Display output of command executed in bash as it is executing

Question

I've got a line in my bash script like this:

echo "`some_command`"

Now, some_command executes for around 1-2mins and it keeps printing alerts/messages in that interval.

Problem is when I execute my bash script, the script has no output for 1-2mins and then directly displays the output after complete execution of some_command.

How can I make my bash script output some_command as it is executing? Is there a way to do so?

Other than just running some_command?

Ignacio Vazquez-Abrams
– Ignacio Vazquez-Abrams

2017-12-04 12:18:43 +00:00
Commented Dec 4, 2017 at 12:18 — Ignacio Vazquez-Abrams
– Ignacio Vazquez-Abrams, Commented Dec 4, 2017 at 12:18
That's a useless use of echo

tripleee
– tripleee

2017-12-04 12:19:20 +00:00
Commented Dec 4, 2017 at 12:19 — tripleee
– tripleee, Commented Dec 4, 2017 at 12:19

chepner · Accepted Answer · 2017-12-04 13:14:56Z

4

The command substitution is capturing everything your command writes to standard output, then displayed all at once by the echo command. Just run the command by itself. (And when you do want a command substitution, use $(...) instead of backticks.)

some_command  # not echo "$(some_command)"

answered Dec 4, 2017 at 13:14

community wiki

chepner

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1 Comment

JRodDynamite Over a year ago

I can't believe I didn't think of this!! I feel like an idiot! Arghhhh! :(

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