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Full Discussion: Shell scripting newbie problem
Top Forums Shell Programming and Scripting Shell scripting newbie problem Post 302527873 by Tibor63 on Monday 6th of June 2011 01:50:18 AM
Old 06-06-2011
Quote:
Originally Posted by agama
Looks like a few of us crossed posts. Glad you figured it out, and you should have some meaningful explanation now too.
Yes, the explanation given in your post and the one above were what I needed.

+thanks to all that helped! Smilie

---------- Post updated at 10:50 PM ---------- Previous update was at 09:31 PM ----------

Quote:
Originally Posted by agama
Easiest solution, is to cd to the desired directory and then run your loop, leaving the $d/ out of the test. And don't forget to test that the directory switch worked and issuing an error if it doesn't --- there's nothing to prevent the user from putting a regular file on the command line.

Since this seems like an assignment, I'm not offering any code.
You're right, it is an assignment, but I'm actually more interested in learning than in "just" getting it done. Even though I had it working with the code I posted earlier in post #7 (changing "for i in *" to "for i in $d/*"), I still am trying to find the most efficient way to do it.

Which is why I appreciate your alternate suggestion above of cd'ing before looping. Funny thing is that I had that thought myself hours before frustration led me to post here... but for whatever reason I didn't try to implement it, or didn't think it would work. But now that you've suggested it, it reminded me I had it, so I tried it out. It is literally as easy as taking the top loop and adding 2 extra lines ("cd $1" before the loop and "cd -" after it). Grr... a 5-second solution that I couldn't find earlier after two hours racking my brains. Oh well, at least I got it working two different ways now. Smilie Thanks again!

EDIT: and you're right I should add a conditional to check it is actually a directory... will get on that soon too!

Code:
if [ $# -eq 0 ] ; then
    d=`pwd`
    for i in * ; do
        if test -d "$d"/"$i" ; then
            echo "$i:"
        fi
    done
else
    if [ $# -eq 1 ] ; then
        cd $1
        d=`pwd`
        for i in * ; do
            if test -d "$d"/"$i" ; then
                echo "$i:"
            fi
        done
        cd -        
    fi
fi

 

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LEARN ABOUT OPENDARWIN

pwd

PWD(1)							    BSD General Commands Manual 						    PWD(1)

NAME

     pwd -- return working directory name

SYNOPSIS

     pwd [-LP]

DESCRIPTION

     The pwd utility writes the absolute pathname of the current working directory to the standard output.

     Some shells may provide a builtin pwd command which is similar or identical to this utility.  Consult the builtin(1) manual page.

     The options are as follows:

     -L      Display the logical current working directory.

     -P      Display the physical current working directory (all symbolic links resolved).

     If no options are specified, the -P option is assumed.

ENVIRONMENT

     Environment variables used by pwd:

     PWD  Logical current working directory.

DIAGNOSTICS

     The pwd utility exits 0 on success, and >0 if an error occurs.

STANDARDS

     The pwd utility conforms to IEEE Std 1003.1-2001 (``POSIX.1'').

SEE ALSO

     builtin(1), cd(1), csh(1), sh(1), getcwd(3)

BUGS

     In csh(1) the command dirs is always faster because it is built into that shell.  However, it can give a different answer in the rare case
     that the current directory or a containing directory was moved after the shell descended into it.

     The -L option does not work unless the PWD environment variable is exported by the shell.

BSD
								 February 4, 2002							       BSD
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