find problem in bash script

Question

I am writing a program printing the file and directory of a specify directory. But I have to ignore certain kind of file. We store them in .ignorefile.

So, to get just the file that I want, I use find function. But, I have the same error every time: No such file or directory.

And, when I try on my terminal it works perfectly with the same arguments.

An example :

In my ignore file I have: *.txt, *.cpp
When I execute my .sh file I put two arguments: the ignore file and a directory.
Then, when I scan all text in my file to construct the argument files then I call my store() function.

Here you can see my code:

    function store(){
      dir="$1" 
      find_arg="$2" # Dir ! -name '*.txt' ! -name '*.cpp' -print
      echo $find_arg
      cmd=`find "$find_arg"`
      echo $cmd
    }

Here the function which build find_args :

function backup()
{
  if [ $1="-i" ] && [ -f $2 ]
  then
    backignore=$2
    if [ $3="-d" ] && [ -d $4 ]
    then
      directory=$4
      find_arg=" $directory"
      while IFS='' read -r line || [[ -n "$line" ]]; do
        find_arg+=" ! -name "
        find_arg+="'"
        find_arg+=$line
        find_arg+="'"
      done < "$backignore"
      find_arg+=" -print"
      store $directory "$find_arg"
    else
      echo "please entrer a right directory"
    fi
  else
    echo "Please enter a right ignore file"
  fi
}

backup $1 $2 $3 $4

I call "sh" file

./file.sh -i ignorefile -d Source

output:

Source ! -name '\*.cpp' ! -name '\*.txt' -print
Source/unix_backup/example_files/backignore   
Source/unix_backup/example_files/Documents/rep1/file3.txt
Source/unix_backup/example_files/Documents/rep1/file4.cpp   
Source/unix_backup/example_files/Documents/rep3 
Source/unix_backup/example_files/Documents/rep3/rep4 
Source/unix_backup/example_files/Documents/rep3/rep4/file7.txt    
Source/unix_backup/example_files/Documents/rep3/rep4/file8.cpp

$dir isn't being used in the find command. try find "$dir" $find_arg. don't use quotes around $find_arg because you want it to add multiple args to the command line, not just one quoted arg. — cas
– cas, Commented Nov 29, 2015 at 11:48
Yes actually i don't use $dir because the directory is already in $find_arg. So now it works but it doesn't exclude *.cpp or *.txt file. I wrote cmd=`find $find_arg` — Alexis
– Alexis, Commented Nov 29, 2015 at 12:05
@l0b0: the output : Source ! -name '*.cpp' ! -name '*.txt' -print Source/unix_backup/example_files/backignore Source/unix_backup/example_files/Documents/rep1/file3.txt Source/unix_backup/example_files/Documents/rep1/file4.cpp Source/unix_backup/example_files/Documents/rep3 Source/unix_backup/example_files/Documents/rep3/rep4 Source/unix_backup/example_files/Documents/rep3/rep4/file7.txt Source/unix_backup/example_files/Documents/rep3/rep4/file8.cpp — Alexis
– Alexis, Commented Nov 29, 2015 at 13:34
@Alexis Please include the output in your question rather than a comment. It's not readable like this. — l0b0
– l0b0, Commented Nov 29, 2015 at 13:49

l0b0 · Accepted Answer · 2015-11-29 12:18:05Z

2

find_arg is passed as a single argument (because of the double quotes), which means find is trying to find all files in a directory called Dir ! -name '*.txt' ! -name '*.cpp' -print (Yes, that's a valid directory name). Try printf '%q\n' "$find_arg" to see the actual arguments as passed to find. Also, if your shell supports arrays, you should probably use them to store arguments. And Use More Quotes™! And don't use eval here - find will print the file names just fine.

edited Nov 29, 2015 at 12:18

answered Nov 29, 2015 at 12:12

l0b0

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Add a comment |

cas · Accepted Answer · 2015-11-29 20:57:39Z

You say that your .ignorefile has*.txt, *.cpp but from the output shown, it looks like it has \*.txt, \*.cpp. Escaping the * is the source of your problem, it tells find to exclude files with a literal asterisk, rather than use it as a glob wildcard. So, don't do that.
This version of your script fixes a few flaws and simplifies it a bit. Amongst other things, it separates the arg processing (and uses getopts, which is built-in to sh) from the main logic of the backup() function.

function store() {
    echo find "$@"
    find "$@"
}   

function backup() {
  directory="$1"
  backignore="$2"

  # skip comments and blank lines
  sed -e 's/#.*// ; /^[[:space:]]*$/d' "$backignore" |
      while IFS= read -r line ; do
          find_arg+=" ! -name '$line'"
      done 
  find_arg+=" -print"
  store "$directory" $find_arg
} 

usage() { 
    echo "$(basename $0) [ -d directory ] [ -i ignorefile ]"
    exit 1
}   

directory=''
backignore='.ignorefile'    # default

while getopts "i:d:h" opt ; do
  case "$opt" in
    d)  directory="$OPTARG" ;;
    i)  backignore="$OPTARG" ;;
    *) usage ;;
  esac 
done

[ -z "$directory" ] && echo "Please provide a directory." && usage

backup "$directory" "$backignore"

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