Shell, compare two script with wildcard

I am trying to compare two "almost" identical strings in a shell script.
They are similar except for one character (A-Z).
In the example below I want the code to return true for all cases of the string from
1-2-3-A-5-6- to 1-2-3-Z-5-6-

Currently my code returns false.

#!/bin/bash
TEMPLATE='1-2-3-*-5-6-'
CASE='1-2-3-A-5-6-'
if [[ "$TEMPLATE" == "$CASE" ]]; then
  echo "It's there."
else
  echo "NO, improve!"
fi

>Solution :

The pattern 1-2-3-*-5-6- is a glob.

This pattern language is described in the Pattern Matching section of the bash manual, and available to both the [[ == ]] test you’re trying to use (manual quoted below), and the case version mentioned in a comment.

[[ expression ]]

… When the == and != operators are used, the string to the right of the operator is considered a pattern and matched according to
the rules
described below under Pattern Matching, as if the extglob shell option were enabled. The = operator is equivalent to
==. If the nocase‐
match shell option is enabled, the match is performed without regard to the case of alphabetic characters.

… Any part of the pattern may be quoted to force the
quoted portion to be matched as a string.

There are only two constraints:

1. you can’t quote the pattern if you want it to be matched as a pattern rather than a literal string

   (see the last quoted paragraph)

2. the pattern must go on the right hand side of the operator

   (first paragraph)

So your code will work fine as

#!/bin/bash
TEMPLATE='1-2-3-*-5-6-'
CASE='1-2-3-A-5-6-'
if [[ $CASE == $TEMPLATE ]]; then
  echo "It's there."
else
  echo "NO, improve!"
fi

NB. Part of the reason for using the built-in [[ ]] conditions instead of the old [ ] ones is specifically that

Word splitting and pathname expansion are not performed on the words between the [[ and ]];

so you don’t generally need to quote variables (and in this particular case, must not quote at least one of them).