extract string portion from filename using sed


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# 1  
extract string portion from filename using sed

Hi All,

I posted something similar before but I now have a another problem.

I have filenames as below

TOP_TABIN240_20090323.200903231830
TOP_TABIN235_1_20090323.200903231830

i need to extract the dates as in bold. Using bash v 3.xx

Im trying to using the print sed command but not getting the result I want

source_file_issue_date=`echo $fname | sed '/_[0-9][0-9]*[.]/p'`
echo "source_file_issue_date: $source_file_issue_date"


Appreciate any ideas?!

Kind Regards
Satnam
# 2  
realised I have to use substitute but but still no joy :-(

sed 's/^.*[a-zA-Z]*[.][0-9][0-9]$//'`
# 3  
Try:

Code:
sed 's/.*_\(.*\)\..*/\1/'

# 4  
works like a treat!

now Im trying to figure the logic behind it!

Thanks again :-)

Satnam
# 5  
Code:
# ls -1 TOP*|awk 'BEGIN{FS="[_.]"}{print $(NF-1)}'

# 6  
BTW, there is no need to invoke sed to parse the filename. It can all be done within bash i.e.
Code:
$ fname="TOP_TABIN240_20090323.200903231830"
$ source_file_issue_date=$(tmp=${fname/*_}; echo ${tmp/\.*})
$ echo "source_file_issue_date: $source_file_issue_date"
source_file_issue_date: 20090323
$

# 7  
yup, I have to say this is all a learning experience!

Thanks to all for feedback!

Regards
Satnam
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