whats wrong with this code

nadman123 · #1 (**permalink**) 04-15-2008

ls -ld | grep $1 /etc/passwd | cut -d: -f6

i need see the content...

DukeNuke2 · #2 (**permalink**) 04-15-2008

$1 is the first argument of a command... i don't understand what you want to achieve with the above...

Franklin52 · #3 (**permalink**) 04-15-2008

If you want to ls for directories of the homedirectory of user $1:

Code:

ls -ld $(grep "$1" /etc/passwd | cut -d: -f6)

krishmaths · #4 (**permalink**) 04-15-2008

Can you please explain a bit what you are trying to achieve. What string are you searching in the passwd file?

cfajohnson · #5 (**permalink**) 04-15-2008

Quote:

Originally Posted by Franklin52

If you want to ls for directories of the homedirectory of user $1:

Code:

ls -ld $(grep "$1" /etc/passwd | cut -d: -f6)

There's no need to look at /etc/passwd or to use an external command (unless your shell doesn't have printf built in):

Code:
eval "printf '%s\n' ~$1/*/"

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