I'm learning Linux scripting and trying to set up a function that finds all files in the current directory. I know I could use ls but I'm wondering if there is a way to get the current directory as a command and pass it to an argument.
#!/bin/bash check_file() { for f in $1: do echo $f done } check_file pwd This just prints out pwd:, which obviously isn't it.
PWD variable does exactly what you want. So just replace pwd with $PWD and you are done
To execute a command and use its output in your script as a parameter, just use back quotes. In your case, it would be:
check_file `pwd` A more modern variant is to use the $(...) notation (which can be nested when needed, and is expanded to the output of the command in between $( and matching )) instead of the backquotes, eg
check_file $(pwd) (but $PWD would work too since shells maintain their PWD variable, see Posix shell).
Read also the advanced bash scripting guide (which can be criticized but is a good start).
$PDW
the present working directory is stored in the variable named $PWD
check_file() { for f in "$(ls $PWD)" do echo "$f" done } check_file Display the full working directory name
echo "$PWD" Display the working directory name
basename "$PWD" Can also use the symbol *
is a short cut to a list of files in the present working directory .
echo * Loop through them :
for item in * do echo "$item" done Function looping through
check_file() { for f in * do echo "$f" done } check_file Store the present directory name in a variable and use it in a function
Var=$(basename "$PWD") func() { echo $1 } func $Var 