Aight, I got this:
'moduleBnet"
ID_JOIN
frmMain.figureout
****
'frmMain
Sub figureout()
Dim itm1 As ListItem, itm2 As ListItem
Dim str1 As String, str2 As String
For Each itm1 In frmAliasManager.lvAliases.ListItems 'loop the first listview
If InStr(1, itm1.Text, ":") Then 'just make sure ":" is in there
str1 = Left$(LCase(itm1.Text), InStr(1, LCase(itm1.Text), ":") - 1) 'get the leftside of ":"
str2 = Mid$(LCase(itm1.Text), InStr(1, LCase(itm1.Text), ":") + 1) 'get the rightside of ":"
For Each itm2 In Form1.ChannelList.ListItems 'loop second listview
If LCase(itm2.Text) = str1 Then 'see if it matches the string
itm2.Text = str2 'if it does, change it
End If
Next itm2
End If
Next itm1
End Sub
The problem is, When someone joins a channel, eveyrthing under ID_JOIN activates exept frmMain.Figureout. When I put Call frmMain.Figureout ina button, it works. But in moduleBnet, it doesent activate. This is gettin frustrating.
Help?
Quote from: GoSu_KaOs on January 27, 2005, 08:15 PMHelp?
Yes: write better code. I'm pretty sure even VB supports splitting text on a delimiter, so that left/mid construct isn't needed.
Quote from: Kp on January 27, 2005, 08:31 PM
Quote from: GoSu_KaOs on January 27, 2005, 08:15 PMHelp?
Yes: write better code. I'm pretty sure even VB supports splitting text on a delimiter, so that left/mid construct isn't needed.
Split() (http://msdn.microsoft.com/library/default.asp?url=/library/en-us/vblr7/html/vafctsplit.asp)
Quote from: Kp on January 27, 2005, 08:31 PM
Quote from: GoSu_KaOs on January 27, 2005, 08:15 PMHelp?
Yes: write better code. I'm pretty sure even VB supports splitting text on a delimiter, so that left/mid construct isn't needed.
What does left/mid have to do with ID_JOIN.
It doesnt activate the code. I tryed making it, Sub figureout, module1.figureout, Call figureout, still not activating.
Quote from: GoSu_KaOs on January 27, 2005, 08:43 PM
Quote from: Kp on January 27, 2005, 08:31 PM
Quote from: GoSu_KaOs on January 27, 2005, 08:15 PMHelp?
Yes: write better code. I'm pretty sure even VB supports splitting text on a delimiter, so that left/mid construct isn't needed.
What does left/mid have to do with ID_JOIN.
It doesnt activate the code. I tryed making it, Sub figureout, module1.figureout, Call figureout, still not activating.
use debug.print and figure out where the problem lies.
Do you have an Error handler somewhere that may be causing it to stop execution?
Also, make a breakpoint at the top of ID_Join and step through to find the problem.
Fixed it! The problem was, I put frmMain.Figureout before the usernames were added to the Listview.
But now, when the userleaves the channel, the renamed username doesent get removed from the listview.
This is my ID_LEAVE code:
Case ID_LEAVE
For X = 1 To Form1.ChannelList.ListItems.Count
If frmMain.ChannelList.ListItems.Item(X).Text = Username Then
frmMain.ChannelList.ListItems.Remove (frmMain.ChannelList.FindItem(Username).Index)
End If
Next X
The problem is , this code compares the Username and the text in the listview. If the username is the same as the listview, it removes the text from the listview. But now, since the listview text is renamed, the code doesnt remove it because the Username and the listview text doesnt match up.
Is there a way to fix this?
I was thinking, mabe using the Split meathod I used before, but in reverse. It loads all the usernames that are stored as "username1:alias1", get the text before and after the ":", and use the text before the ":" to find the text after ":" in the listview. Then remove the text.
I cant figure out how to write this.
Quote from: Kp on January 27, 2005, 08:31 PM
Quote from: GoSu_KaOs on January 27, 2005, 08:15 PMHelp?
Yes: write better code. I'm pretty sure even VB supports splitting text on a delimiter, so that left/mid construct isn't needed.
VB5 doesn't. Have to use the Left/Mid thing.
Quote from: Adron on January 28, 2005, 03:26 AM
Quote from: Kp on January 27, 2005, 08:31 PM
Quote from: GoSu_KaOs on January 27, 2005, 08:15 PMHelp?
Yes: write better code. I'm pretty sure even VB supports splitting text on a delimiter, so that left/mid construct isn't needed.
VB5 doesn't. Have to use the Left/Mid thing.
Then the suggestion would be to use a more up-to-date language. Java 1.3 also doesn't have split(), so I update to Java 1.4 :)
I need this done in a few days, can anyone figure out how to remove the changed username from the listview?
I've been tying but all I get is errors.
Case ID_LEAVE
For X = 1 To Form1.ChannelList.ListItems.Count
If frmMain.ChannelList.ListItems.Item(X).Text = Username Then
frmMain.ChannelList.ListItems.Remove (frmMain.ChannelList.FindItem(Username).Index)
End If
Next X
One problem is you're using a 1 based counting system on a zero based index. Also, I don't see why you need to find the username again to remove it, the index x should contain it if it fulfills the IF condition. Also you should either convert everything to upper case or lower case to compare usernames.
Case ID_LEAVE
For X = 0 To Form1.ChannelList.ListItems.Count - 1
If UCase(frmMain.ChannelList.ListItems.Item(X).Text) = UCase(Username) Then
frmMain.ChannelList.ListItems.Remove (X)
X = X - 1 'Compensate for the removed item
End If
Next X
Aight.. This works..
Sub RemoveUser()
On Error GoTo error
Dim itm1 As ListItem, itm2 As ListItem
Dim str1 As String, str2 As String
For Each itm1 In frmManagers.lvAliases.ListItems 'loop the first listview
If InStr(1, itm1.Text, ":") Then 'just make sure ":" is in there
str1 = Left$(LCase(itm1.Text), InStr(1, LCase(itm1.Text), ":") - 1) 'get the leftside of ":"
str2 = Mid$(LCase(itm1.Text), InStr(1, LCase(itm1.Text), ":") + 1) 'get the rightside of ":"
For Each itm2 In Form1.ChannelList.ListItems 'loop second listview
If LCase(itm2.Text) = str2 Then 'see if it matches the string
'itm2.Text = str2 'if it does, change it
Form1.ChannelList.ListItems.Remove (Form1.ChannelList.FindItem(str2).Index)
End If
Next itm2
End If
Next itm1
error:
End Sub
BUT, If I take off the error handler, I get an error, " Control's collection has been modified." and it highlights Next itm2. But when I press play again, it sais," For loop not initialized.".
Will this make a problem or can I just leave it with the errors.
Quote from: GoSu_KaOs on January 28, 2005, 04:31 PMWill this make a problem or can I just leave it with the errors.
You should always catch and properly handle your errors.
QuoteOn Error GoTo Hell
Can someone tell me why the error is showing up ("Control's collection has been modified")?
Quote from: GoSu_KaOs on January 29, 2005, 12:39 AM
Can someone tell me why the error is showing up ("Control's collection has been modified")?
Because you're changing the control's collection while inside the For Each statement, which operates on an enumerator.
Quote from: Adron on January 28, 2005, 03:26 AM
Quote from: Kp on January 27, 2005, 08:31 PM
Quote from: GoSu_KaOs on January 27, 2005, 08:15 PMHelp?
Yes: write better code. I'm pretty sure even VB supports splitting text on a delimiter, so that left/mid construct isn't needed.
VB5 doesn't. Have to use the Left/Mid thing.
He could always MAKE one.
So how can I prevent this?
Quote from: GoSu_KaOs on January 30, 2005, 08:22 PMSo how can I prevent this?
Use only functions, not member methods. Or if you must use a member method, make it static.