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Access VBA exit a called sub and skip the rest of the sub



2019 Community Moderator ElectionHow Can You Delete All Color Categories?Excel Macro (save to Sharepoint Library) no longer accepting full stop?Excel VBA ignores my IF statement to exit the subExcel Find a sheet based on nameKeep Excel sub runing after calling access sub procedureHow to stop a sub which calls upon multiple subsShow only selected table column after filter to new worksheetsExit all subs in that buttonHow can I go out from a sub when another sub that is inside this sub outputs a Msgbox?Error Handling - Where to enter 'Exit Sub'?










0















I have a sub CreaNewT as below to create a new table. In CreatNewT I call a public sub ChecTabl to check if this table exists already.



Sub CreaNewT()

Dim ...
Dim ...

Call ChecTabl("TableName")
...


In ChecTabl I have



Dim TS As TableDefs
Dim T As TableDef

Set TS = CurrentDb.TableDefs

For Each T In TS
If T.Name = Str_Tabl Then
MsgBox "This table already exists. Please choose another table name.", vbOKOnly
Exit Sub
End If
Next


I wrote Exit Sub because I would like to exit ChecTabl and the sub that calls it if this table already exists. However, it only stops executing ChecTabl and continues with the rest of CreaNewT. How I can code so that it stops executing ChecTabl and the sub that calls it? Thanks










share|improve this question

















  • 1





    Make ChecTabl a function and return False if the table already exists.

    – Tim Williams
    Mar 7 at 5:30















0















I have a sub CreaNewT as below to create a new table. In CreatNewT I call a public sub ChecTabl to check if this table exists already.



Sub CreaNewT()

Dim ...
Dim ...

Call ChecTabl("TableName")
...


In ChecTabl I have



Dim TS As TableDefs
Dim T As TableDef

Set TS = CurrentDb.TableDefs

For Each T In TS
If T.Name = Str_Tabl Then
MsgBox "This table already exists. Please choose another table name.", vbOKOnly
Exit Sub
End If
Next


I wrote Exit Sub because I would like to exit ChecTabl and the sub that calls it if this table already exists. However, it only stops executing ChecTabl and continues with the rest of CreaNewT. How I can code so that it stops executing ChecTabl and the sub that calls it? Thanks










share|improve this question

















  • 1





    Make ChecTabl a function and return False if the table already exists.

    – Tim Williams
    Mar 7 at 5:30













0












0








0








I have a sub CreaNewT as below to create a new table. In CreatNewT I call a public sub ChecTabl to check if this table exists already.



Sub CreaNewT()

Dim ...
Dim ...

Call ChecTabl("TableName")
...


In ChecTabl I have



Dim TS As TableDefs
Dim T As TableDef

Set TS = CurrentDb.TableDefs

For Each T In TS
If T.Name = Str_Tabl Then
MsgBox "This table already exists. Please choose another table name.", vbOKOnly
Exit Sub
End If
Next


I wrote Exit Sub because I would like to exit ChecTabl and the sub that calls it if this table already exists. However, it only stops executing ChecTabl and continues with the rest of CreaNewT. How I can code so that it stops executing ChecTabl and the sub that calls it? Thanks










share|improve this question














I have a sub CreaNewT as below to create a new table. In CreatNewT I call a public sub ChecTabl to check if this table exists already.



Sub CreaNewT()

Dim ...
Dim ...

Call ChecTabl("TableName")
...


In ChecTabl I have



Dim TS As TableDefs
Dim T As TableDef

Set TS = CurrentDb.TableDefs

For Each T In TS
If T.Name = Str_Tabl Then
MsgBox "This table already exists. Please choose another table name.", vbOKOnly
Exit Sub
End If
Next


I wrote Exit Sub because I would like to exit ChecTabl and the sub that calls it if this table already exists. However, it only stops executing ChecTabl and continues with the rest of CreaNewT. How I can code so that it stops executing ChecTabl and the sub that calls it? Thanks







vba exit






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Mar 7 at 5:17









davidzxc574davidzxc574

679




679







  • 1





    Make ChecTabl a function and return False if the table already exists.

    – Tim Williams
    Mar 7 at 5:30












  • 1





    Make ChecTabl a function and return False if the table already exists.

    – Tim Williams
    Mar 7 at 5:30







1




1





Make ChecTabl a function and return False if the table already exists.

– Tim Williams
Mar 7 at 5:30





Make ChecTabl a function and return False if the table already exists.

– Tim Williams
Mar 7 at 5:30












1 Answer
1






active

oldest

votes


















0














Thanks Tim, I changed ChecTabl into a boolean function. I can check its returned value and decide whether to exit the sub that calls this function.



Public Function ChecTabl(Str_Tabl As String) As Boolean

Dim TS As TableDefs
Dim T As TableDef

Set TS = CurrentDb.TableDefs

For Each T In TS
If T.Name = Str_Tabl Then
MsgBox "This table already exists. Please choose another table name.", vbOKOnly
ChecTabl = True
Exit Function
End If
Next

ChecTabl = False

Exit_Func:
Set TS = Nothing
Set T = Nothing

End Function





share|improve this answer






















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    0














    Thanks Tim, I changed ChecTabl into a boolean function. I can check its returned value and decide whether to exit the sub that calls this function.



    Public Function ChecTabl(Str_Tabl As String) As Boolean

    Dim TS As TableDefs
    Dim T As TableDef

    Set TS = CurrentDb.TableDefs

    For Each T In TS
    If T.Name = Str_Tabl Then
    MsgBox "This table already exists. Please choose another table name.", vbOKOnly
    ChecTabl = True
    Exit Function
    End If
    Next

    ChecTabl = False

    Exit_Func:
    Set TS = Nothing
    Set T = Nothing

    End Function





    share|improve this answer



























      0














      Thanks Tim, I changed ChecTabl into a boolean function. I can check its returned value and decide whether to exit the sub that calls this function.



      Public Function ChecTabl(Str_Tabl As String) As Boolean

      Dim TS As TableDefs
      Dim T As TableDef

      Set TS = CurrentDb.TableDefs

      For Each T In TS
      If T.Name = Str_Tabl Then
      MsgBox "This table already exists. Please choose another table name.", vbOKOnly
      ChecTabl = True
      Exit Function
      End If
      Next

      ChecTabl = False

      Exit_Func:
      Set TS = Nothing
      Set T = Nothing

      End Function





      share|improve this answer

























        0












        0








        0







        Thanks Tim, I changed ChecTabl into a boolean function. I can check its returned value and decide whether to exit the sub that calls this function.



        Public Function ChecTabl(Str_Tabl As String) As Boolean

        Dim TS As TableDefs
        Dim T As TableDef

        Set TS = CurrentDb.TableDefs

        For Each T In TS
        If T.Name = Str_Tabl Then
        MsgBox "This table already exists. Please choose another table name.", vbOKOnly
        ChecTabl = True
        Exit Function
        End If
        Next

        ChecTabl = False

        Exit_Func:
        Set TS = Nothing
        Set T = Nothing

        End Function





        share|improve this answer













        Thanks Tim, I changed ChecTabl into a boolean function. I can check its returned value and decide whether to exit the sub that calls this function.



        Public Function ChecTabl(Str_Tabl As String) As Boolean

        Dim TS As TableDefs
        Dim T As TableDef

        Set TS = CurrentDb.TableDefs

        For Each T In TS
        If T.Name = Str_Tabl Then
        MsgBox "This table already exists. Please choose another table name.", vbOKOnly
        ChecTabl = True
        Exit Function
        End If
        Next

        ChecTabl = False

        Exit_Func:
        Set TS = Nothing
        Set T = Nothing

        End Function






        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Mar 8 at 2:31









        davidzxc574davidzxc574

        679




        679





























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