find -exec not working in bash script but working in terminal

Question

I am trying to write some bash script to replace a command I quite often use. Here is the code from my file test.sh

#!/bin/bash
echo -e "\n"

i=0
args[i++]=$0
for arg in $@ ; do
  args[i++]=$arg
done

where="."
what="-type f"
gcase=
str=

while getopts "d:f:F:ih" opt ; do
  case $opt in
    h)
      echo -e "This is the help of this search function."
      echo -e "\t $0 [-d <dir>] [-f|-F <pattern>] [-i] string"
      echo -e "This will output the result of"
      echo -e "\t find dir -[i]name pattern -exec grep --color -Hn[i] string {} \;"
      echo -e "Default is"
      echo -e "\t find . -type f -exec grep --color -Hn string {} \;\n"
      exit 0
      ;;
    d)
      OPTIND=$(($OPTIND-1))
      where=
      tmp=${args[$OPTIND]}
      while [[ $OPTIND -lt $# ]] && [[ "${tmp:0:1}" != "-" ]] ; do
        where="$where"" $tmp"
        OPTIND=$(($OPTIND+1))
        tmp=${args[$OPTIND]}
      done
      ;;
    F)
      what="-iname "
      what="$what""\"$OPTARG\""
      ;;
    f)
      what="-name "
      what="$what""\"$OPTARG\""
      ;;
    i)
      gcase="-i"
      ;;
    \?)
      echo "Invalide option, use option -h for help." >&2
      exit 0
      ;;
  esac
done

str=${args[$OPTIND]}

command="find $where $what -exec grep --color -Hn $gcase \"$str\" {} \;"
echo "$command"
$command

Now, from my terminal, I do ./test.sh -d auto-avoid -F "TEST*" "main" and I get

find  auto-avoid -iname "TEST*" -exec grep --color -Hn  "main" {} \;
find: missing argument to `-exec'

(auto-avoid is a directory with a small c++ program I wrote for fun.)

Then, in my terminal I copy-paste the command find auto-avoid -iname "TEST*" -exec grep --color -Hn "main" {} \; and I get

auto-avoid/test.cpp:26:int main(int argc, char **argv)

which is the expected result.

The question is: what did I miss?

For now I wrote it as an independent script to test it, but the goal is to have it as a function in my .bash_aliases.

I have found some similar topic but nothing that could help me. If you find that this is a duplicated question, I will gladly take the solution.

I am pretty sure some people will tel me to use grep -r, but I would at least want to understand why my script does not work. This is a minimal "not"-working example, I will exclude some directory from my find later.

Answer 1

Use set -x to see what the shell really tries to run:

$ command='find foo -iname "TEST*" -exec grep --color -F -Hn "main" {} \;'
$ echo "$command"
find foo -iname TEST* -exec grep --color -F -Hn main {} \;
$ set -x
$ $command
+ find foo -iname 'TEST*' -exec grep --color -F -Hn main '{}' '\;'
find: missing argument to `-exec'

Note the '\;': you're giving find a literal backslash, which isn't what it expects.

The double quotes accomplish the same function as the backslash would, escaping the semicolon so that it's taken as a character, and not a command separator.

These should be equivalent:

$ foo="something ;"
$ foo=something\ \;

Also, note that running a command line with $command is a bit hairy: if you have spaces in any of the arguments going to the resulting command (e.g. in the pathname you have in $where), they will get split. Shell arrays give a more robust way to do that.