How to pass parameter to a bash function from shell

Question

I have the following function defined in .bash_aliases

findphp () {
    find -name '*.php' -exec grep -H  --color "$@" "{}" \;
}

It allows me to run something like findphp 'function x' to find all references to function x in php files

Now I want to pass the file type to search, example js, css, ctp as a parameter in bash

I defined the follwing function and sourced .bash_aliases

myfind () {
    find -name '*.$1' -exec grep -H  --color "$2" "{}" \;
}

But it's not working, I tried single quotes, double quotes around the positional parameters with no avail.

How could I define a function that takes the file extension and search string as arguments from the command line?

Karoly Horvath · Accepted Answer · 2011-09-15 22:56:44Z

2

Single quotes prevent variable expansion. Use "*.$1"

answered Sep 15, 2011 at 22:56

Karoly Horvath

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Comments

w00t · Accepted Answer · 2011-09-16 12:47:43Z

2

Try this:

findphp () {
    find . -name "*.$1" -exec grep -H --color "${@:2}" "{}" \;
}

or

findphp () {
    find . -name "*.$1" -print0 | xargs -0 grep -H --color "${@:2}"
}

if you want the filenames too

answered Sep 16, 2011 at 12:47

w00t

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