Regex for validate repeating slash

Question

I am trying to write a regexp (javascript) to validate a string where slashes are repeated.

Must be:

/ - valid string

/abs - valid string

/abs/ - valid string

/abs/a - valid string

// - invalid string

//asd//sd/ -invalid string

/as//a - invalid string

a/a - invalid string

abs - invalid string

////////sdf///// - invalid string

I am trying this regexp:

(\/){1}[\w-]*

^(\/){1}[\w-]*

^(\/\w+)+(\.)?\w+(\?(\w+=[\w\d]+(&\w+=[\w\d]+)*)+){0,1}$

How do I write an expression to pass the validation criteria?

You don't need regex for such a simple task. If I understood your examples correctly, the input string is invalid if it contains two consecutive slashes. Use String.includes() to find if the input string contains the substring '//'. — axiac
– axiac, Commented Aug 10, 2020 at 16:37
It doesn't work for me. Validation occurs with a field on the form that is not available to me. I can just set the validation parameters. In this case, a regular expression — Jo Po
– Jo Po, Commented Aug 10, 2020 at 16:43
If I could I would just do it - "////asd///sd//".replace(/\/+/g, '/') — Jo Po
– Jo Po, Commented Aug 10, 2020 at 16:45
Why are "a/a" and "abs" invalid? You need define "valid". I suggest you add a sentence such as the following after your first sentence: "A string is valid if it meets the following conditions:...". Your first sentence needs to be clarified, as you are not validating strings containing repeated forward slashes. — Cary Swoveland
– Cary Swoveland, Commented Aug 10, 2020 at 17:41

A.Hristov · Accepted Answer · 2020-08-10 17:38:27Z

1

This is a pretty simple regex and should work for any engine:

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

The idea is that it should always start with a forward slash and no two forward slashes should touch. You can test it here: https://regex101.com/r/1BCQs3/2/

answered Aug 10, 2020 at 17:38

A.Hristov

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2 Comments

Cary Swoveland Over a year ago

This assumes (perhaps correctly) that strings that do not begin with a forward slash are invalid. That is consistent with the examples given in the question but was not stated as a requirement of valid strings. You therefore should make that assumption explicit. Easier would be ^(?!.*\/\/)\/.

Jo Po Over a year ago

@CarySwoveland Nice! This works for my examples. Thank!

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