Bash script - single & double digit problem.

[David Wall]

I know it's not COBOL but I tried to put it in a bash forum with nil result (couldn't get access).
I'm trying to use a variable that ranges from 1 to 99 as part of a filename in a script as follows:
1. the variable - dw=0
2. Increment it- ((dw++))
3. Now use it - filename=david$dw

I 'would have' expected david01, david 02 etc: but obviously get david1, david2 etc:
I've tried adding printf -v dw "%02d" $dw with no change
I've tried also filename=david\$(( 10#\$dw )) also with no change

I get names like david7, david8 & david9 followed by david10 so it's not thinking the variable is octal.

Any smart ideas as to how to make the variable retain the leading 0 ??.

[Simon Sobisch]

As you want a series of of files with a sequence you may want to use a loop of for dw in {01..99} or for dw in $(seq -f "%02g" 1 99), also see https://stackoverflow.com/questions/8789729/how-to-zero-pad-a-sequence-of-integers-in-bash-so-that-all-have-the-same-width

Your printf trick also should work:

sh dw=0 ((dw++)) filename=david$(printf "%02d" $dw)

[David Wall]

Yep - Thanks Simon - the last one worked.
Keeping it all in the one statement was the trick - I was doing the printf & THEN using the $dw in the filename which resulted in going back to the single digit.