Match group before nth character and after that

Question

I want to match everything before the nth character (except the first character) and everything after it. So for the following string

/firstname/lastname/some/cool/name

I want to match

Group 1: firstname/lastname
Group 2: some/cool/name

With the following regex, I'm nearly there, but I can't find the correct regex to also correctly match the first group and ignore the first /:

([^\/]*\/){3}([^.]*)

Note that I always want to match the 3rd forward slash. Everything after that can be any character that is valid in an URL.

Answer 1

Your regex group are not giving proper result because ([^\\/]*\\/){3} you're repeating captured group which will overwrite the previous matched group Read this

You can use

^.([^/]+\/[^/]+)\/(.*)$

 let str = `/firstname/lastname/some/cool/name` let op = str.match(/^.([^/]+\\/[^/]+)\\/(.*)$/) console.log(op)

Answer 2

The quantifier {3} repeats 3 times the capturing group, which will have the value of the last iteration.

The first iteration will match / , the second firstname/ and the third (the last iteration) lastname/ which will be the value of the group.

The second group captures matching [^.]* which will match 0+ not a literal dot which does not take the the structure of the data into account.

If you want to match the full pattern, you could use:

^\/([^\/]+\/[^\/]+)\/([^\/]+(?:\/[^\/]+)+)$

Explanation

• ^ Start of string
• ( Capture group 1
  • [^\\/]+/[^\\/]+ Match 2 times not a / using a negated character class then a /
• ) Close group
• \\/ Match /
• ( Capture group 2
  • [^\\/]+ Match 1+ times not /
  • (?:\\/[^\\/]+)+ Repeat 1+ times matching / and 1+ times not / to match the pattern of the rest of the string.
• ) Close group
• $ End of string

Regex demo

Answer 3

Ignoring the first / , then capturing the first two words, then capturing the rest of the phrase after the / .

^(:?\/)([^\/]+\/[^\/]+)\/(.+)

See example