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I've used the array formula
--> {=RIGHT(A1,LEN(A1)-MIN(IF(MID(A1,ROW($A$1:$A$100),1)<>"0",ROW($A$1:$A$100)))+1)}
However, it is too complicated.
I would like to solve with regular expressions.
Snap85.jpg
I've used the array formula
--> {=RIGHT(A1,LEN(A1)-MIN(IF(MID(A1,ROW($A$1:$A$100),1)<>"0",ROW($A$1:$A$100)))+1)}
However, it is too complicated.
I would like to solve with regular expressions.
It works on the attached. If the pattern changes it might not.
Hi Chief,
Give this a try - a little trickery, but it works.
001A 01A
002B 02B
010A 10A
100C 100C
=IF(LEFT(C7,1)="0",RIGHT(C7,LEN(C7)-1),C7)
Hope this helps,
Snuggles
Last edited by Snuggles; 11-21-2015 at 01:10 PM.
If Regular expression, should be
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Last edited by jindon; 11-21-2015 at 02:15 PM.
How about a little less horsepower?
Row\Col 001a 1a B2: =LZTrim(A2) 002b 2b 010a 10a 100c 100c
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Entia non sunt multiplicanda sine necessitate
How about =IF(LEFT(A1,2)="00",RIGHT(A1,LEN(A1)-2),IF(LEFT(A1,1)="0",RIGHT(A1,LEN(A1)-1),A1))
Hi! AB33
Thank you so much!Have a nice day.
Thank you very much!
Hi shg
Thank you so much!
Hi! Jimbo77
Why use an array formula ... 0 --> can increase
I appreciate your interest.
Hi! Snuggles
I wrote in another article...Why use an array formula ... 0 --> can increase
I appreciate your interest.
Thanks for responding to me, Chief.
I learned something for sure with this thread.
First, I had never heard the term "regular expression."
There are definitely many ways to do the same thing !
Regards,
Snuggles
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