Regex to find at least 4 occurrences of forward slash in string

Question

I have some urls and I would like to make sure they have at least 4 forward slashes as some of my URLS have less. For example:

Pass: http://localhost:2000/machine/my-test-machine/3

Fail: http://localhost:2000/my-test-machine

Any help would be much appreciated. Thanks Dave

Answer 1

You can try following regex,

\/\/(.*\/){3}

Working Demo

Answer 2

Just match it 4 times:

(?:.*?/){4}

See live demo.

The reluctant quantifier *? will ensure that slashes are not skipped over when matching (eliminating back tracking)

_{Your regex engine (unspecified) may require the forward slash to be escaped, ie \/}

Answer 3

Use a lookahead such as:

^(?=(?:[^\/]*\/){4,})(.*)

Demo