On Mon, Aug 19, 2013 at 1:56 PM, Mark Haney <mark.haney@xxxxxxxxx> wrote: > I've hit a problem I can't quite figure out which a bash script I'm writing. > I'm trying to copy backup files in the format 2013-August-18--1123.zip to an > NFS share. I want to have the script copy the file with just the date. In > bash I've setup vars that get the current date: > > # Date variables > log_year=`date "+%Y"` > log_month=`date "+%B"` > log_day=`date "+%d"` > > # Filename format YYYY_MM_DD--HHMM.zip > filename=$log_year"-"$log_month"-"$log_day"--"/*".zip" > > The problem is I don't really care about the stuff after the '--'. I.e. from > the CLI I'd just 'ls 2013-August-18--*.zip' to get all the files with that > date in the file name. How can I do that in a bash script? If you just want the list you can do basically the same thing you do from the CLI. ls $log_year"-"$log_month"-"$log_day"--*.zip If you want to act on each file individually then maybe a loop. Assuming your script is running in the same directory as the files something like this maybe? for f in $log_year"-"$log_month"-"$log_day"--*.zip; do // do whatever you want with $f here done John -- users mailing list users@xxxxxxxxxxxxxxxxxxxxxxx To unsubscribe or change subscription options: https://admin.fedoraproject.org/mailman/listinfo/users Fedora Code of Conduct: http://fedoraproject.org/code-of-conduct Guidelines: http://fedoraproject.org/wiki/Mailing_list_guidelines Have a question? Ask away: http://ask.fedoraproject.org
