Bash-scripting, usage of the 'for' loop

Question

I am trying to write a simple bash script that displays the contents of files.

#!/bin/bash

echo 'Input the path of a file or directory...'
read File

if [ -e $File ] && [ -f $File ] && [ -r $File ]
    then
    echo 'Displaying the contents of the file '$File
    cat $File


elif [ -d $File ] && [ -r $File ]
then
echo 'Displaying the contents of the directory '$File       
    for FILE in `ls -R $File`
        do
               cd $File/$FILE
               echo 'Displaying the contents of the file '$FILE
               cat $FILE
        done

else 
echo 'Oops... Cannot read file or directory !'
fi

The user should input a file or directory path. If the user inputs a file the program displays it with cat. If the user inputs a directory it should display the content of all files including those in the subdirectories. That part of the program doesn't work very well. I would like to get a result that doens't display errors like 'No such file or directory' but only the content of files. Can you help me ? Thanks in advance.

Never use for i in $(command) . See this answer for more details : stackoverflow.com/questions/19606864/ffmpeg-in-a-bash-pipe/… Otherwise, I think that the best solution is to use find as described in some answer of this thread. — Idriss Neumann
– Idriss Neumann, Commented Oct 28, 2013 at 21:20

evil otto · Accepted Answer · 2013-10-30 22:53:14Z

6

ls -R is the wrong tool to find all files in all subdirectories. find is a much better choice:

echo "displaying all files under $File"
find "$File" -type f -printf "Displaying contents of %p\n" -exec cat {} \;

edited Oct 30, 2013 at 22:53

answered Oct 28, 2013 at 16:41

evil otto

10.6k28 silver badges38 bronze badges

Sign up to request clarification or add additional context in comments.

5 Comments

buster Over a year ago

Thanks for your response ! The code works fine, but is there a way to print the name of the file just before displaying its content ?

chepner Over a year ago

… -exec echo {}; cat {} \;

buster Over a year ago

Thanks chepner, but I get '' find: missing argument to `-exec' ''.

evil otto Over a year ago

... -exec echo "Filename: {}" \; -exec cat {} \; --- edited answer

gniourf_gniourf Over a year ago

Don't use -exec echo ... it's silly. Use -printf "Displaying contents of %p\n" (or %f) instead.

LeartS · Accepted Answer · 2013-10-28 16:41:12Z

3

You can print all files in the current directory just doing

for f in * do
    cat $f;
done

answered Oct 28, 2013 at 16:41

LeartS

2,9162 gold badges27 silver badges46 bronze badges

3 Comments

LeartS Over a year ago

Yes but in the for loop he can add other commands (like the "debug" echo he has) easily. I just gave him the structure. Why wrong?

jlliagre Over a year ago

You are correct the loop allows more flexibility. The main issue is a for loop is the wrong approach. The OP uses ls -R which clearly means subdirectories are expected. Neither his loop nor yours handle this case properly. findis the way to go.

glenn jackman Over a year ago

You can do recursive with shopt -s nullglob globstar; for subdir in **/*/ but then it gets tedious to separate out file and directories. find is certainly easier if you just want to cat files.

SilverCode · Accepted Answer · 2013-10-28 22:25:44Z

2

the find command will save you a lot of the logic:

#!/bin/bash 

echo 'Input the path of a file or directory...'
read File
DirName="."

if  echo $File | grep '/' ;  then
  DirName=$(dirname $File)
  File=$(basename $File)
fi

find "$DirName" -type f -name "$File" -exec cat {} \;
find "$DirName" -type d -name "$File" -exec ls {}

The first find will look for all 'regular' (-type f) files name $File and cat them The second find will look for all 'directories' (-type d) and list them.

If they do not find any then the -exec portion will not execute. The grep will split the path is there is a slash in there.

edited Oct 28, 2013 at 22:25

answered Oct 28, 2013 at 16:43

SilverCode

2091 silver badge4 bronze badges

2 Comments

buster Over a year ago

Thanks for your response ! When I try to run your script I get an error ''find: warning: Unix filenames usually don't contain slashes...''.

SilverCode Over a year ago

to get around that you can split the name using basename and dirname e.g. find $(dirname $File) -name "$(basename $File)"

Collectives™ on Stack Overflow

Bash-scripting, usage of the 'for' loop

3 Answers 3

5 Comments

3 Comments

2 Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

5 Comments

3 Comments

2 Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related