使用正则表达式匹配括号顺序和条件

Question

I wanted to apply the regular expressions to find and replace the unwanted parenthesis and operators in the input string.

Here is the possible input for me: 4 types from a through d. [Invalid Inputs]

a). 1 and (2 or 3) ()
b). ( and 2)
c). (or 4)
d). ()

all these 4 are invalid cases , the valid ones should be as [Valid Inputs]

a). 1 and 2
b). (1 and 2)
c). 1 and (2 or 4)

Based on this requirement, i have written the regex, but i have written in 2 parts and need help in joining them to a single regex.

a). ([(]+[\s]*[)]+) -> to find the empty parenthesis
b). (([(]+[\s]*[and|or]+[\s]*)) -> to find cases like b or c in invalid inputs.

Kindly suggest a way to combine the above. further i want to do removal of the invalid parts in the inputs, which i can do in javascript like string.replace(regex).

Kindly analyze and give comments on this process.

Answer 1

/\((\s*|\s*(?:and|or)\s*\d+\s*|\s*\d+\s*(?:and|or)\s*|\s*(?:and|or)\s*)\)/

is a Regex that checks for the content of a bracket pair: either empty, missing operand on the left, missing operand on the right or no operands at all.

But watch out! This neither checks for validity of un-bracketed expressions, nor is it rekursive, as Colin Fine already mentioned. If you liked to check for that, I'd propose to replace from inside up:

var s = string;
var oneoperator = /^\s*\d+\s*(and|or)\s*\d+\s*$/;
while (true) {
    s = s.replace(/\(([^)])\)/, function(all, inner) {
        if (inner.match(oneoperator)
            return "0"; // or any other valid operand
        else
            throw new SyntaxError("Math Syntax mismatch");
    });
    if (s.match(oneoperator))
        break; // return true
}
// to be improved