How can I substitute a quote in bash script

Question

I have this bash script

 #!/bin/bash
 find . -type f > /home/wschrabi/filenames
 while read filename; do
       stripped="$(printf '%s\n' "$filename" | tr -d -C '[[:alnum:]][[:space:]][!\"\#\$\%\&\(\)\*\+\,\-\.\/\:\;\<\=\>\?\@\[\\\]\^\_`{|}~]')";
       ohne="$(sed -e 's/[\d126-\d255\"*:<>\?\\\|]/_/g' <<<$filename)";
       test "$filename" = "$stripped" || printf "mv '%s' '%s'\n " "${filename//[\']/'\\''}" "${ohne//[\'\"\`]/_}"; 
 done < /home/wschrabi/filenames

and at the part ${filename//[\']/'\\''} I would like to substitute a single quote in the filename. The purpose should be that when I pipe the output in a bash script I would like to rename all filenames automatically. But the single quotes make problems.

Thanks a lot for any advices. Walter

EDIT: Here is for completion my chekc NTFS for Filename script.

     #!/bin/bash
     #
     # Quote a string, wrapping it in tick marks ('), and escaping any
     # embedded ticks.
     #
     # Inputs:
     #   $1: The string to quote
     #
     # Returns:
     #   (stdout) - The quoted string
     #
     quote() {
         if [ $(echo "$@" | tr -d "\n" | wc -c) -eq 0 ]; then
             echo "''"
         elif [ $(echo "$@" | tr -d "[a-z][A-Z][0-9]:,.=~_/\n-" | wc -c) -gt 0 ]; then
             echo "$@" | sed -e "s/'/\'\"\'\"\'/g" | sed -e "s/^/'/g" -e "s/$/'/g"
         else
             echo "$@"
         fi
     }

     if [ $# -eq 0 ]
       then echo "No arguments supplied: USAGE: checkNTFS_filenames.sh <file_to_check>" 
     fi

     if [ $# -eq 1 ]
     then

     #find . -type f > /home/wschrabi/filenames2
     while read filename; do
           stripped="$(printf '%s\n' "$filename" | tr -d  '\"\*\:\<\>\?\\\|')";
           ohne="$(sed -e 's/[\d126-\d255\"*:<>\?\\\|]/_/g' <<<$filename)";
           if [ "$filename" != "$stripped" ]  ;
           then  printf "mv %s %s\n" "$(quote "$filename")" "$(quote "$stripped")" 
           fi
     done < $1
     fi

score 1 · Accepted Answer · answered Nov 17 '16 at 16:54

If you put your filenames between tick marks, escaping characters with a backslash won't work. There is another way to do it though. Here's a quote function I've used in the past that might be useful to you.

#
# Quote a string, wrapping it in tick marks ('), and escaping any
# embedded ticks.
#
# Inputs:
#   $1: The string to quote
#
# Returns:
#   (stdout) - The quoted string
#
quote() {
    if [ $(echo "$@" | tr -d "\n" | wc -c) -eq 0 ]; then
        echo "''"
    elif [ $(echo "$@" | tr -d "[a-z][A-Z][0-9]:,.=~_/\n-" | wc -c) -gt 0 ]; then
        echo "$@" | sed -e "s/'/\'\"\'\"\'/g" | sed -e "s/^/'/g" -e "s/$/'/g"
    else
        echo "$@"
    fi
}

When you run it you get outputs like these.

$ quote abc
abc
$ quote "abc'def"
'abc'"'"'def'
$ quote \"abc\"
'"abc"'
$ quote "abc def"
'abc def'

The idea here is that you can replace the final printf in your script with this:

printf "mv %s %s\n" "$(quote "$filename")" "$(quote "$ohne")"

Using a filename of abc'def with your script with the above change I got this as my output.

mv 'abc'"'"'def' abcdef

and that successfully renamed the file when executed.

MariusMatutiae · Answer 2 · 2016-11-18T10:39:50.797

Virtex answer is of course perfectly correct. I would just like to point out that this kind of operation is easier to carry out with Bash arrays: just to make it clear, let us assume we wish to strip the initial characters ./ which find leaves on the file names in this simple command:

find -maxdepth 1 -type f

This can be done as follows:

#!/bin/bash

declare -a mylist
mylist=(`find -maxdepth 1 -type f `)
mylist=(${mylist[@]/\.\//})
echo "${mylist[@]}"

Bash search and replace works on arrays as one would hope: by modifying the array elements one by one. This can be modified to include your example.

EDIT:

what if I have millions of files?

How about 10 millions?

$ time /bin/bash -c 'unset array &&  declare -a array && array=($(seq 1 10000000)) && echo "${array[6]}"; unset array'
7

real    0m11.110s
user    0m10.412s
sys     0m0.724s

$ inxi -C
CPU:       Dual core Intel Core i7-5500U (-HT-MCP-) cache: 4096 KB 
           clock speeds: max: 3000 MHz 1: 2509 MHz 2: 2471 MHz 3: 2631 MHz 4: 2471 MHz

How can I substitute a quote in bash script

2 Answers2