Stream filter.findfirst

Question

Stream filter.findfirst

try {
    Stream<String> stringStream = Files.lines(path, StandardCharsets.UTF_8);
    stringStream.forEach((String e) -> {
        if (e.contains("12")) {
            System.out.println(path);
            break;   // как остановить работу если нашёл нужное
        }
     });
} catch (IOException e) {
    e.printStackTrace();
}

I can't figure out how to use filter.findfirst show my example

0

java filter потоки-данных

Author: Sergey Gornostaev, 2018-08-15

Source

4 answers

score 1 · Answer 1

From the question, it is not quite clear what you want, but I will assume:

Stream<String> stringStream = Files.lines(path, StandardCharsets.UTF_8);
Optional<String> firstStr = stringStream.filter(i -> i.contains("12")).findFirst();
firstStr.ifPresent(System.out::println);

score 0 · Answer 2

It will stop at the first found element when using filter and findFirst. How to check? Like this:

Stream<String> stringStream = Stream.of("345", "12", "789", "101112");
Optional<String> found = stringStream.filter(s -> {
        System.out.println(s);
        return s.contains("12");
    }).findFirst();

345

12

If you need to additionally output values that are suitable for the filter, I recommend writing this (in this case, only "12" will be output, since peek is already after the filter.):

Optional<String> found = stringStream
        .filter(s -> s.contains("12"))
        .peek(System.out::println)
        .findFirst();

score 0 · Answer 3

You really need it .filter And then check if something is in optional.

Stream<Integer> stringStream = Arrays.stream(new Integer[]{5, 7, 15 , 7});
Optional<Integer> result = stringStream.filter(integer -> integer == 7).findFirst();

Expanded lambda

new Predicate<Integer>() {
        @Override
        public boolean test(Integer integer) {
            return integer == 7;
        }
    }

score 0 · Answer 4

I suspect that path is responsible for the names of the files that are viewed, and if the file has the desired substring ("12"), then the name of this file (path) is output. If the author wanted such logic, it is better to use .anyMatch(): For example, there are 3 files: 1.txt, 2.txt, 3.txt

    for (int i = 0; i < 3; i++) {
        Path path = new File((i + 1) + ".txt").toPath();
        try {
            Stream<String> stringStream = Files.lines(path, StandardCharsets.UTF_8);
            if (stringStream.anyMatch(s -> s.contains("12"))) {
                System.out.println(path);
            }
        } catch (IOException e) {
            e.printStackTrace();
        }
    }