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Contributor II

ltrim(replace(Individual_Item_Number,'77','')) as Dupe I used this to try and identify individual_items where they start with '77%' and return it's non 77 equivalent: Where ind_item = 12345, Dupe is 12345 as it should be. Where ind_item = 7712345, Dupe is 12345 as it should be...however, if ind_item = 1277345, it returns 12345 when I would like it to remain 1277345. I only want it to act if the two left most digits meet the '77' starting criteria.  Thoughts?

1 Solution

Accepted Solutions
MVP

## Re: Remove leading characters IF

Hi,

Try with this

if(Left(Trim(Individual_Item_Number),2)='77',Right(Trim(Individual_Item_Number),len(Trim(Individual_Item_Number))-2,Trim(Individual_Item_Number)) as Dupe

Hope it helps

Celambarasan

6 Replies
New Contributor III

I think you'd better to use the below.

If(WildMatch(Ltrim(Individual_Item_Number), '77*') > 0

, ltrim(replace(Individual_Item_Number,'77',''))

, Ltrim(Individual_Item_Number)

) as Dupe

Good luck!

Steve Kang.

Contributor II

Not quite, still strips all instances of 77.  7717708 returns 108 as opposed to 17708.

MVP

## Re: Remove leading characters IF

Hi,

Try with this

if(Left(Trim(Individual_Item_Number),2)='77',Right(Trim(Individual_Item_Number),len(Trim(Individual_Item_Number))-2,Trim(Individual_Item_Number)) as Dupe

Hope it helps

Celambarasan

New Contributor III

## Re: Remove leading characters IF

I hope it achieves what u want.

If(WildMatch(Ltrim(Individual_Item_Number), '77*') > 0

,mid(Individual_Item_Number,3),Individual_Item_Number

) as Dupe,

Goodluck

MVP & Luminary

Hi,

This works as you expect, in this we are checking the two numbers from left if it is 77 then we truncate the left 77 and fetch the remaining numbers by using Right().

=If(Left(Trim(Individual_Item_Number), 2) = '77',

Right(Trim(Individual_Item_Number), len(Trim(Individual_Item_Number))-2),

Trim(Individual_Item_Number)) as Dupe

Hope this helps you.

Regards,

Jagan.

Contributor II