Regex pattern needed for 123.12/23

Question

I am trying to check my string having 123.12/23 with pattern \\d+(.\\d+)*\\/\\d+(.\\d+)* but it is not working, it is passing 123.12/23/24 also.

I need below scenarios to be covered :

Strings to be passed : 12/23 , 12.23/23 , 12/23.33

Strings to be failed : 12/13/14 , 12.23/2/4

See [Java RegEx meta character (.) and ordinary dot?](http://stackoverflow.com/questions/3674930/java-regex-meta-character-and-ordinary-dot) — Wiktor Stribiżew, Jan 11 '16 at 08:36

score 3 · Answer 1 · answered Jan 11 '16 at 06:06

3

^\d+(?:\.\d+)?\/\d+(?:\.\d+)?$

You were close.Escape the ..See demo.

https://regex101.com/r/iJ7bT6/1

For java it would be

^\\d+(?:\\.\\d+)?\\/\\d+(?:\\.\\d+)?$

answered Jan 11 '16 at 06:06

vks

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In Java, it is enough to use `^\\d+(?:\\.\\d+)?/\\d+(?:\\.\\d+)?$`. No need escaping the forward slash. – Wiktor Stribiżew Jan 11 '16 at 08:37

Regex pattern needed for 123.12/23

1 Answers1