Hi,
I have series of number in column L, some of it are in Text format and some not.
Can anyone help me with the VBA which converts entire row to text format along with ensuring all numbers are of & digit only?
Thanks
Hi,
I have series of number in column L, some of it are in Text format and some not.
Can anyone help me with the VBA which converts entire row to text format along with ensuring all numbers are of & digit only?
Thanks
Try this. It wasn't clear what you wanted to do about the cells that weren't 7 characters so I just put them in a message box.
![]()
Please Login or Register to view this content.
Hi,
I am sorry, If I confused you. I have L column with numbers only. It will never have characters but just numbers.
1) I want to ensure that all numbers are of 7 digit and if not then add 0 ( zeros ) in the beginning of number.
2) Convert the cell format to text so would not loose the added 0(zeros)
Please advice
thanks
Try:
![]()
Please Login or Register to view this content.
Here is the file.
I tired your code, but I am gettin error message " Cell not found"
Last edited by shiva_reshs; 09-18-2013 at 02:20 PM.
Keep the Forum clean :
1. Use [ code ] code tags [ /code ]. It keeps posts clean, easy-to-read, and maintains VBA formatting.
2. Show appreciation to those who have helped you by clicking * Add Reputation below their posts.
3. If you are happy with a solution to your problem, mark the thread as [SOLVED] using the tools at the top.
![]()
Please Login or Register to view this content.
Do you want it like tis?
![]()
Please Login or Register to view this content.
Hey stnkynts & jindon,
Both of your code works perfectly.
Jindon >> Where in your code it says that to look for column L? Though your code works, but just for my learning I wanted to know.
Thanks
Reps added.
Regards,
?![]()
Please Login or Register to view this content.
so 16 means coloum L?
So if I want to make it look for column B, then would it be column 2?
Thanks
It's not 16, l6 , L6 in upper case.
![]()
Please Login or Register to view this content.
ohh!! I am sorry
Thanks!!!
There are currently 1 users browsing this thread. (0 members and 1 guests)
Bookmarks