project, while others may require code changes. <br />
<br />
The service System.Windows.Forms.Design.ISelectionUIService already exists in the service container. Parameter name: serviceType <br />
Hide     <br />
<br />
at System.ComponentModel.Design.ServiceContainer.AddService(Type serviceType, Object serviceInstance, Boolean promote)<br />
at System.ComponentModel.Design.ServiceContainer.AddService(Type serviceType, Object serviceInstance)<br />
at System.ComponentModel.Design.DesignerHost.System.ComponentModel.Design.IServiceContainer.AddService(Type serviceType, Object serviceInstance)<br />
at System.Windows.Forms.Design.ComponentDocumentDesigner.Initialize(IComponent component)<br />
at System.ComponentModel.Design.DesignerHost.AddToContainerPostProcess(IComponent component, String name, IContainer containerToAddTo)<br />
at System.ComponentModel.Design.DesignerHost.Add(IComponent component, String name)<br />
at System.ComponentModel.Design.DesignerHost.System.ComponentModel.Design.IDesignerHost.CreateComponent(Type componentType, String name)<br />
at System.ComponentModel.Design.Serialization.DesignerSerializationManager.CreateInstance(Type type, ICollection arguments, String name, Boolean addToContainer)<br />
at System.ComponentModel.Design.Serialization.DesignerSerializationManager.System.ComponentModel.Design.Serialization.IDesignerSerializationManager.CreateInstance(Type type, ICollection arguments, String name, Boolean addToContainer)<br />
at System.ComponentModel.Design.Serialization.TypeCodeDomSerializer.Deserialize(IDesignerSerializationManager manager, CodeTypeDeclaration declaration)<br />
at System.ComponentModel.Design.Serialization.CodeDomDesignerLoader.PerformLoad(IDesignerSerializationManager manager)<br />
at Microsoft.VisualStudio.Design.Serialization.CodeDom.VSCodeDomDesignerLoader.PerformLoad(IDesignerSerializationManager serializationManager)<br />
at System.ComponentModel.Design.Serialization.BasicDesignerLoader.BeginLoad(IDesignerLoaderHost host) <br />

Afterwards, I will not be able to view the original designer but I think the designer related to MyPictureBox which is blank.

But if I put the body of MyPictureBox after the body of Form as shown below, everything is fine!

<br />
//file Form1.cs<br />
<br />
public partial class Form1 : Form<br />
{<br />
 // many functions/events related to UI controls are here<br />
}<br />
<br />
public class MyPictureBox : System.Windows.Forms.PictureBox<br />
{<br />
  // some special code will be put here<br />
}<br />

Pls let me know what is the procedure in adding a derived UI control class?

Thanks

Martin#30-Jul-07 20:02

30-Jul-07 20:02

Hello,

The Problem is that both classes have design support.
It is not good practice to but them all in one file.
If you put the picturebox in a special file you will also be able to open it over the designer.

All the best,

Martin

Vikas K.30-Jul-07 20:12