Cut last 7 characters of a variable


 
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# 1  
Old 12-16-2010
Cut last 7 characters of a variable

I need to cut the last seven characters of a variable length variable. The variable may be 7 characters or 70. I need to always be able to grab the last 7 characters. I looked at the cut command but it always seems to want to start at the beginning of a line, not the end and count backwards.

Anyone know how to do this with cut? Awk? Sed? something else?
# 2  
Old 12-16-2010
Code:
 
echo "${var}' | awk '{print substr($0,length($0)-7,7)}'

This User Gave Thanks to prasad111 For This Post:
# 3  
Old 12-16-2010
Code:
 
value=$(echo ${orig_value} | sed 's/.*\(.......\)$/\1/')

# 4  
Old 12-16-2010
Code:
bash --version
GNU bash, version 4.1.5(1)-release (i486-pc-linux-gnu)
echo ${Variable:(-7)}

will return the 7 last chars of Variable
# 5  
Old 12-16-2010
The awk command didn't give me the last 7 characters. It gave me the first seven characters. The sed command worked though. Thanks all
# 6  
Old 12-16-2010
here is working example with awk

Code:
echo $var | awk '{print substr($0,length($0)- 6,length($0))}'

echo 1ddddddddddddddddddddddd1111111234567 | awk '{print substr($0,length($0)- 6,length($0))}'
1234567

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