RegEx to match stuff between parentheses

2019-01-13 05:43发布

问题:

I'm having a tough time getting this to work. I have a string like:

something/([0-9])/([a-z])

And I need regex or a method of getting each match between the parentheses and return an array of matches like:

[
  [0-9],
  [a-z]
]

The regex I'm using is /\((.+)\)/ which does seem to match the right thing if there is only one set of parenthesis.

How can I get an array like above using any RegExp method in JavaScript? I need to return just that array because the returned items in the array will be looped through to create a URL routing scheme.

回答1:

You need to make your regex pattern 'non-greedy' by adding a '?' after the '.+'

By default, '*' and '+' are greedy in that they will match as long a string of chars as possible, ignoring any matches that might occur within the string.

Non-greedy makes the pattern only match the shortest possible match.

See Watch Out for The Greediness! for a better explanation.

Or alternately, change your regex to

\(([^\)]+)\)

which will match any grouping of parens that do not, themselves, contain parens.



回答2:

Use this expression:

/\(([^()]+)\)/g

e.g:

function()
{
    var mts = "something/([0-9])/([a-z])".match(/\(([^()]+)\)/g );
    alert(mts[0]);
    alert(mts[1]);
}


回答3:

If s is your string:

s.replace(/^[^(]*\(/, "") // trim everything before first parenthesis
 .replace(/\)[^(]*$/, "") // trim everything after last parenthesis
 .split(/\)[^(]*\(/);      // split between parenthesis


回答4:

var getMatchingGroups = function(s) {
  var r=/\((.*?)\)/g, a=[], m;
  while (m = r.exec(s)) {
    a.push(m[1]);
  }
  return a;
};

getMatchingGroups("something/([0-9])/([a-z])"); // => ["[0-9]", "[a-z]"]