Using regex, Conditionally replace with $1 or empty string

Question

I want to create a function in javascript which accepts two arguments : a regex object and a string.
The regex can have at most one capturing group.
If there is a capturing group, it shall return $1, otherwise the remaining string;

So myfunc("hello123", /\d+/) shall return hello and
myfunc("hello123", /\w+(\d+)/) shall return 123

How to identify if there's a capturing group or not?

In your first example: myfunc("hello123", /\d+/) shall return hello ..you are replacing the second argument with nothing and returning the string. In your second example: myfunc("hello123", /\w+(\d+)/) shall return 123 you are matching and returning match[1]. There has to be a uniform behaviour to build a function on. — erosman
– erosman, Commented May 22, 2014 at 10:57
There's a logic : If there is a capturing group in the regex, it shall return $1, otherwise it shall replace the string with the given pattern. First example : myfunc("hello123", /\d+/), no matching group, so the string matching the given pattern /\d+/ shall be removed. second example myfunc("hello123", /\w+(\d+)/), there's a matching group, so I need the part that matched! — GrSrv
– GrSrv, Commented May 26, 2014 at 13:19

plalx · Accepted Answer · 2014-05-22 10:58:34Z

1

When using match if the pattern couldn't be matched at all null is returned. If there was a match, the string at index 0 will be the full match while subsequent indexes will contain capturing groups. Basically the first capturing group will be at index 1 and so on.

answered May 22, 2014 at 10:58

plalx

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Comments

anubhava · Accepted Answer · 2014-05-22 10:58:12Z

1

You can use this function:

function myfunc(str, re) {
   m = str.match(re);
   if (m && m.length == 2)
      return m[1];
   else
      return str.replace(m[0], '');
}

Testing:

myfunc("hello123", /\d+/);
hello

myfunc("hello123", /\w+?(\d+)/)
123

PS: Use m.length to figure out how many matced groups are there in the regex being passed as argument.

answered May 22, 2014 at 10:58

anubhava

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2 Comments

GrSrv Over a year ago

fails in this case : myfunc("hello", /\w+?(\d+)/)

anubhava Over a year ago

It returns 123 for that case and that is your expected output also.

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