Test whether a string contains a literal asterisk

chepner's helpful answer explains the problem with your original approach well.

The following prompts until a literal * is found anywhere in the user input:

until [[ $string == *'*'* ]]; do  echo "just give me a string with an asterisk, for crying out loud"  read -r stringdone

• [[ ... ]] rather than [ ... ] must be used in order to enable pattern matching on the RHS of == (=).

  • Unless you're writing POSIX-compliant scripts (for /bin/sh), where only [ ... ] is supported, consider using [[ ... ]] routinely, because it offers more features and fewer surprises: see this answer of mine.

• *'*'* is a pattern that says: match a literal * - the single-quoted part - anywhere in the string; the unquoted * instances represent any run of characters.

  • These patterns are the same as the ones used in filename globbing and are not regular expressions.
  • However, Bash also offers regular expression matching, via [[ ... =~ ... ]]. Thus, the conditional could also be written as [[ $string =~ '*' ]] or [[ $string =~ \* ]]

• -r should be used with read as a matter of habit, to prevent unexpected processing of \ chars.

@mklement's answer actually provides a solution; I am primarily describing the problem.

You aren't comparing the value of the parameter string; you are comparing the literal string string. This has nothing to do with globs (you are otherwise quoting things properly.

until [ "$string" = "[asteriskstuff]" ]; do  read stringdone

You don't need to quote the argument to read, since it is a literal string that needs to be a valid identifier, which can only consist of _, letters, and numbers. Quoting it doesn't hurt, though.