Pass wildcards as parameters.

[TheJanitor]

Hey, so i have a function in my bashrc that i use to shred files recursively. it's something like this.

find "$1" -name "$2" -exec shred -uzvn 6 '{}' \;

it works fine if i pass it parameters like:

$ function ~/ foo*

but when i pass it say a parameter like

$ function ~/ *

it will lift a file name out of my current working directory and pass it as a parameter. so essentially, it tries to find that erroneous file name in the directory specified. how do i prevent this without using any special input?

EDIT: it appears if i pass the parameter as a literal, ie with 'single quotes' all works fine, but is there any way to achieve this solely inside the script?

thanks.

[JohnGraham]

EDIT: it appears if i pass the parameter as a literal, ie with 'single quotes' all works fine, but is there any way to achieve this solely inside the script?

No - the * token is expanded by the shell before your program gets hold of it, so you cannot see it. Furthermore, you probably don't really want to do this - as you've found out, there's a way to explicitly bypass this behaviour, and subverting what users expect the shell to do normally will only lead to confusion.

[TheJanitor]

yeah thanks, it was a stupid question really. thank you.