Shell Programming and Scripting

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

realised I have to use substitute but but still no joy :-(

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

Try:

Code:

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

works like a treat!

now Im trying to figure the logic behind it!

Thanks again :-)

Satnam

Code:

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

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
$

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

Thanks to all for feedback!

Regards
Satnam