How to remove a fixed starting characters

[Thanh]

Thanks for all the help. This is a great forums.

When I scan a part number it sometimes come up with 1Pxxxxxxxxx, the rest of characters real part number. The real part number can have 8 - 10 characters.

How can I remove the first 2 characters (1P) if they exixt?

Again Thanks!

[mr_vodka]

Well you can use the left, right, position functions but maybe you should supply us with more information so that we can help you in greater detail.

Is there always 2 characters in the beginning that you want to get rid of? If so is the second character always a Alpha-character? etc etc.

The more detail the better.

[Thanh]

Here's some examples:

1PP1234LPMM - will need to trim the starting 1P

1PM123456PMM9 - will need to trim the starting 1P

M123456PMM9 - No need to do any trim

P12345 - No need to do any trim

Thanks

[mr_vodka]

Ok...

so how does one determine if something needs to be trimmed or not? You still have not answered my earlier questions. We would need the rule to determine if it needs to be trimmed or not, otherwise how can we know when to trim it or not? Does it need to be trimmed only when it starts with 1P ?

[Lee Smith]

Have you tried?

Substitute(Text, "1P", "")

[Ted S]

Should work... unless the character sequence "1P" occurs elsewhere in the string and truly belongs there.

[LaRetta]

It seems pretty clear that 1P should only be removed if it exists as the first two characters. In which case, Lee, your calc would also remove 1P if it exists later in the string. This should protect from that possibility ...

If (

Left ( PartNumber ; 2 ) = "1P" ;

Substitute ( PartNumber ; Left ( PartNumber ; 2 ) ; "" ) ;

PartNumber )

LaRetta :wink2:

Edited January 6, 2006 by Guest
Corrected for vs. 7

[Lee Smith]

Darn, I missed that it was for v7, I responded to his profile of v6.

Yes, Ted, I'm aware of that too.

so now, Thanh has a couple of choices.

Lee

[LaRetta]

Darn is right, Lee. I saw version 6!! Change the commas to semi-colons for vs. 7.

Update: I corrected the above calc

Edited January 6, 2006 by Guest
Added Update

[comment]

I'm afraid your calc suffers from the same flaw as Lee's:

If (

Left ( PartNumber ; 2 ) = "1P" ;

Substitute ( PartNumber ; Left ( PartNumber ; 2 ) ; "" ) ;

PartNumber )

If the test is true, the substitution part evaluates as:

Substitute ( PartNumber ; "1P ; "" )

Suppose the PartNumber is entered as "1P55AA1P". Your result will be "55AA", instead of "55AA1P".

Try:

Case (

Left ( PartNumber ; 2 ) = "1P" ;

Right ( PartNumber ; Length ( PartNumber ) - 2 ) ;

PartNumber

)

[LaRetta]

Thank you for catching it. I missed that logic about Substitute(). :wink2:

[Thanh]

Thank You All. I really appreciated.