Bash exiftool command

Question

I am trying to run an if statement on a exiftool variable set in a bash script. Looking to trigger an action if the ProductVersion returns a particular version. I get the below error when running the bash script sh testing.sh

testing.sh: 4: [: exiftool: unexpected operator

Below is the script I am attempting to run:

#!/bin/bash
ExifCommand="exiftool -ProductVersion /serverdata/serverfiles/test/Server.exe"

if [ $ExifCommand =~ *3.7* ]; then
echo "testing success"
fi

Another thing to note, if I put the command in ExifCommand=$(exiftool -ProductVersion /serverdata/serverfiles/test/Server.exe) I get the below error:

testing.sh: 3: Product: not found

Answer 1

From comments:

how do I go about putting the output of that command into its own variable

The way to capture the standard output of a command is to use a command substitution, which you already demonstrate in the question: $(some command). That can be protected against word splitting by double-quoting it, and the result assigned to a variable if that's what you want. For example:

ExifOutput="$(exiftool -ProductVersion /serverdata/serverfiles/test/Server.exe)"

You show a similar attempt in the question, which could well have been foiled by the absence of quotation marks.

The old-school / POSIX way of testing for a substring in that would be via a case statement:

case ${ExifOutput} in
  *3.7*) echo "testing success" ;;
esac

Note that there, the *3.7* is a glob pattern, not a regular expression.

There is a bash-specific alternative involving its internal conditional-evaluation command, [[, and a regular-expression matching operator =~. That could be spelled like so:

if [[ "$ExifOutput" =~ .*3[.]7.* ]]; then
  echo "testing success"
fi

Note well that pathname expansion is not performed on the arguments of [[, else the regular expression would need to be quoted. Note also that it is a bona fide regular expression (POSIX flavor), not a glob pattern.