On Thu, 2012-11-29 at 23:33 +0000, John Horne wrote: > Hello, > > I have a bash script in which a variable is set to one or more lines of > text. What I want is to remove any lines up to and including a blank > line (or alternatively to echo all the lines after the last blank line). > There may be zero or more blank lines, and the blank lines need not be > consecutive. If there is no blank line, then all the lines should be > shown. If the last line is blank, then nothing should be shown. So for > example the variable may contain: > > ============ (the '=' are not part of the variable) > abc def > > hijk > xyz > ============ > > So in this case what is wanted is: > > ============ > hijk > xyz > ============ > > to be shown. > > I tried something like: > > echo "$XX" | sed -e '/./,/^$/d' > > but this didn't display anything. (Where XX is the variable.) > I also tried using a 'for' loop but again this displayed nothing: > > opt="" > IFS=$'\n' > for n in $XX; do test -z "$n" && opt="" || opt="$opt $n"; done > > (Echoing $opt after this shows that it contains nothing.) I'm not sure > why but even using a for loop just to show it had seen a blank line > didn't work either (using something like 'test -z "$n" && echo found'). > My understanding was that by setting IFS to a newline, then the 'for' > loop should see the blank line and just set '$n' to the null string. We > should then be able to test on that. > > Ideally what I am looking for is a snappy one line 'sed' or 'awk' > command to handle this :-) Unfortunately at the moment I seem to be > getting nowhere though, even with the 'for' loop. echo $FOO | grep . poc -- users mailing list users@xxxxxxxxxxxxxxxxxxxxxxx To unsubscribe or change subscription options: https://admin.fedoraproject.org/mailman/listinfo/users Guidelines: http://fedoraproject.org/wiki/Mailing_list_guidelines Have a question? Ask away: http://ask.fedoraproject.org
