Matching character sequence at end of string with regex in Java

Asked May 06 '17 at 08:29

Active May 06 '17 at 08:29

Viewed 21 times

I have been trying to match the following sequence "/1" in various ways, but none of them are working. I am matching division by 1 at the end of a string. Here's what I have tried:

String m = "1204/1"
pattern = "[\\/1]+$"; \\doesn't work
pattern = "(\\/1)$"; \\doesn't work
pattern = "[\\1,1]{2}$"; \\doesn't work
if(m.matches(pattern)System.out.println("Ham");
else System.out.println("cheese");

all I have been getting as output is "cheese". Any ideas would be great.

thanks

asked May 06 '17 at 08:29

i_o

`pattern = "(\\/1)$";` works withg `Matcher#find()` – Wiktor Stribiżew May 06 '17 at 08:31
Try this regex `(\d+)\/1$` – t0mm13b May 06 '17 at 08:32
Do you know of any pattern that would work without find() @WiktorStribiżew – i_o May 06 '17 at 08:33
it did not work @t0mm13b – i_o May 06 '17 at 08:35
1

Wih `matches`? Just use a pattern that will match the whole input as is described in the linked post. `s.matches("(?s).*/1")` (a `/` is not a special regex char and needs no escaping). – Wiktor Stribiżew May 06 '17 at 08:36
what does the * do? what does the "s" represent? It works. Thank you. @WiktorStribiżew – i_o May 06 '17 at 08:42
* is a quantifier matching zero or more occurrences. s makes dot match line breaks. – Wiktor Stribiżew May 06 '17 at 11:35

Matching character sequence at end of string with regex in Java

0 Answers0