Testing a string containing only spaces (tabs, or “ ”)?

Question

My code below doesn't work:

stringZ="    "

if [[ "$stringZ" == ^[[:blank:]][[:blank:]]*$ ]];then
  echo  string is  blank
else
  echo string is not blank
fi 

Result:

string is not blank   # wrong

How can I test this?

Answer 1

No need for bash specific code:

case $string in
  (*[^[:blank:]]*) echo "string is not blank";;
  ("") echo "string is empty";;
  (*) echo "string is blank"
esac

Answer 2

From man bash:

An additional binary operator, =~, is available, with the same precedence as == and !=. When it is used, the string to the right of the operator is considered an extended regular expression and matched accordingly (as in regex(3)).

So the regular expression matching operator should be =~:

if [[ "$stringZ" =~ ^[[:blank:]][[:blank:]]*$ ]];then
  echo  string is  blank
else
  echo string is not blank
fi

You can reduce the verbosity of the regular expression by using + quantifier (meaning previous entity 1 or more times):

if [[ "$stringZ" =~ ^[[:blank:]]+$ ]]; then