Simple Shell Scripting

Question

An argument example:

../path/cse/lab3/remove

Right now, it's printing out all the directory and files in 'lab3'. I want it to print out all the files in 'remove'. I'm not sure how to do that. I want to use a for loop.

Code:

if test -d $1    #check if argument is a directory.
then

    for fileName in *
    do  
        echo "what is this::: $fileName"

    done

fi

Also how do I print out all the files as strings to look something like this:

remove/test.out
remove/test2.out

See the basename command for the formatting.

Some programmer dude
– Some programmer dude

2012-11-26 02:57:57 +00:00
Commented Nov 26, 2012 at 2:57 — Some programmer dude
– Some programmer dude, Commented Nov 26, 2012 at 2:57

Alexey Feldgendler · Accepted Answer · 2012-11-26 02:54:31Z

2

for fileName in $1/*

...is probably what you mean.

answered Nov 26, 2012 at 2:54

Alexey Feldgendler

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