How to make a regex to find if a string contains a placeholder like %1$s

Question

I failed after several tries to make a regex to find if a string contains a placeholder like %1$s or %3$d in a message like:

Hello, %1$s! You have %2$d new messages.

Can somebody give me, any help ?

How can I count how many placeholders are there ?

score 0 · Answer 1 · answered Jan 27 '14 at 17:29

0

You can use this regex in String#matches method call:

".*?%\\d+\\$[sd].*"

Code:

boolean found = string.matches( ".*?%\\d+\\$[sd].*" );

answered Jan 27 '14 at 17:29

anubhava

nKn · Accepted Answer · 2014-01-27T17:38:52.170

0

Try with this pattern:

Pattern.compile("^.*%[0-9]+\\$[sd].*");

To count all matches, you could use Matcher and count like this:

int count = 0;
while (matcher.find())
  count++;

I'm not sure there's a more efficient way.

edited Jan 27 '14 at 17:38

answered Jan 27 '14 at 17:30

nKn

I'll take your word :-) Glad it helped. – nKn Jan 27 '14 at 21:33
IF somebody reaches here and has to count the matches it may have to use this version: Pattern.compile("^*%[0-9]+\\$[sd]*"); – Buda Florin Jan 28 '14 at 17:45

score 0 · Answer 3 · edited May 23 '17 at 11:51

0

This will capture both cases. If you are using Java, I've provided a link on how to count matches.

(%1\$s|%2\$d)

edited May 23 '17 at 11:51

Community

answered Jan 27 '14 at 17:33

dthartman

In lieu of changing my regex to look like the other posts, let me just say the other regex answers provided are good, since they are more generic. – dthartman Jan 27 '14 at 17:44

3 Answers3