Can We Search Regexp From The Middle Of A Text Back To Beginning?

I have a text, and a 'marker' (regexp = 'error'). I can find position of the 'marker', but base target is number of article, that stands before the 'marker'. In short, I need to fi

Solution 1:

You may use

/\d{2}\/\d{2}\/\d{4}(?=(?:(?!\d{2}\/\d{2}\/\d{4})[^])*?error)/g

See the regex demo.

To match the pattern as whole word, add word boundaries:

/\b\d{2}\/\d{2}\/\d{4}\b(?=(?:(?!\b\d{2}\/\d{2}\/\d{4}\b)[^])*?\berror\b)/g

Details

• \d{2}\/\d{2}\/\d{4} - two digits, /, two digits, /, four digits
• (?=(?:(?!\d{2}\/\d{2}\/\d{4})[^])*?error) - immediately to the right from the current location, there should be a match of
  • (?:(?!\d{2}\/\d{2}\/\d{4})[^])*?- any char ([^], you may also use [\s\S]), 0 or more repetitions but as few as possible (*?), that does not start the \d{2}\/\d{2}\/\d{4} pattern described above
  • error - an error substring.

JS demo:

var text = "harum voluptatibus laboriosam blanditiis similique commodi" + 
 "labore 09/09/4567 repellat error quasi animi nostrum magnam, ab asperiores unde porro! "+
 "ipsum dolor sit amet, consectetur adipisicing elit. Velit, delectus esse aperiam quod " +
 "aliquid sunt iure ducimus. Nesciunt eveniet, possimus 09/09/4568 adipisci accusamus " + 
 "reiciendis , quos pariatur, sapiente rem quaerat cumque.\n" +
 "one 01/01/1111 two error 02/02/2222 three four 03/03/3333 five error";
var rx = /\d{2}\/\d{2}\/\d{4}(?=(?:(?!\d{2}\/\d{2}\/\d{4})[^])*?error)/g;
console.log(text.match(rx));

Solution 2:

In a comment I asked:

What two results do you want from "one 01/01/1111 two error 02/02/2222 three four 03/03/3333 five error"? Do you want 01/01/1111 and 02/02/2222, or 01/01/1111 and 03/03/3333? (Note that 'error' only appears twice in that string.)

and you answered

i need [01/01/1111, 03/03/3333]

I can't do that with a single regular expression. I tried /.*(\d\d\/\d\d\/\d\d\d\d).*?error/ but that gets just 03/03/3333.

Doing it by finding error and then looking for the nearest digits to it works:

const text = "one 01/01/1111 two error 02/02/2222 three four 03/03/3333 five error blah blah";
const rexError = /error/g;
const rexDigits = /.*(\d\d\/\d\d\/\d\d\d\d)/;
let result;
let last = 0;
while (result = rexError.exec(text)) {
  result = rexDigits.exec(text.substring(last, result.index))
  if (result) {
    console.log(result[1]);
  }
}

The .* at the beginning is what skips the first set of digits and lets the match reach the last set instead.