Menu
I want to run some executables with the time command
time myexec -args
How can I store only the time output to a variable in bash? Thats the only part I care about for this script, not the output of the executable. Is there a way to get that value, or will I have to parse the text of the entire command?
931
Using Bash Shell's TIMEFORMAT The TIMEFORMAT is a string format that will be printed after the execution of the block code inside the time{} wrapper finishes. The %R specifies to print the elapsed time in seconds with milliseconds precision. Let's test our script: $ ./elapsed_time.sh It took 12.008 seconds.
The echo command is used to display a line of text that is passed in as an argument. This is a bash command that is mostly used in shell scripts to output status to the screen or to a file.
Again, $() is a command substitution which means that it “reassigns the output of a command or even multiple commands; it literally plugs the command output into another context” (Source).
bash [filename] runs the commands saved in a file. $@ refers to all of a shell script's command-line arguments. $1 , $2 , etc., refer to the first command-line argument, the second command-line argument, etc. Place variables in quotes if the values might have spaces in them.
See BashFAQ/032.
All output (stdout, stderr and time) captured in a variable:
var=$( { time myexec -args; } 2>&1 )
Output to stdout and stderr go to their normal places:
exec 3>&1 4>&2
var=$( { time myexec -args 1>&3 2>&4; } 2>&1 ) # Captures time only.
exec 3>&- 4>&-
If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
Donate Us With
