javascript – Regex Doubt

Question:

Good night.

I need a regular expression that can capture the numbers/expressions before and after a certain character, like this ^ .

For example, in the string below I need it to return me (5 * 77 + 4)^6 and 7^3 :

5! + 8 - 5(ln(3)) + (5 * 77 + 4)^6 - 7^3

The expressions I created so far were these:

• /((\()?(.)+(\))?)+(\^)+((\()?(.)+(\))?)/g

• /\(*\w+\)*(\^)+\(*\w+\)*/g

It is important that he check that there is only one ^ .

In this example there is a functional application of the second case in a string, however I am not able to apply this in javascript.

Answer:

Well, what I wanted with capturing these numbers/expressions was to simulate the action of the exponential function with the ^ sign.

I got a way to do this, but it wasn't just regex .

First, I had to capture everything before the ^ so that it syntactically fits into the exponential function. For example:

In the string below, I would capture the (2 - 9) , ignoring the whitespace, regardless of their amount:

log(10) + (2 - 9) ^ 5

And then I would have to do the same with what was after the ^ , which in the example above would be the 5 .

After that I should apply these captures in the javascript Math, pow() function Math, pow() that performs the exponentiation, according to what has already been said, it would be:

Math.pow((2 - 9), 5)

And finally, I should repeat this capture>capture>concatenation process until there is no more ^ left in the middle of the string . So I could have an exponentiation within another exponentiation:

(2 ^ 3)^7

the Javascript

To capture what was before the ^ , the code was this:

function captureBefore(string) {
  var result;
  var n = string.indexOf("^");
  var i = n - 1;
  while (!(string.charAt(i) != " ")) {
    --i;
  }
  if (!(/\d/.test(string.charAt(i)))) {
    var nOpen = 1;
    var nClose = 0;
    while (nOpen != nClose) {
      if (string.charAt(i - 1) == ")") {
        ++nOpen;
      } else if (string.charAt(i - 1) == "(") {
        ++nClose;
      }
      i -= 1;
    }
    if (/[a-z]/.test(string.charAt(i - 1))) {
      while ((/[a-zA-Z.]/.test(string.charAt(i - 1)))) {
        i -= 1;
      }
    }
    result = string.substring(i, n);
  } else {
    while (/\d/.test(string.charAt(i)) || /[-|+|.]/.test(string.charAt(i))) {
      i -= 1;
    }
    result = string.substring(i + 1, n);
  }
  return result;
}

And for what was after:

function captureAfter(string) {
  var result;
  var n = string.indexOf("^");
  var i = n + 1;
  while (string.charAt(i) == " " || /[a-zA-Z.]/.test(string.charAt(i))) {
    ++i;
  }
  if (/[-|+]/.test(string.charAt(i))) {
    ++i;
  }
  if (!(/(\d)/.test(string.charAt(i)))) {
    var nOpen = 1;
    var nClose = 0;
    while (nOpen != nClose) {
      if (string.charAt(i + 1) == "(") {
        ++nOpen;
      } else if (string.charAt(i + 1) == ")") {
        ++nClose;
      }
      i += 1;
    }
    result = (string.substring(n + 1, i + 1));
  } else {
    while (/\d/.test(string.charAt(i)) || /[-|+|.]/.test(string.charAt(i))) {
      i += 1;
    }
    result = (string.substring(n + 1, i));
  }
  return result;
}

And so that it could be repeated and the function reached all the signs:

if (/\^/g.test(string)) {
  string = string.replace(/\s*\^\s*/g, "^")
  while (/\^/g.test(string)) {
    string = string.replace(captureBefore(string) + "^" + captureAfter(string), "Math.pow(" + captureBefore(string) + ", " + captureAfter(string) + ")");
  }
}

With that I was able to execute the function successfully. But if anyone has an idea that does the same thing with less code, I'm still open to possibilities.

working example