Question 322933 in Javascript Question posted in
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Matching two path elements using javascript regex to make the right hand side greedy

I am trying to match this.

/1234/5678
/1234

Into A and B, where the B is optional using this regular expression.

/.*/(?<A>d+)(/(?<B>.+?))?$/.exec("/1234/5678").groups;

I want A=1234, and B=5678 when the path contains two parts, and when there’s only one part, then just match A.

How do I make the optional /5678 greedy so that it matches if it’s there, but still matches A if it’s not.

/1234/5678 (A=1234, B=5678)
/1234 (A=1234, B=undefined)
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2

Answers


  1. 0

    You can try this.

    console.log(/(?:/(?<A>[^/]+))?(?:/(?<B>[^/]+))?/.exec("/1234/5678").groups);
console.log(/(?:/(?<A>[^/]+))?(?:/(?<B>[^/]+))?/.exec("/1234").groups);
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  2. 0

    There’s no need in a regular expression, you can just split the string and use array destructuring:

    {
 const [_, A, B] = '/1234/5678'.split('/');
 console.log({A, B});
}
{
 const [_, A, B] = '/1234'.split('/');
 console.log({A, B});
}
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