Is there a way to retrieve the groups of match/matchAll similar to accessing an array?

Question

Is there a way to retrieve the groups of match/matchAll similar to accessing an array?

My goal is to convert a version string to a number using the following code:

function convertVersionToNumber(line) {
  const groups = line.matchAll(/^# ([0-9]).([0-9][0-9]).([0-9][0-9])\s*/g);

  return parseInt(groups[1] + groups[2] + groups[3]);
}

convertVersionToNumber("# 1.03.00")

I encountered an issue where groups is an

IterableIterator<RegExpMatchArray>

and not an array. I tried using Array.from, but it didn't work as expected. Is there a simple way to convert groups into an array, ideally in a concise manner?

The API for

IterableIterator<RegExpMatchArray>

seems inconvenient, especially when trying to skip the first element in a for...of loop. I am looking for a more concise solution to handle this without adding excessive lines of code. I am working with TypeScript, so any syntactic sugar that can simplify this task would be greatly appreciated.

javascript regex typescript

Answer 1

Answer №1

1) matchAll function will return an Iterator object

Iterator [RegExp String Iterator]

result will hold the Iterator and by using the spread operator, all matches can be accessed. In this case, as there is only one match, the result will contain a single element.

[ '# 1.03.00', '1', '03', '00', index: 0, input: '# 1.03.00', groups: undefined ]

Lastly, a spread operator is utilized to retrieve all values and encapsulate them in an array

[...result]

function convertVersionToNumber(line) {
  const result = line.matchAll(/^# ([0-9]).([0-9][0-9]).([0-9][0-9])\s*/g);
  const groups = [...result][0];
  return parseInt(groups[1] + groups[2] + groups[3]);
}

console.log(convertVersionToNumber("# 1.03.00"));

Since the regex pattern used is

/^# ([0-9]).([0-9][0-9]).([0-9][0-9])\s*/

https://i.sstatic.net/xHSzE.png

2) If there are multiple matches, results can be spread in an array and then looped over using for..of

function convertVersionToNumber(line) {
  const iterator = line.matchAll(/# ([0-9]).([0-9][0-9]).([0-9][0-9])\s*/g);
  const results = [...iterator];
  for (let arr of results) {
    const [match, g1, g2, g3] = arr;
    console.log(match, g1, g2, g3);
  }
}

convertVersionToNumber("# 1.03.00 # 1.03.00");

Alternate approach: The same result can also be achieved using a simple match method

function convertVersionToNumber(line) {
  const result = line.match(/\d/g);
  return +result.join("");
}

console.log(convertVersionToNumber("# 1.03.00"));

Answer 2

1) matchAll function will return an Iterator object

Iterator [RegExp String Iterator]

result will hold the Iterator and by using the spread operator, all matches can be accessed. In this case, as there is only one match, the result will contain a single element.

[ '# 1.03.00', '1', '03', '00', index: 0, input: '# 1.03.00', groups: undefined ]

Lastly, a spread operator is utilized to retrieve all values and encapsulate them in an array

[...result]

function convertVersionToNumber(line) {
  const result = line.matchAll(/^# ([0-9]).([0-9][0-9]).([0-9][0-9])\s*/g);
  const groups = [...result][0];
  return parseInt(groups[1] + groups[2] + groups[3]);
}

console.log(convertVersionToNumber("# 1.03.00"));

Since the regex pattern used is

/^# ([0-9]).([0-9][0-9]).([0-9][0-9])\s*/

https://i.sstatic.net/xHSzE.png

2) If there are multiple matches, results can be spread in an array and then looped over using for..of

function convertVersionToNumber(line) {
  const iterator = line.matchAll(/# ([0-9]).([0-9][0-9]).([0-9][0-9])\s*/g);
  const results = [...iterator];
  for (let arr of results) {
    const [match, g1, g2, g3] = arr;
    console.log(match, g1, g2, g3);
  }
}

convertVersionToNumber("# 1.03.00 # 1.03.00");

Alternate approach: The same result can also be achieved using a simple match method

function convertVersionToNumber(line) {
  const result = line.match(/\d/g);
  return +result.join("");
}

console.log(convertVersionToNumber("# 1.03.00"));

Answer 3

Answer №2

If you're looking to match a string in a specific format and reformat it by keeping only three captured substrings, you can achieve this without using .matchAll. Instead, you can use .replace:

function convertVersionToNumber(line) {
  return parseInt(line.replace(/^# (\d{1})\.(\d{2})\.(\d{2})[\s\S]*/, '$1$2$3'));
}
console.log( convertVersionToNumber("# 1.03.00") );

If needed, you can check if the original string matches the new string after replacement to verify if a match occurred.

Remember to escape dots in the regex pattern to match them as literal characters.

The pattern ^# (\d)\.(\d{2})\.(\d{2})[\s\S]* matches the following:

^ - start of string
# - # preceded by a space
(\d) - Group 1: a single digit
\. - a dot
(\d{2}) - Group 2: two digits
\. - a dot
(\d{2}) - Group 3: two digits
[\s\S]* - match any character (including newlines) zero or more times

The replacement pattern $1$2$3 concatenates the values of Group 1, 2, and 3.

Answer 4

If you're looking to match a string in a specific format and reformat it by keeping only three captured substrings, you can achieve this without using .matchAll. Instead, you can use .replace:

function convertVersionToNumber(line) {
  return parseInt(line.replace(/^# (\d{1})\.(\d{2})\.(\d{2})[\s\S]*/, '$1$2$3'));
}
console.log( convertVersionToNumber("# 1.03.00") );

If needed, you can check if the original string matches the new string after replacement to verify if a match occurred.

Remember to escape dots in the regex pattern to match them as literal characters.

The pattern ^# (\d)\.(\d{2})\.(\d{2})[\s\S]* matches the following:

^ - start of string
# - # preceded by a space
(\d) - Group 1: a single digit
\. - a dot
(\d{2}) - Group 2: two digits
\. - a dot
(\d{2}) - Group 3: two digits
[\s\S]* - match any character (including newlines) zero or more times

The replacement pattern $1$2$3 concatenates the values of Group 1, 2, and 3.

Is there a way to retrieve the groups of match/matchAll similar to accessing an array?

Answer №1

Answer №2

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