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adding a 6 digit number retaining 0s on the left

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# 1  
Old 03-08-2007
Question adding a 6 digit number retaining 0s on the left
i am new to shell scripting. i want to keep on increamenting a 6 digit number. For eg. 000000 + 1 = 000001 But instead of 000001 i get only 1. How do i do this ? Pls help.
# 2  
Old 03-08-2007
Code:
x=$( printf "%06d" $( expr 000000 + 1 ) )

# 3  
Old 03-08-2007
to change one of the tokens of a filename
Thanks so much. It works fine Smilie
One more thing, I have a file like “uuu.1.2.090909.gz.111”.
I just want to change the 4th token, like “uuu.1.2.050505.gz.111”. How can I do in minimum steps?
# 4  
Old 03-08-2007
Code:
mv uuu.1.2.090909.gz.111 $( echo "uuu.1.2.090909.gz.111" | awk -F"." -v OFS="." ' $4 = 050505 ' )

# 5  
Old 03-08-2007
Code:
echo uuu.1.2.090909.gz.111 | sed 's/\(.*\)\.\(.*\)\.\(.*\)\.\(.*\)\.\(.*\)\.\(.*\)/\1.\2.\3.050505.\5.\6/'

Code:
echo uuu.1.2.090909.gz.111 | sed 's/9/5/g'

# 6  
Old 03-08-2007
not working
i need to use variable for the filename(uuu.1.2.090909.gz.111) and for the 4the token also (050505).it works with the actual values but when i replace with variables it doesnt.can u pls help ???
Last edited by kanchan_cp; 03-08-2007 at 09:11 AM..
# 7  
Old 03-08-2007
Code:
src=uuu.1.2.090909.gz.111
replace=050505
echo $src | sed "s/\(.*\)\.\(.*\)\.\(.*\)\.\(.*\)\.\(.*\)\.\(.*\)/\1.\2.\3.$replace.\5.\6/"
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