获取符合条件的第一个元素

Question

如何获取与流中的条件匹配的第一个元素？我已经尝试过了但是没用

this.stops.stream().filter(Stop s-> s.getStation().getName().equals(name));

该条件不起作用，在除Stop之外的其他类中调用filter方法。

public class Train {

private final String name;
private final SortedSet<Stop> stops;

public Train(String name) {
    this.name = name;
    this.stops = new TreeSet<Stop>();
}

public void addStop(Stop stop) {
    this.stops.add(stop);
}

public Stop getFirstStation() {
    return this.getStops().first();
}

public Stop getLastStation() {
    return this.getStops().last();
}

public SortedSet<Stop> getStops() {
    return stops;
}

public SortedSet<Stop> getStopsAfter(String name) {


    // return this.stops.subSet(, toElement);
    return null;
}
}


import java.util.ArrayList;
import java.util.List;

public class Station {
private final String name;
private final List<Stop> stops;

public Station(String name) {
    this.name = name;
    this.stops = new ArrayList<Stop>();

}

public String getName() {
    return name;
}

}

Answer 1

这可能是您要寻找的：

yourStream
    .filter(/* your criteria */)
    .findFirst()
    .get();

一个例子：

public static void main(String[] args) {
    class Stop {
        private final String stationName;
        private final int    passengerCount;

        Stop(final String stationName, final int passengerCount) {
            this.stationName    = stationName;
            this.passengerCount = passengerCount;
        }
    }

    List<Stop> stops = new LinkedList<>();

    stops.add(new Stop("Station1", 250));
    stops.add(new Stop("Station2", 275));
    stops.add(new Stop("Station3", 390));
    stops.add(new Stop("Station2", 210));
    stops.add(new Stop("Station1", 190));

    Stop firstStopAtStation1 = stops.stream()
            .filter(e -> e.stationName.equals("Station1"))
            .findFirst()
            .get();

    System.out.printf("At the first stop at Station1 there were %d passengers in the train.", firstStopAtStation1.passengerCount);
}

输出为：

At the first stop at Station1 there were 250 passengers in the train.