I keep getting this msg
VBA
Compile Error:
Label not defined
Yes, Label starts with a letter.
Label Has No spaces
Label Has Nothing but letters.
Label’s First letter is in column 1 ( per the editor in Access VBA code editor )
Label Does end with a “ : “ at end.
Not with the extra spaces or quotes.
i Did comment the line out with the Label, ( only label on the line). On a new line I retyped the same label.
i Did save, do a RefreshAll, did a close of VBA editor, then restarted VBA editor, clicked on reset, then debug got same msg.
i did this between every time i tried a different way.
I used GoTo ErrHandler
then added On Error GoTo ErrHander
I copied the label from the Goto line, then pasted it at spot I want, with first letter in column 1, added “
: “ at the end, so no way I could have misspelled the label.
I also did the same using 57
as the label, I did put label on line 57 just in case. as in goto a line number.
Every time I get the same msg.
Only thing I noticed that was different is in Tools / References I see under Available References the below
EditionUpgradeHelperLib
Location C:WindowsSystems32 EditionUpgradeHelperLib.dll
Language: Standard
That is the only reference I do NOT ever recall seeing before. If definitely was no picked before.
I did search net, only info I got came to this, download third party app(s) that is supposed to check all “dll” files , fix them, and remove them if you want to, but that doc had a LONG list of things to do before run program, and hope fixes, not make worse.
and how to reinstall the dll if it turns out you MUST have it.
Was never able to find out what that file is supposed to do.
Never before had this msg, unless I made a typo.
I do have under Available references checked
Visula Basic For Applications
Microsoft Access 16.0 Object Library
OLE Automation
Microsoft Office 16.0 Access database engine Object Library
Microsoft Office 16.0 Object Library
i thank you for any help you can give.
mark J
Mark J
Asked
7 years, 7 months ago
Viewed
38k times
I have a VBA macro which gave me that error message.
Sub Function1()
' Give the user macro options based on how fast or slow the computer
' is using advanced conditional compiling
vuserChoice = MsgBox("This macro by default treats all numbers as decimals for maximum precision. If you are running this macro on an old computer, you may want to declare numbers as singles, to speed up the macro.")
MsgBox ("Decimal: recommended for maximum precision. Also slower." & vbNewLine & "Long: not recommended. Rounds to nearest integer." & vbNewLine & "Single: not recommended. A lightweight double." & vbNewLine & "Integer: not recommended. Quick and low-precision.")
If vuserChoice = "Decimal" Or "decimal" Then
GoTo FunctionDecimal
ElseIf vuserChoice = "Double" Or "double" Then
GoTo FunctionDouble
ElseIf vuserChoice = "Single" Or "single" Then
GoTo FunctionSingle
ElseIf vuserChoice = "Long" Or "long" Then
GoTo FunctionLong
Else
GoTo FunctionNotValidVarType
End If
' MEeff = measure of efflux due to crudely purified HDL in scintillation
MsgBox "For additional information about this macro:" & vbNewLine & "1. Go to tab Developer" & vbNewLine & "2. Select Visual Basic or Macro." & vbNewLine & "See the comments or MsgBoxes (message boxes)."
End Sub
The offending line is:
GoTo FunctionNotValidVarType
I have the function FunctionNotValidVarType below this code. I have it as:
Public Sub FunctionNotValidVarType()
MsgBox "VarType " & VarType & " is not supported. Please check spelling."
End Sub
What do I need to do to let the first function recognize FunctionNotValidVarType? Thanks.
asked Jun 24, 2015 at 15:05
GoTo will try and transfer the code execution to a different position in the current Subroutine with the given label.
Specifically, GoTo FunctionNotValidVarType will try and execute the line:
FunctionNotValidVarType: 'Do stuff here
which doesn’t exist in your current code.
If you want to call another function use Call FunctionNotValidVarType
answered Jun 24, 2015 at 15:09
kaybee99kaybee99
4,4082 gold badges36 silver badges42 bronze badges
6
Remove the word GoTo
GoTo tells the code to jump to a label, you want it to enter a new procedure, not go to a label
answered Jun 24, 2015 at 15:14
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SierraOscarSierraOscar
17.4k5 gold badges41 silver badges68 bronze badges
Remove Goto from the call to your Sub()
If you really wanted to use a Goto (and you shouldn’t), you would
goto Label
Label:
where the label is defined by the trailing colon :
answered Jun 24, 2015 at 15:10
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FreeManFreeMan
5,5711 gold badge27 silver badges53 bronze badges
3
GoTo transitions to a label, a label is defined with :
For example:
Sub G()
On Error GoTo err_handling
a=1/0
Exit Sub
err_handling:
MsgBox "Holy Shit, an error occurred !"
End Sub
To apply GoTo on a Sub you need call it and exit:
Call FunctionNotValidVarType
Exit Sub
(Technically, it is not the same as GoTo if you take the call stack into consideration, but the end result is the same)
GoTo is not considered a good practice, but if that doesn’t concern you, take a look also at GoSub at the official docs.
answered Jun 24, 2015 at 16:29
Uri GorenUri Goren
13k6 gold badges56 silver badges106 bronze badges
This is the whole code:
Private Sub CommandButton1_Click()
If ListBox1.Selected(0) Then GoTo errhandler1
ListBox1.Selected(0) = False
If ListBox1.Selected(1) Then GoTo errhandler2
ListBox1.Selected(1) = False
If ListBox1.Selected(2) Then GoTo errhandler3
ListBox1.Selected(2) = False
If ListBox1.Selected(3) Then GoTo errhandler4
ListBox1.Selected(3) = False
If ListBox1.Selected(4) Then GoTo errhandler5
ListBox1.Selected(4) = False
If ListBox1.Selected(5) Then GoTo errhandler6
ListBox1.Selected(5) = False
If ListBox1.Selected(6) Then GoTo errhandler7
ListBox1.Selected(6) = False
errhandler1:
Sheet19.Range(«a7:t45»).PrintOut
Sheet21.Range(«a6:aa42»).PrintOut
Sheet22.Range(«a6:aa69»).PrintOut
Sheet23.Range(«a6:aa27»).PrintOut
Sheet24.Range(«a6:aa66»).PrintOut
Sheet25.Range(«a6:aa21»).PrintOut
Sheet26.Range(«a6:aa39»).PrintOut
Sheet27.Range(«a6:aa54»).PrintOut
Sheet28.Range(«a6:aa66»).PrintOut
Sheet29.Range(«a6:aa36»).PrintOut
Sheet30.Range(«a6:aa81»).PrintOut
Sheet31.Range(«a6:aa27»).PrintOut
End Sub
errhandler2:
Sheet19.Range(«a46:t84»).PrintOut
Sheet21.Range(«a43:aa79»).PrintOut
Sheet22.Range(«a70:aa133»).PrintOut
Sheet23.Range(«a28:aa49»).PrintOut
Sheet24.Range(«a67:aa127»).PrintOut
Sheet25.Range(«a22:aa37»).PrintOut
Sheet26.Range(«a40:aa73»).PrintOut
Sheet27.Range(«a55:aa103»).PrintOut
Sheet28.Range(«a67:aa127»).PrintOut
Sheet29.Range(«a37:aa67»).PrintOut
Sheet30.Range(«a82:aa157»).PrintOut
Sheet31.Range(«a28:aa49»).PrintOut
End Sub
errhandler3:
Sheet19.Range(«a84:t123»).PrintOut
Sheet21.Range(«a80:aa116»).PrintOut
Sheet22.Range(«a134:aa197»).PrintOut
Sheet23.Range(«a50:aa71»).PrintOut
Sheet24.Range(«a128:aa188»).PrintOut
Sheet25.Range(«a38:aa53»).PrintOut
Sheet26.Range(«a74:aa107»).PrintOut
Sheet27.Range(«a104:aa152»).PrintOut
Sheet28.Range(«a128:aa188»).PrintOut
Sheet29.Range(«a68:aa98»).PrintOut
Sheet30.Range(«a158:aa233»).PrintOut
Sheet31.Range(«a50:aa71»).PrintOut
End Sub
errhandler4:
Sheet19.Range(«a124:t162»).PrintOut
Sheet21.Range(«a117:aa153»).PrintOut
Sheet22.Range(«a198:aa261»).PrintOut
Sheet23.Range(«a72:aa93»).PrintOut
Sheet24.Range(«a189:aa249»).PrintOut
Sheet25.Range(«a54:aa69»).PrintOut
Sheet26.Range(«a108:aa141»).PrintOut
Sheet27.Range(«a153:aa201»).PrintOut
Sheet28.Range(«a189:aa249»).PrintOut
Sheet29.Range(«a99:aa129»).PrintOut
Sheet30.Range(«a234:aa309»).PrintOut
Sheet31.Range(«a72:aa93»).PrintOut
End Sub
errhandler5:
Sheet19.Range(«a163:t201»).PrintOut
Sheet21.Range(«a154:aa190»).PrintOut
Sheet22.Range(«a262:aa325»).PrintOut
Sheet23.Range(«a94:aa115»).PrintOut
Sheet24.Range(«a250:aa310»).PrintOut
Sheet25.Range(«a70:aa85»).PrintOut
Sheet26.Range(«a142:aa175»).PrintOut
Sheet27.Range(«a202:aa250»).PrintOut
Sheet28.Range(«a250:aa310»).PrintOut
Sheet29.Range(«a130:aa160»).PrintOut
Sheet30.Range(«a310:aa385»).PrintOut
Sheet31.Range(«a94:aa115»).PrintOut
End Sub
errhandler6:
Sheet19.Range(«a202:t240»).PrintOut
Sheet21.Range(«a191:aa227»).PrintOut
Sheet22.Range(«a326:aa389»).PrintOut
Sheet23.Range(«a116:aa137»).PrintOut
Sheet24.Range(«a311:aa371»).PrintOut
Sheet25.Range(«a86:aa101»).PrintOut
Sheet26.Range(«a176:aa209»).PrintOut
Sheet27.Range(«a251:aa299»).PrintOut
Sheet28.Range(«a311:aa371»).PrintOut
Sheet29.Range(«a161:aa191»).PrintOut
Sheet30.Range(«a386:aa461»).PrintOut
Sheet31.Range(«a116:aa137»).PrintOut
End Sub
errhandler7:
Sheet19.Range(«a241:t279»).PrintOut
Sheet21.Range(«a228:aa264»).PrintOut
Sheet22.Range(«a390:aa453»).PrintOut
Sheet23.Range(«a138:aa159»).PrintOut
Sheet24.Range(«a372:aa432»).PrintOut
Sheet25.Range(«a102:aa117»).PrintOut
Sheet26.Range(«a210:aa243»).PrintOut
Sheet27.Range(«a300:aa348»).PrintOut
Sheet28.Range(«a372:aa432»).PrintOut
Sheet29.Range(«a192:aa222»).PrintOut
Sheet30.Range(«a462:aa537»).PrintOut
Sheet31.Range(«a138:aa159»).PrintOut
End Sub
End Sub
-
Apr 24th, 2006, 07:15 PM
#1

Thread Starter
New Member

Label not defined error for a GoTo Statement
How do I solve a «Label not defined» for a goto Statement like below
These are the exact lines
Public Sub txtSell_KeyPress(KeyAscii As Integer)
If KeyAscii = vbKeyReturn Then
GoTo Trick1
End If
End SubHence Trick1: would be located in another part of the code.
I’d really appreciate some help.
-
Apr 24th, 2006, 07:20 PM
#2
Re: Label not defined error for a GoTo Statement
Welcome to the forums
VB Code:
Public Sub txtSell_KeyPress(KeyAscii As Integer)
If KeyAscii = vbKeyReturn Then
GoTo Trick1
End If
'Maybe more code etc....
Trick1:
End Sub
-
Apr 24th, 2006, 07:24 PM
#3
Re: Label not defined error for a GoTo Statement
A GoTo statement can only move to something within the current Sub/Function. Perhaps you want to call a different Sub?
VB Code:
Public Sub txtSell_KeyPress(KeyAscii As Integer)
If KeyAscii = vbKeyReturn Then
Trick1
End If
End Sub
Private Sub Trick1()
' Some Code
End Sub
-
Apr 24th, 2006, 07:24 PM
#4
Re: Label not defined error for a GoTo Statement
Trick1 has to be in the same procedure as the GoTo, but using GoTo like that is usually not a good idea.
-
Apr 24th, 2006, 07:30 PM
#5

Thread Starter
New Member

Re: Label not defined error for a GoTo Statement
Then, how do I send it to a line that is not within the same sub process.
What statement can I use?
-
Apr 24th, 2006, 07:32 PM
#6
Re: Label not defined error for a GoTo Statement
Use bushmobiles example if you want to call another sub routine from your current one.
-
Apr 24th, 2006, 07:33 PM
#7
Re: Label not defined error for a GoTo Statement
Your code shouldn’t be designed so that you need to jump into the middle of another sub.
-
Apr 24th, 2006, 07:35 PM
#8

Thread Starter
New Member

Re: Label not defined error for a GoTo Statement
Some example code would be nice!
-
Apr 24th, 2006, 07:36 PM
#9
Re: Label not defined error for a GoTo Statement
-
07-24-2018, 02:51 AM
#1

Registered User

Very urgent — Getting error Label not defined even after defining it.
Hi,
When I tried to compile my VBA code I’m getting an error that says «Label not defined «. The label that it is referring to is getNextRowStudentVIIandVIII.
I’m attaching the excel with the macro and vba code. Kindly help me. Its quite urgent. Thanks friends.
-
07-24-2018, 03:54 AM
#2
Re: Very urgent — Getting error Label not defined even after defining it.
If this is a label as in
then the label needs a colon ( : ) on the end of it.
Trevor Shuttleworth — Excel Aid
I dream of a better world where chickens can cross the road without having their motives questioned
‘Being unapologetic means never having to say you’re sorry’ John Cooper Clarke

-
07-24-2018, 06:08 AM
#3
Re: Very urgent — Getting error Label not defined even after defining it.
Hi !
As a good code does not need any label — except for error handling — since more than 20 years for VBA ! …
-
07-24-2018, 06:55 AM
#4

Registered User

Re: Very urgent — Getting error Label not defined even after defining it.
Hi , I have given a colon. I rechecked. Still getting the same error
-
07-24-2018, 07:10 AM
#5
Re: Very urgent — Getting error Label not defined even after defining it.
In this case, it is because you have put an «End Sub» after each of the Return statements. You will need to comment them out except for the last, the actual end of the subroutine. It will then compile. GoSub is not an approach that I use so I cannot comment on its effectiveness.
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07-24-2018, 07:17 AM
#6
Re: Very urgent — Getting error Label not defined even after defining it.
Gosub and Goto are statements from more than 30 years ago of the VBA grand pa’ : the BASIC !
But they were left with the coming of some advanced version like Quick Basic and Turbo Basic more than 20 years ago.
So since the beginning of VBA (more than 20 years) a good code does not need any Gosub neither any Goto statement …
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07-24-2018, 07:36 AM
#7
Re: Very urgent — Getting error Label not defined even after defining it.
@Marc L: I agree with you, but the OP is where the OP is, so it is perhaps more helpful to get the code to compile and, hopefully, work rather than criticising the approach. I’d probably use called subroutines or functions but, that said, I wouldn’t know where to start to address this problem … so the OP is one or two steps up on me.
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07-25-2018, 06:56 AM
#8

Registered User

Re: Very urgent — Getting error Label not defined even after defining it.
Thanks a lot friends. So many end subs !! I agree. It is compiling now. Thank you very much
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07-25-2018, 08:37 AM
#9
Re: Very urgent — Getting error Label not defined even after defining it.
You’re welcome.