regex in bash expression

Question

I have 2 questions about regex in bash expression.

1.non-greedy mode

local temp_input=\'\"a1b\", \"d\" , \"45\"\'
if [[ $temp_input =~ \\\".*?\\\         


        

Answer 1

1. Your regular expression matches the string starting with the first quotation mark (before ab) and ending with the last quotation mark (after ef). This is greedy, even though your intention was to use a non-greedy match (*?). It seems that bash uses POSIX.2 regular expression (check your man 7 regex), which does not support a non-greedy Kleene star.

   If you want just "ab", I'd suggest a different regular expression:

   if [[ $temp_input =~ \"[^\"]*\" ]]
   

   which explicitly says that you don't want quotation marks inside your strings.

2. I don't understand what you mean. If you want to find all matches (and there are two occurrences of b here), I think you cannot do it with a single ~= match.