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Print/cut/grep/sed/ date yyyymmdd on the filename only.


 
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# 1  
Old 12-12-2013
Print/cut/grep/sed/ date yyyymmdd on the filename only.

I have this filename "RBD_EXTRACT_a3468_d20131118.tar.gz" and I would like print out the "yyyymmdd" only. I use this command below, but if different command like cut or print....etc. Thanks

Code:
 
ls RBD_EXTRACT* |  sed 's/.*\(........\).tar.gz$/\1/' > test.txt

# 2  
Old 12-12-2013
Using parameter substitution:
Code:
for file in RBD_EXTRACT*
do
        fname="${file##*_[a-z]}"
        printf "%s\n" "${fname%%.*}"
done

# 3  
Old 12-12-2013
Really Nice....and any different command in one line? Thanks
# 4  
Old 12-12-2013
Code:
echo "RBD_EXTRACT_a3468_d20131118.tar.gz" | awk -F'[_.]' '{d=$(NF-2);gsub(/[a-z]/,"",d);print d}'
20131118

# 5  
Old 12-12-2013
Nice...but seem not work for me. What's gusb?

Code:
ls "RBD_EXTRACT_a3468_d20131118.tar.gz" | awk -F'[_.]' '{d=$(NF-2);gsub(/[a-z]/,"",d);print d}'

awk: syntax error near line 1
awk: illegal statement near line 1
# 6  
Old 12-12-2013
You never mentioned what OS, shell, etc., that you're using.

The gsub function removes the 'd' in front of the date string by replacing it with an empty value ("").
# 7  
Old 12-12-2013
uname -a
SunOS scorpion 5.10 Generic_147440-19 sun4v sparc sun4v
 

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