Is there anyway to choose specific parts in strnig but ignore others?

user_11261622 0
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I'm trying to take only expressions like 'A==1' or 'D1 >= 2' from a string (including spaces). For example:
From - '(A == 3 AND B == 4) OR ( A==1 AND B==2)'
I expect to get: [A == 3, B == 4, A==1, B==2].
Here's my code:

let myString = '(A == 3 AND B == 4) OR ( A==1 AND B==2)';
const result = myString.match(/[a-z0-9\s]+(>|<|==|>=|<=|!=|\s)\d/gi);  
console.log(result); //result => [A == 3 ,AND B == 4,A==1 ,AND B==2]

I want my regex to take only the specific pattern of {param}{operator}{param} but with blank spaces. I tried many ways, but none was successful.
I would appreciate any help.

javascript regex
2 Answers

You might write the pattern with the case insensitive flag as:

\b[a-z0-9]+\s*(?:[=!]=|[<>]=?)\s*\d+\b

Explanation

  • \b A word boundary to prevent a partial word match
  • [a-z0-9]+ Match 1+ chars a-z or a digit 0-9
  • \s* Match optional whitspace cahrs
  • (?:[=!]=|[<>]=?) Match either =! == < > <= >=
  • \s* Match optional whitespace chars
  • \d+\b 1+ digits and a word boundary

Regex demo

For only leading uppercase chars:

\b[A-Z]+\s*(?:[=!]=|[<>]=?)\s*\d+\b


const regex = /\b[A-Z]+\s*(?:[=!]=|[<>]=?)\s*\d+\b/g;
const s = `(A == 3 AND B == 4) OR ( A==1 AND B==2)`;
console.log(s.match(regex))

Regex demo

user_54249880

Something like this would work:

[a-z\d]+\s*[=<>!]+\s*[a-z\d]

Explanation:

  • [a-z\d]+\s* - left side of comparison
  • [=<>!]+ - comparison operators
  • \s*[a-z\d] - right side of comparison

let myString = '(A == 3 AND B == 4) OR ( A==1 AND B==2)';

const result = myString.match(/[a-z\d]+\s*[=<>!]+\s*[a-z\d]/gi);

console.log(result);

user_21915720
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