get values from 'time' command via bash script [duplicate]

All output (stdout, stderr and time) captured in a variable:

var=$( { time myexec -args; } 2>&1 )

Output to stdout and stderr go to their normal places:

exec 3>&1 4>&2var=$( { time myexec -args 1>&3 2>&4; } 2>&1 )  # Captures time only.exec 3>&- 4>&-

bash shell scripting time

Something like this?

TIME="$(sh -c "time myexec -args &> /dev/null" 2>&1)"

bash shell scripting time

Probably closer to what I was looking for

CodeHunter