Regex to match a digit in android java

Question

I'm trying to use regex to match digits in java, something like:

Pattern p = Pattern.compile("(\d+) / (\d+)");
String myRunway = "12 / 1234";
Matcher m = p.matcher(myRunway);
int nrGroups = m.groupCount();
String rwData = m.group(1); //should have 12
String rwLen = m.group(2); //should have 1234

The compiler doesn't like \d (for any digit), it says the only valid escapes are \b \t \n \f \r \" \' \\

Just for yucks I then tried (\\d+) / (\\d+) and it compiles, but does not match. However, nrGroups is 2 which doesn't make sense if there was no match. How do I parse groups of digits in java? In searching the forum, I found only C# postings on this.

Actually, I eventually want to be able to match "12R / 1234" using (\d+).* / (\d+) to get "12" and "1234" as the two groups, but I simplified the above to try to get it working.

Thanks!

Peter Boughton · Accepted Answer · 2011-09-06 02:18:22Z

5

Java strings need any backslashes in them to be escaped.

So you need to use \\d in order to get \d passed through for the regex pattern.

Similarly you'd need to use \" if you wanted a double-quote - to be clear this is on the Java string side, not the regex side of things.

Your (\\d+) / (\\d+) version should have matched... I think the problem was that you didn't do an m.find() so it didn't populate the groups. i.e. try this:

Pattern p = Pattern.compile("(\\d+) / (\\d+)");
String myRunway = "12 / 1234";
Matcher m = p.matcher(myRunway);
m.find();  // <----- this 'executes' the matcher, and populates the group info
int nrGroups = m.groupCount();
String rwData = m.group(1); //should have 12
String rwLen = m.group(2); //should have 1234

Also, regarding:

I eventually want to be able to match "12R / 1234" using (\d+).* / (\d+) to

Don't use .* to match R - identify clearly what you are matching. If there is only a small number of characters, use a lazy quantifier instead (that's *? instead of *), or if there can't be a backslash then use a negated group, e.g. (\d+)[^/]*/ (\d+)

Also, consider perhaps splitting on a series of non-digits instead, something like:

String myRunway = "12R / 1234";
String[] Groups = myRunway.split('\\D+');
String rwData = Groups[0];
String rwLen = Groups[1];

No messing about with matchers that way - assuming your string is in a predictable/suitable format.

edited Sep 6, 2011 at 2:18

answered Sep 6, 2011 at 2:03

Peter Boughton

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3 Comments

Paul Kinzelman Over a year ago

The missing m.find() was the key, I couldn't really find any examples on the net of what I wanted to do. What I did find didn't tell me I had to call .find to populate the groups, thanks so much! I also didn't describe what I wanted well with respect to the extra character, but I took your suggestion about not using .* and so now I'm using "(\\d+)\\D{0,1} / (\\d+)" because the "R" in the example could be any letter or it might not be there at all.

Peter Boughton Over a year ago

Well if it can be any letter than [A-Z]? might be a better choice - the ? is a common shorthand for {0,1}, whilst [A-Z] says any character between A and Z (inclusive). - not a big issue since what you have works, but might make it more obvious for someone looking at the expression in six months time, wondering if there was an edge case that required \D instead of [A-Z] or whatever.

Paul Kinzelman Over a year ago

Yes, [A-Z] would work too, that's a good idea on making it more self-documenting. No complex edge case. They're all just airport runway identifiers, and actually, the only possibilities are R, C, and L (right, center, left) so I could use [RCL], although if the database where I got the runway data is corrupted, then to be more defensive I should leave it [A-Z] or \D.

Julian · Accepted Answer · 2011-09-06 02:03:48Z

2

You need to escape the backslash for Java:

Pattern p = Pattern.compile("(\\d+) / (\\d+)");

answered Sep 6, 2011 at 2:03

Julian

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1 Comment

Paul Kinzelman Over a year ago

I had tried both ways and what was missing was the call to m.find(); Thanks!

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