I'd like to extract only a specific value from command output.
The string that the command returns is something like this:
Result: " 5 Secs (11.2345%) 60 Secs (22.3456%) 300 Secs (33.4567%)"
And I want to filter only the "60 Secs" value between ()
22.3456%
How can I do that?
If that is the exact string that the command returns, then sed will work.
command_output | sed 's/.*60 Secs..\(.*\)..300.*/\1/'
That prints everything between 60 Secs ( and ) 300.
Result:
22.3456%
. before 300. One of them represents the ) and the other represents the space.
You can do it in sed, yes. Just search for 60 Secs ( and then as many non-% characters as possible until the first %:
$ echo "5 Secs (11.2345%) 60 Secs (22.3456%) 300 Secs (33.4567%)" |
sed 's/.* 60 Secs (\([^%]*%\).*/\1/'
22.3456%
Or, in perl:
$ echo "5 Secs (11.2345%) 60 Secs (22.3456%) 300 Secs (33.4567%)" |
perl -lne '/.* 60 Secs \(([^%]*%)/; print "$1"'
22.3456%
Or, if the format is standard and you know you always want the 6th field, you can do:
$ echo "5 Secs (11.2345%) 60 Secs (22.3456%) 300 Secs (33.4567%)" |
awk '{gsub(/[()]/,""); print $6}'
22.3456%
or:
$ echo "5 Secs (11.2345%) 60 Secs (22.3456%) 300 Secs (33.4567%)" |
cut -d' ' -f6 | tr -d '()'
22.3456%
.* 60?
360 Secs, for example. The space makes sure I only match 60 and not other things that include 60.
60 Secs (22.3456%) ... some of your solutions will fail leaving the blank before "60".
sed 's/.*[[:blank:]]*60 Secs (\([^%]*%\).*/\1/'?
You can use GNU grep instead of sed:
$ foo='Result: "5 Secs (11.2345%) 60 Secs (22.3456%) 300 Secs (33.4567%)"'
$ echo "$foo" | grep -Po "(?<= 60 Secs \()[0-9\.%]*"
22.3456%
And with sed:
$ echo "$foo" | sed 's/^.* 60 Secs (\([0-9\.%]*\).*$/\1/'
22.3456%
Both commands capture the string with 0-9.% characters after "60 secs (".
With awk:
$ echo "$foo" | awk -F'[() ]' '{ print $10 }'
22.3456%
You can also use grep -P's \K to reset the start of the match:
$ echo "$result" | grep -Po '60 Secs \(\K[^)]*'
22.3456%
()after5 Secs.