选择正斜杠之前但任何空格之后的第一个字符

Question

I have the following string patterns which I need to match as described.

I need only the first char/digit on each of the following examples. Before the '/' and after any space:

12/5 <--match on 1
x23/4.5 match on x
234.5/7 match on 2
2 - 012.3/4 match on 0

regex something like the following is obviously not enough:

\d(?=\d\/))

To make Clear I'm actauly using the regex with js split so it's some mpping function which takes each one of the strings and split it on the match. So for example 2 - 012.3/4 would be split to [ 2 - 0, 12.3/4] and 12/5 to 1, [2/5] and so on.

See example (with non working regex) here:

https://regex101.com/r/N1RbGp/1

Answer 1

Try a regular expression like this:

(?<=^|\s)[a-zA-Z0-9](?=[^\s]*[/])

Breaking it down:

• (?<=^|\\s) is a zero-width (non-capturing) positive lookbehind that ensures that the match will begin only immediately following start-of-text or a whitespace character.

• [a-zA-Z0-9] matches a single letter or digit.

• (?=\\S*[/]) is a zero-width (non-capturing) positive lookahead that requires the matched letter or digit to be followed by zero or more non-whitespace characters and a solidus (' / ') character.

Here's the code:

const texts = [
  '12/5',
  'x23/4.5',
  '234.5/7',
  '2 - 012.3/4',
];
texts.push( texts.join(', ') );

for (const text of texts) {
  const rx = /(?<=^|\s)[a-zA-Z0-9](?=\S*[/])/g;

  console.log('');
  console.group(`text: '${text}'`);
  for(let m = rx.exec(text) ; m ; m = rx.exec(text) ) {
    console.log(`matched '${m[0]}' at offset ${m.index} in text.`);
  }
  console.groupEnd();

}

This is the output:

text: '12/5'
  matched '1' at offset 0 in text.

text: 'x23/4.5'
  matched 'x' at offset 0 in text.

text: '234.5/7'
  matched '2' at offset 0 in text.

text: '2 - 012.3/4'
  matched '0' at offset 4 in text.

text: '12/5, x23/4.5, 234.5/7, 2 - 012.3/4'
  matched '1' at offset 0 in text.
  matched 'x' at offset 6 in text.
  matched '2' at offset 15 in text.
  matched '0' at offset 28 in text.

Answer 2

If you want to be able to scan the whole document:

/(?<=(^|\s))\S(?=\S*\/)/g

https://regex101.com/r/rN08sP/1

 s = `12/5 x23/4.5 234.5/2 534/5.6 - 49.55/6.5 234.5/7`; console.log(s.match(/(?<=(^|\\s))\\S(?=\\S*\\/)/g));

But if you want to extract that character in a short string: (did you mean there is a space in front?)

It'd be /\\s(\\S)\\S*\\//

 const arr = [ " 12/5", " x23/4.5", " 234.5/7", " 2 - 012.3/4" ]; arr.forEach(s => { let result = s.match(/\\s(\\S)\\S*\\//); if (result) console.log("For", s, "result: ", result[1]) });

But if "beginning of line" is ok... so no space is needed in front, then /(^|\\s)(\\S)\\S*\\// :

 const arr = [ "12/5", "x23/4.5", "234.5/7", "2 - 012.3/4" ]; arr.forEach(s => { let result = s.match(/(^|\\s)(\\S)\\S*\\//); if (result) console.log("For", s, "result: ", result[2]) });

But actually, if you don't mean literally a space but just boundary in general:

 const arr = [ "12/5", "x23/4.5", "234.5/7", "2 - 012.3/4" ]; arr.forEach(s => { let result = s.match(/\\b(\\S)\\S*\\//); if (result) console.log("For", s, "result: ", result[1]) });

Answer 3

The first group in this regex matches the character you're asking for:

([^\s])[^\s]*/

You could also just use:

[^\s]+/

And use the first character of the match (or perhaps you need the rest anyway).