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Hi,
If you know the answer for my question, pls let me know, else where can i find solution?
I didnt intendly put the question twice, I clicked submit button two times so it posted the same question twice.
Thankfully,
jm
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Hi,
I have created a sample Windows Service, in which I want to provide a notify icon on the system tray.
I have placed a notify icon and a contextmenustrip on the Service's Designer form, and on the OnStart() method of the Service I have assigned a value to the contextmenustrip property of the notifyicon. The following code snippet is written on the OnStart() event of the service,
contextMenuStrip1.Items.Add("Login");<br />
contextMenuStrip1.Items.Add("ViewRecentDocuments");<br />
<br />
notifyIcon1.ContextMenuStrip = contextMenuStrip1;<br />
notifyIcon1.Text = "Notifier";<br />
notifyIcon1.Visible = true;<br />
After that I have installed the Service in the system, and started the service. The notifyicon get displayed, but while right-clicking on it, contextmenu didnt displayed.
Is my code is correct one, if not what change should I do, to make the contextmenu to be displayed on notifyicon's right-click event..
Anyone knows the solution, please reply me.
Thankfully,
jm
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Hi everyone,
I'm pretty new to the programming process. I've never done a complete program until now. How do I create an install file for my program, in order to distribute and install my program to multiple computers? Right now I'm just running it from the debug file...
Thanks for your help
Eric
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Web : copy all the files to the server and create a virtual directory.
windows : compile in release mode and install the exe in all the comp.
I think you need to complete the program first before you can do all this 
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Thank you, I'll give it a try..
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Just for a follow-up. I think it has worked for now. But I won't know until tomorrow. It didn't create an install file, but an application file. Hopefully this will be sufficient.
Thanks again
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Great attitude mate, u sound like a complete MVP
1. The guy has not even completed a single app 1000 comp installation will be a bit less.
2. I think Bill Gates runs to his customers with a floppy disk in his hand (for may be Windows XP).
-- modified at 1:57 Tuesday 19th June, 2007
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Did you notice the quote was yours? It was in response your suggestion, it just got mixed up in the DB.
only two letters away from being an asset
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Yeah I did notice that. May be u did not read what i wrote. Try to solve problems more that making smilies
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Hi guys, I need your help in preventing multiple instances of the same window from opening in an MDI parent child window application ? Its urgent ...thanks ...
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What is the question, where are you stuck?
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My question is simple and straight..I want to prevent multiple instances of the same child window from opening i.e. only one instance of a window should open up when i select a certain menu option from the main menu. I hope i made myself clear this time..!!
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One way can be:
use frm.ShowDialog();
This way, unless you have not closed the current child form, you cannot open anyother.
This might be a quick fix for your problem. If you don't want this you might have to google intensly
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yeah..I have tried that approach, but as u say, i was not able to to open up any other window until i closed this one..but tht should not happen in an MDI windows application ..right ?
Can Dispose() method help in this case?
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yeah..I have tried that approach, but as u say, i was not able to to open up any other window until i closed this one..but tht should not happen in an MDI windows application ..right ?
Can Dispose() method help in this case?
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MDI Parent forms contain a collection of it's children (this.MdiChildren). Before you open a new instance of a form you could check this collection if your form exist and simply activate this instance.
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Can you elaborate a bit more...thanks
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I found the solutions mate:
Form2 frm = new Form2();
if (MdiChildren.Length != 0)
{
for (int i = 0; i < this.MdiChildren.Length; i++)
{
if (this.MdiChildren[i].GetType().Name != "Form2")
{
frm.Show();
frm.MdiParent = this;
}
}
}
else
{
frm.Show();
frm.MdiParent = this;
}
PS: you will have to modify this the above is real quick and dirty.
You will have to check this for each and every form you open.
Cheers.
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Can you fire an event inside of an event that is already executing?
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Is there anything special you have to do to get this to work because the event inside the executing event is not firing?
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It is possible that you are not firing the event properly, can you pass me the code.
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This is the code:
protected void btnNext_Clicked(object sender, EventArgs e)
{
switch (m_teststage)
{
case "STEP1":
hidMode.Value = "STEP2";
break;
case "STEP2":
hidMode.Value = "STEP3";
break;
}
m_teststage = hidMode.Value;
FormatPageChange();
}
private void FormatPageChange()
{
if (m_teststage == "STEP2") InitialiseDMRGrid();
if (m_teststage == "STEP3") InitialiseTestGrid();
SetPanelVisibility();
SetButtonVisibility();
SetValidatorsEnabled();
}
private void InitialiseDMRGrid()
{
WebGridHelper m_webgridhelperDMR = new WebGridHelper(wgDMR, "", "ID");
wgDMR.PrepareDataBinding += new DataSourceEventHandler(wgDMR_PrepareDataBinding);
wgDMR.InitializeDataSource += new DataSourceEventHandler(wgDMR_InitializeDataSource);
m_webgridhelperDMR.SortColumns += new SortColumnsEventHandler(m_webgridhelperDMR_SortColumns);
wgDMR.InitializePostBack += new PostBackEventHandler(wgDMR_InitializePostBack);
}
If I click the Next button the event fires but when it gets to the InitialiseDMRGrid method it goes in but does not fire the events inside.
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