Making quotation marks survive to a command run inside a bash script

Question

The following code saves my_string in the file my_filename and then checks whether pattern1 and pattern2 are in the file. The first is; the second isn't.

#!/bin/bash

my_string="One two three four" 
pattern1="two three"
pattern2="two four"

my_filename="my_temp_file"

safeGrepCommand() {
    typeset command_to_run="$*"
    typeset ret_code

    # echo command_to_run=$command_to_run
    eval $command_to_run
    ret_code=$?
    if [ $ret_code != 0 ]; then
        printf "Pattern %s is not in the file %s.\n" "${my_pattern}" "${my_filename}"
        exit $ret_code
    fi
}

echo $my_string > $my_filename

grep_command1="grep --quiet ${pattern1} ${my_filename}"
safeGrepCommand "$grep_command1"

grep_command2="grep --quiet ${pattern2} ${my_filename}"
safeGrepCommand "$grep_command2"

rm -f $my_filename

I am expecting to see the output

Pattern two four is not in the file my_temp_file.

Instead I see

grep: three: No such file or directory
grep: four: No such file or directory

As you'll see if you uncomment the echo line inside the function, the problem is that the quotation marks are not seen by grep.

command_to_run=grep --quiet two three my_temp_file
command_to_run=grep --quiet two four my_temp_file

How do I get bash to pass the quotation marks to grep? Alternatively, how do I specify "two three" as a regexp with [:space] and not worry about quotation marks.

Do not stick commands in strings. If you need to do this sort of thing use an array. You cannot get this to work reliably. See http://mywiki.wooledge.org/BashFAQ/050 for more details and discussion. In your case this is even simpler as you can just not use `grep_command1` at all and just pass the arguments to `safeGrepCommand` directly. (Though assigning `$*` to a variable will still bite you. Don't do that. Just use "$@" where you need the expansion (and you don't even need eval for that). — Etan Reisner, Nov 13 '14 at 16:21
possible duplicate of [Preserving escapes in bash arguments $@](http://stackoverflow.com/questions/21820240/preserving-escapes-in-bash-arguments) — rici, Nov 13 '14 at 16:24
possible duplicate of [Passing arguments to a command in Bash script with spaces](http://stackoverflow.com/questions/4753993/passing-arguments-to-a-command-in-bash-script-with-spaces) — Etan Reisner, Nov 13 '14 at 16:25
I would suggest that using `eval` is the *opposite* of "safe" — glenn jackman, Nov 13 '14 at 16:32
@rici Indeed. I picked the one with the highest vote count I could find quickly. — Etan Reisner, Nov 13 '14 at 16:34

rici · Answer 1 · 2014-11-13T18:02:05.417

Here's one way of resending arguments in a shell function:

doit_or_complain() {
  # The arguments are captured only to show how to do it. In practice,
  # I would just use "${@}" to run the command.
  local -a command=("$@")
  "${command[@]}"
  local rc=$?
  if ((rc)); then
    echo "The command failed" >> /dev/stderr
  fi
  return $rc
}

That will work with any random arguments:

doit_or_complain grep "a  b" "./my file with spaces in its name"

If you want to pass a complicated command to the shell function, you should use an array:

theCommand=(grep "a  b" "./my file with spaces in its name")
#...
doit_or_complain "${theCommand[@]}"

Note: The function in the OP used exit $rc if the command failed. That will exit the current shell, not just the function, which might be considered a tad unexpected, if not unsafe.

score 0 · Accepted Answer · answered Nov 13 '14 at 16:21

0

Wrap the pattern variable in double quotes as

grep_command1="grep --quiet \"${pattern1}\" \"${my_filename}\""
safeGrepCommand "$grep_command1"

grep_command2="grep --quiet \"${pattern2}\" \"${my_filename}\""
safeGrepCommand "$grep_command2"

answered Nov 13 '14 at 16:21

nu11p01n73R

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Thanks for your reply, but it seems something else is still wrong. If I replace "twofour" with "twothree" (i.e., with two spaces), I am expecting that the grep will fail. Yet after using the double quotes it succeeds. What is happening? – Calaf Nov 13 '14 at 16:57
@Calaf its because the multiple spaces within double quotes are treated as a single space. If you want to force double space, escape the spaces as well. Which will work – nu11p01n73R Nov 13 '14 at 17:00
1

@Calaf: this is not the way to go. Please see any of the answers using arrays in the linked duplicate questions, and/or read the bash faq http://mywiki.wooledge.org/BashFAQ/050 . (However, the problem is that you are not quoting the argument to `eval`.) – rici Nov 13 '14 at 17:01
@nu11p01n73R I had tried that, but instead of getting "Pattern twothree is not in the file my_temp_file." I got the two lines "Pattern two is not in the file my_temp_file.\nPattern three is not in the file my_temp_file." – Calaf Nov 13 '14 at 17:05
@rici I tried to make the question as brief as possible. This seems like a trivial exercise. Still, my use of a function was intended only for safety. How would you write it? – Calaf Nov 13 '14 at 17:10
@Calaf but in the script in question you havnt' set values for `my_pattern` or `my_filename` – nu11p01n73R Nov 13 '14 at 17:14
@Calaf: what safety issue are you thinking of? It seems to me that your function is less safe. Put my response in an answer. – rici Nov 13 '14 at 17:54

Making quotation marks survive to a command run inside a bash script

2 Answers2