Grab certain pattern from string using bash

Use sed:

$ cat input | sed -e 's/^\([0-9]\+\) \+\([0-9]\+\) .*\<END=\([0-9]\+\).*/\1 \2 \3/'

You can use a perl script for this, something like:

pax> perl -ne '{
         @arr=split;
         if (@arr[7] =~ /^END=/) {
             @arr[7] =~ s/^END=//;
         } else {
             @arr[7] =~ s/^.*;END=//;
         }
         @arr[7] =~ s/;.*$//;
         printf "%s %s %s\n", @arr[0], @arr[1], @arr[7];
     }' <qq.in
2 118610455 118610566
1 859214 860180

I've formatted that script for readability but you can just as easily use the one-liner:

perl -ne '{@arr=split;if (@arr[7] =~ /^END=/) {@arr[7] =~ s/^END=//;} else {@arr[7] =~ s/^.*;END=//;} @arr[7] =~ s/;.*$//; printf "%s %s %s\n", @arr[0], @arr[1], @arr[7];}' <qq.in

The way it works is simple once you understand it. The split gives you an array of the elements on the line and you just have to modify number 7 slightly.

If it starts with END=, just get rid of that bit. Otherwise get rid if everything up to and including ;END=.

Then get rid of everything after the first ; (in the already-modified version which has the N bit of END=N at the start).

Then just print out the three desired values.

Having thought about it some more, it may be better off as something a little simpler, like:

pax> perl -ne '{
        ($a,$b,$x,$x,$x,$x,$x,$c,$x) = split;
        $c = ";$c";
        $c =~ s/^.*;END=//;
        $c =~ s/;.*$//;
        print "$a $b $c\n";
    }' <qq.in

or the equivalent one-liner:

perl -ne '{($a,$b,$x,$x,$x,$x,$x,$c,$x)=split;$c=";$c";$c=~s/^.*;END=//;$c=~s/;.*$//;print "$a $b $c\n";}' <qq.in