How to output both command and output of the command to both console and log?

Question

I wrote the following bash function:

function log {
    echo "[$(date -u +'%Y-%m-%dT%H:%M:%S.000')]: $*" | tee -ai  $logfile
}

Intending that the output of the commands will be printed to both the console and the log.

Edit: This is how I'm using the function in the script:

log ls -l

In reality, the log contains the commands and not their output, while I want the output to go both to the console and the log while adding the date and time only to the line of the command itself in the log.

How can it be done?

I think you did not provide the complete content of your log funcition — Adam Siemion
– Adam Siemion, Commented Dec 13, 2017 at 10:14
In the log, I see only the command itself but not the output and no output is going in terminal. — Itai Ganot
– Itai Ganot, Commented Dec 13, 2017 at 10:38

Würgspaß · Accepted Answer · 2017-12-13 11:32:49Z

2

I want the output to go both to the console and the log while adding the date and time only to the line of the command itself in the log

I think you ask for some function which executes and logs commands. It could look like this:

function executeAndLog {
    echo "[$(date -u +'%Y-%m-%dT%H:%M:%S.000')]: $@" | tee -ai  logfile.txt
    "$@" | tee -ai  logfile.txt
}

executeAndLog ls logfile.txt

which outputs this both to log and console

[2017-12-13T10:38:40.000]: ls logfile.txt
logfile.txt

edited Dec 13, 2017 at 11:32

answered Dec 13, 2017 at 10:40

Würgspaß

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1 Comment

tripleee Over a year ago

Using "$*" instead of "$@" is wrong unless you specifically want to lose quoting. The added call is also a useless use of echo -- you can simply "$@" | tee -ai logfile.txt (sic)

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