VBA Access code that checks if a String has “[”, “]” characters
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In the main part of my program I have this code:
If Me!Proveedor = "BIMBO" Then
Me!NombreProducto = Left(BuscaNombre("BIMBO", Mid(Me!Producto, InStr(1, Me!Producto, "[") + 1, (InStr(1, Me!Producto, "]")) - (InStr(1, Me!Producto, "[")) - 1)), 50)
Nevertheless, when the value of Me!Proveedor has not '[ ]' the code fails.
What I've tried so far is this piece of code:
Me!NombreProducto = iif(InStr(1, Me!Producto, "["), Left(BuscaNombre("BIMBO", Mid(Me!Producto, InStr(1, Me!Producto, "[") + 1, (InStr(1, Me!Producto, "]")) - (InStr(1, Me!Producto, "[")) - 1)), 50), Left(BuscaNombre("BIMBO", Me!Producto.text), 50)
But Still, if fails. Any idea of why?
vba ms-access
edited Nov 16 '18 at 13:01
Erik A
20.2k62441
add a comment |
In the main part of my program I have this code:
If Me!Proveedor = "BIMBO" Then
Me!NombreProducto = Left(BuscaNombre("BIMBO", Mid(Me!Producto, InStr(1, Me!Producto, "[") + 1, (InStr(1, Me!Producto, "]")) - (InStr(1, Me!Producto, "[")) - 1)), 50)
Nevertheless, when the value of Me!Proveedor has not '[ ]' the code fails.
What I've tried so far is this piece of code:
Me!NombreProducto = iif(InStr(1, Me!Producto, "["), Left(BuscaNombre("BIMBO", Mid(Me!Producto, InStr(1, Me!Producto, "[") + 1, (InStr(1, Me!Producto, "]")) - (InStr(1, Me!Producto, "[")) - 1)), 50), Left(BuscaNombre("BIMBO", Me!Producto.text), 50)
But Still, if fails. Any idea of why?
vba ms-access
edited Nov 16 '18 at 13:01
Erik A
20.2k62441
add a comment |
In the main part of my program I have this code:
If Me!Proveedor = "BIMBO" Then
Me!NombreProducto = Left(BuscaNombre("BIMBO", Mid(Me!Producto, InStr(1, Me!Producto, "[") + 1, (InStr(1, Me!Producto, "]")) - (InStr(1, Me!Producto, "[")) - 1)), 50)
Nevertheless, when the value of Me!Proveedor has not '[ ]' the code fails.
What I've tried so far is this piece of code:
Me!NombreProducto = iif(InStr(1, Me!Producto, "["), Left(BuscaNombre("BIMBO", Mid(Me!Producto, InStr(1, Me!Producto, "[") + 1, (InStr(1, Me!Producto, "]")) - (InStr(1, Me!Producto, "[")) - 1)), 50), Left(BuscaNombre("BIMBO", Me!Producto.text), 50)
But Still, if fails. Any idea of why?
vba ms-access
edited Nov 16 '18 at 13:01
Erik A
20.2k62441
In the main part of my program I have this code:
If Me!Proveedor = "BIMBO" Then
Me!NombreProducto = Left(BuscaNombre("BIMBO", Mid(Me!Producto, InStr(1, Me!Producto, "[") + 1, (InStr(1, Me!Producto, "]")) - (InStr(1, Me!Producto, "[")) - 1)), 50)
Nevertheless, when the value of Me!Proveedor has not '[ ]' the code fails.
What I've tried so far is this piece of code:
Me!NombreProducto = iif(InStr(1, Me!Producto, "["), Left(BuscaNombre("BIMBO", Mid(Me!Producto, InStr(1, Me!Producto, "[") + 1, (InStr(1, Me!Producto, "]")) - (InStr(1, Me!Producto, "[")) - 1)), 50), Left(BuscaNombre("BIMBO", Me!Producto.text), 50)
But Still, if fails. Any idea of why?
vba ms-access
vba ms-access
edited Nov 16 '18 at 13:01
Erik A
20.2k62441
edited Nov 16 '18 at 13:01
Erik A
20.2k62441
edited Nov 16 '18 at 13:01
Erik A
20.2k62441
edited Nov 16 '18 at 13:01
Erik A
20.2k62441
edited Nov 16 '18 at 13:01
Erik A
20.2k62441
20.2k62441
asked Nov 16 '18 at 12:50
user10423230
add a comment |
add a comment |
1 Answer
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Don't use IIF if you're returning something that's not valid for one of the conditions! Or better yet, don't use IIF in VBA at all. Just use a normal If statement.
IIF evaluates the condition, the True part and the False part regardless of if the condition is true. The True part is invalid if the condition is false in your case.
If InStr(1, Me!Producto, "[") Then
Me!NombreProducto = Left(BuscaNombre("BIMBO", Mid(Me!Producto, InStr(1, Me!Producto, "[") + 1, (InStr(1, Me!Producto, "]")) - (InStr(1, Me!Producto, "[")) - 1)), 50)
Else
Me!NombreProducto = Left(BuscaNombre("BIMBO", Me!Producto.text), 50)
End If
answered Nov 16 '18 at 13:03
Erik AErik A
20.2k62441
add a comment |
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Don't use IIF if you're returning something that's not valid for one of the conditions! Or better yet, don't use IIF in VBA at all. Just use a normal If statement.
IIF evaluates the condition, the True part and the False part regardless of if the condition is true. The True part is invalid if the condition is false in your case.
If InStr(1, Me!Producto, "[") Then
Me!NombreProducto = Left(BuscaNombre("BIMBO", Mid(Me!Producto, InStr(1, Me!Producto, "[") + 1, (InStr(1, Me!Producto, "]")) - (InStr(1, Me!Producto, "[")) - 1)), 50)
Else
Me!NombreProducto = Left(BuscaNombre("BIMBO", Me!Producto.text), 50)
End If
answered Nov 16 '18 at 13:03
Erik AErik A
20.2k62441
add a comment |
Don't use IIF if you're returning something that's not valid for one of the conditions! Or better yet, don't use IIF in VBA at all. Just use a normal If statement.
IIF evaluates the condition, the True part and the False part regardless of if the condition is true. The True part is invalid if the condition is false in your case.
If InStr(1, Me!Producto, "[") Then
Me!NombreProducto = Left(BuscaNombre("BIMBO", Mid(Me!Producto, InStr(1, Me!Producto, "[") + 1, (InStr(1, Me!Producto, "]")) - (InStr(1, Me!Producto, "[")) - 1)), 50)
Else
Me!NombreProducto = Left(BuscaNombre("BIMBO", Me!Producto.text), 50)
End If
answered Nov 16 '18 at 13:03
Erik AErik A
20.2k62441
add a comment |
Don't use IIF if you're returning something that's not valid for one of the conditions! Or better yet, don't use IIF in VBA at all. Just use a normal If statement.
IIF evaluates the condition, the True part and the False part regardless of if the condition is true. The True part is invalid if the condition is false in your case.
If InStr(1, Me!Producto, "[") Then
Me!NombreProducto = Left(BuscaNombre("BIMBO", Mid(Me!Producto, InStr(1, Me!Producto, "[") + 1, (InStr(1, Me!Producto, "]")) - (InStr(1, Me!Producto, "[")) - 1)), 50)
Else
Me!NombreProducto = Left(BuscaNombre("BIMBO", Me!Producto.text), 50)
End If
answered Nov 16 '18 at 13:03
Erik AErik A
20.2k62441
Don't use IIF if you're returning something that's not valid for one of the conditions! Or better yet, don't use IIF in VBA at all. Just use a normal If statement.
IIF evaluates the condition, the True part and the False part regardless of if the condition is true. The True part is invalid if the condition is false in your case.
If InStr(1, Me!Producto, "[") Then
Me!NombreProducto = Left(BuscaNombre("BIMBO", Mid(Me!Producto, InStr(1, Me!Producto, "[") + 1, (InStr(1, Me!Producto, "]")) - (InStr(1, Me!Producto, "[")) - 1)), 50)
Else
Me!NombreProducto = Left(BuscaNombre("BIMBO", Me!Producto.text), 50)
End If
answered Nov 16 '18 at 13:03
Erik AErik A
20.2k62441
answered Nov 16 '18 at 13:03
Erik AErik A
20.2k62441
answered Nov 16 '18 at 13:03
Erik AErik A
20.2k62441
answered Nov 16 '18 at 13:03
Erik AErik A
20.2k62441
20.2k62441
add a comment |
add a comment |
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