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April 19th, 2012, 03:36 PM

Kry56

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Shell Scripting Help

This is an assignment for a class I'm taking

You need to write a shell script that calculates the following information for the contents of a given directory:

1. The total number of directories that are in the given directory.
2. The total number of files in the given directory.
3. The number of items (files / directories) in the current directory that are readable.
4. The number of items (files / directories) in the current directory that are writable.
5. The number of items (files / directories) in the current directory that are executable.

It also has to check for the correct number of arguments (one argument) and check if the command line argument is actually a directory.

April 19th, 2012, 03:55 PM

aitken325i

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Hi Kry56 and welcome to Dev Shed.

OK - now, can you show us what you have so far ? Don't mind helping you, but not doing it for you!

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April 19th, 2012, 04:06 PM

Kry56

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Quote:

Originally Posted by aitken325i

Hi Kry56 and welcome to Dev Shed.

OK - now, can you show us what you have so far ? Don't mind helping you, but not doing it for you!

That's the thing, I have no idea where to start

April 19th, 2012, 04:31 PM

Kry56

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So far I've found that using the find command I can output how many files are in a directory, but how can I alter the command to echo "Number of files: #" as well as the number of readables etc.?

April 19th, 2012, 09:27 PM

SimonJM

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If you are going to use the find command for all of these you may want to check out the -perm operand.

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April 20th, 2012, 08:51 AM

LKBrwn_DBA

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Quote:

Originally Posted by Kry56

So far I've found that using the find command I can output how many files are in a directory, but how can I alter the command to echo "Number of files: #" as well as the number of readables etc.?

Also check out the 'ls', 'grep' and 'cut' commands, they may be of some help...

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April 20th, 2012, 09:53 AM

Lucretiuss

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I'm on the same assignment and have a question about counts.

He gives us this example:
count=`wc -w <$1`
# the value of count is assigned the number of words in file $1

But we need to count directories and files in a directory (non recursively), so this example doesn't really help other than tell me I need to use `'s. I've tried count=`ls -d` (as a logical jump from `ls -1` returning all files, but this returns nothing (I already set the cd to $1 so I wouldn't need to do <$1, correct?). Also, how do I output the value of count? $count returns all the names of the files, but #count does nothing. ("$count" returns all the files one line each and "#count" still returns nothing).

Thanks.

EDIT: Ok I found a lot of mistakes on my own thanks to this: (google ls commands unix...being blocked from posting links..) if anybody else needs help.

April 20th, 2012, 11:20 AM

LKBrwn_DBA

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Here (for starters) is count of directories:

Code:

ls -l|grep ^d|wc -l

April 20th, 2012, 11:27 AM

SimonJM

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Don't forget that if you are actually on the *nix machine you will have man pages available.

Let's walk through some basic examples ...

Code:

ls

That will list the contents of the current directory.

Code:

ls -1

That will do the same thing but will list them one per line.

Code:

ls -1p

That will list directoru contents, one per line, but will append a / character to the name of any directory.
Can you see where we are going with this? If we count the number of words (the wc -w command) we might get the right numbers - but consider files and directories that have spaces in them: suddenly we will get a wrong answer. Now, with the details being one per line we can use the wc -l command to count the number of lines - if we wish.
So, we could use the following command to list the directory contents, extract only the lines ending in / and count them:

Code:

ls -1p | grep "\$" | wc -l

A better way would be to combine the filtering and counting, which we can do with grep:

Code:

ls -1p | grep -c "\$"

If we want to count files we would need to look for lines that do not end in a \ - effectively inverting the filter:

Code:

ls -1p | grep -vc "\$"

So, we now have a number being returned. Just assign that to a variable:

Code:

dir_count=$(ls -1p | grep -c "\$")
file_count=$(ls -1p | grep -vc "\$")
echo "We found $dir_count directories and $file_count files within the current directory $(pwd)"

#10

April 20th, 2012, 01:16 PM

Lucretiuss

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Ok, I managed to make it output the number of files and directories, and now I have to check if the files are readable, writable, or executable. I'm thinking about doing something like

Code:

excount=0
for filenames in -x
do    
     excount=`expr $excount + 1`
done
     (do this twice more, but for 2 other counts for reading and writing)
echo "The number of executable files: " $excount
     (do this twice more, again for reading and writing)

2nd attempt:

Code:

excount=0
if [ -x filename ]
then
     excout=`expr $excount + 1`
elif [ for read and write...]
fi
echo "The number of executable files: " $excount
    (twice more for reading and writing)

2nd attempt returns 0 for all numbers, which is wrong.

EDIT: Trimmed it down a little, still doesn't count correctly.

#11

April 20th, 2012, 03:05 PM

LKBrwn_DBA

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This is one way to do it:

Code:

r0=0; w0=0; x0=0;
for f in `ls -l|grep -v ^d|grep -v ^total`
do
  r=`echo $f|cut -c2`
  w=`echo $f|cut -c3`
  x=`echo $f|cut -c4`
  [ "$r" = 'r' ]&& ((r0+=1))
  [ "$w" = 'w' ]&& ((w0+=1))
  [ "$x" = 'x' ]&& ((x0+=1))
done
echo "Readable: $r0, Writable: $w0, Executable: $x0"

#12

April 20th, 2012, 04:44 PM

aitken325i

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Looks like you guys (Lucretiuss and Kry56) are getting somewhere, excellent.

Out of curiosity, what class are you guys taking ?

#13

April 20th, 2012, 05:45 PM

Lucretiuss

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Intro to Unix. A lot of non CS kids take it though as it's a 1 credit hour class at 3000 level and above. I'm surprised how helpful you guys are, typically when I ask for c++ help I get "Well, your problem is in 500linesofcode.cpp".

Thanks for the help!

#14

April 20th, 2012, 05:54 PM

Lucretiuss

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Quote:

Originally Posted by LKBrwn_DBA

This is one way to do it:

Code:

r0=0; w0=0; x0=0;
for f in `ls -l|grep -v ^d|grep -v ^total`
do
  r=`echo $f|cut -c2`
  w=`echo $f|cut -c3`
  x=`echo $f|cut -c4`
  [ "$r" = 'r' ]&& ((r0+=1))
  [ "$w" = 'w' ]&& ((w0+=1))
  [ "$x" = 'x' ]&& ((x0+=1))
done
echo "Readable: $r0, Writable: $w0, Executable: $x0"

Took me a few minutes but to figure this out:

Code:

r=`echo $f | cut -c2`

Quite clever...didn't think of that.

What do the ampersands do in:

Code:

[ "$r" = 'r' ]&& ((r0+=1))

Is that just shorthand for if [ "$r" = 'r'], then r0+=1 ?

EDIT: I figured out it was just shorthand, finished the assignment. Thanks again for the help guys.

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