I have following text
foo-11.11-fo.o-foo-bar
bar-foo-11.11-22.11
I want to remove last strings that have two "-".
Expected output:
foo-11.11-fo.o
bar-foo
I have tried multiple methods using cut, but nothing works.
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2 Answers
Using sed:
$ sed 's/-[^-]*-[^-]*$//' file
foo-11.11-fo.o
bar-foo
This will remove -X-X at the end of every line in file, where X is any string that does not include a -.
If the strings are in a shell variable:
$ s='foo-11.11-fo.o-foo-bar'
$ printf '%s\n' "${s%-*-*}"
foo-11.11-fo.o
$ s='bar-foo-11.11-22.11'
$ printf '%s\n' "${s%-*-*}"
bar-foo
or in a bash array:
$ s=( 'foo-11.11-fo.o-foo-bar' 'bar-foo-11.11-22.11' )
$ printf '%s\n' "${s[@]%-*-*}"
foo-11.11-fo.o
bar-foo
This removes anything matching the pattern -*-* at the end of the string in the variable s through a suffix pattern match/removal. In the case where s in an array, the removal is done on all elements of the array.
$ cat ip.txt
foo-11.11-fo.o-foo-bar
bar-foo-11.11-22.11
$ rev ip.txt
rab-oof-o.of-11.11-oof
11.22-11.11-oof-rab
$ rev ip.txt | cut -d- -f3- | rev
foo-11.11-fo.o
bar-foo
Reverse each line, then use cut to select all fields except first two and then reverse the output again
You can also use perl, but would print empty lines if any input line has less than 3 fields
$ perl -F'-' -lane 'print join "-", @F[0..$#F-2]' ip.txt
foo-11.11-fo.o
bar-foo
Specify - as input delimiter and then print all but last two fields
-
rev that's what I missed. thanks a lot, working great.Buvanesh Kumar– Buvanesh Kumar2017-09-25 12:18:03 +00:00Commented Sep 25, 2017 at 12:18