Get date from filename

Shell Programming and Scripting
1

Hi all,
I have this files:
aaa20080714.log
bbbb20080714.log
ccccccc20080714.log

Can i get the 20080714 from each file?

2 
sed 's/^[a-z]*\([0-9]*\).log/\1/g' infile
20080714
20080714
20080714
3

ls .log | awk -v d=$(date +%Y) -F"." '{ pat="."d; gsub(pat,"",$1);print d$1}'

4

Just for the fun, and to give a good reason why one should learn awk and sed to survive in shell scripting:

ant:/tmp $ ll ccc*
-rw-r--r-- 1 vbe sys 0 Jul 15 15:26 ccccccc20080714.log
ant:/tmp $ ls|grep ccc*.log|cut -c$(echo $(expr $(ls|grep ccc*.log|wc -c) - 12)"-"$(expr $(ls|grep ccc*.log|wc -c) - 5))
20080714

5

Or... try tr

echo aaa20080714.log| tr -dc '[:digit:]'
# if the filenames are in a file
tr -dc '[:digit:]'  < infile
6

ls *.log | awk -F'.' '{print substr($1,length($1)-7)}'

7

It's working, thanks alot guys.
I'm just learning awk & sed and got this trouble. Gonna dig deeper.