提取正则表达式Java的一部分

Question

I need help to know how to extract just a part of regular expression in Java with reference. Example: I have a text like:

"Jean Moulin - Porte d'Orléans - Paris 14e Jean Moulin. Apartment 37 m² renovated in 2013. Materials and quality services. Large living room with kitchen, separate bedroom area, dressing room, toilet / sink, large shower, toilet with sink and washing machine equipment, marble floor, double glazed window sound. Quiet, functional, spacious. On the first floor with open views, elevator, guards. Furniture and equipment new appliances. € 310,000.

Rooms: 2 Bedrooms: 1 Surface: 37 m² "

And I want only to take the number of rooms with the mark "rooms" How I can do? I know just that:

Pattern p;
Matcher m;
p= Pattern.compile("Rooms: +[0-9]");

but I want only the number, I want to delete "Rooms: " from the result.

And the 2nd question, How to take the name of the city with a regular expression, because if I have a lot of text with different name of city: like Paris, London, Lyon, Rome etc... How I can do to make one regular expression that does this?

Answer 1

I want only the number, I want to delete "Rooms: " from the result.

Use Matcher#group() to get the matched group that is enclosed inside the Parenthesis () that is used to enable grouping of regex phrases.

Here group(1) returns the (\\\\d+) if found.

Sample code:

    Pattern pattern = Pattern.compile("Rooms:\\s+(\\d+)");
    Matcher matcher = pattern.matcher("Rooms: 2 Bedrooms: 1 Surface: 37 m² ");

    if (matcher.find()) {
        System.out.println(matcher.group(1)); // output 2 only
    }

Please have a look at Java Regex Pattern where all the pattern are explained in details. Find sample code here on Java Tutorials - Quantifiers

Answer 2

使用带有反向引用的replaceAll()在一行中提取它：

String rooms = input.replaceAll("(?is).*\\brooms\\s+(\\d+).*", "$1");