Creating a Shell command to execute another command plus more

Question

I have a question regarding the creation of a utility that executes another command.

My script called notify will be placed in my /usr/local/bin directory and will do the following:

Execute the command that it was told to execute, then play a beep.

An example use case is the following:

> notify grep -r "hard_to_find_word" /some/huge/directory/

This is just an example, but could involve some other slower commands.

Essentially, notify will execute the grep, and then play a sound.

I know how to play a sound, but I do not know how to execute the provided command.

How do I execute the command that follows the call of notify?

Thank you for any input!

Also, don't abuse the word "function"; it has a specific meaning in bash, which neither of these are. — Ignacio Vazquez-Abrams
– Ignacio Vazquez-Abrams, Commented Mar 19, 2012 at 22:59
Sorry! I was a little confused about that but wasn't sure. So this would be more of a utility, correct? — Kaushik Shankar
– Kaushik Shankar, Commented Mar 19, 2012 at 23:14

Ignacio Vazquez-Abrams · Accepted Answer · 2012-03-19 22:56:33Z

5

"$@" is all the arguments properly separated.

"$@"

answered Mar 19, 2012 at 22:56

804k160 gold badges1.4k silver badges1.4k bronze badges

Sign up to request clarification or add additional context in comments.

4 Comments

Oh! That is great! I knew that $@ provides all the parameters, but I didn't know that you could just call it by itself! Thank you!

@Kaushik , play close attention to the use of quotes here -- very important.

Hmm I tried it with and without quotes and it seems to be working the same. What do the quotes do?

The quotes make sure that arguments containing whitespace are handled properly.