number of tokens in bash variable

Question

how can I know the number of tokens in a bash variable (whitespace-separated tokens) - or at least, wether it is one or there are more.

Answer 1

For a robust, portable sh solution, see @JoSo's functions using set -f.

(Simple bash-only solution for answering (only) the "Is there at least 1 whitespace?" question; note: will also match leading and trailing whitespace, unlike the awk solution below:

 [[ $v =~ [[:space:]] ]] && echo "\$v has at least 1 whitespace char."

)

Here's a robust awk-based bash solution (less efficient due to invocation of an external utility, but probably won't matter in many real-world scenarios):

# Functions - pass in a quoted variable reference as the only argument.
# Takes advantage of `awk` splitting each input line into individual tokens by
# whitespace; `NF` represents the number of tokens.
# `-v RS=$'\3'` ensures that even multiline input is treated as a single input 
# string.
countTokens() { awk -v RS=$'\3' '{print NF}' <<<"$1"; }
hasMultipleTokens() { awk -v RS=$'\3' '{if(NF>1) ec=0; else ec=1; exit ec}' <<<"$1"; }

# Example: Note the use of glob `*` to demonstrate that it is not 
# accidentally expanded.
v='I am *'

echo "\$v has $(countTokens "$v") token(s)."

if hasMultipleTokens "$v"; then
  echo "\$v has multiple tokens."
else
  echo "\$v has just 1 token."
fi