如何在流中获取与条件匹配的第一个元素?我试过这个但不起作用
this.stops.stream().filter(Stop s-> s.getStation().getName().equals(name));
该标准不起作用,过滤器方法是在 Stop 之外的其他类中调用的。
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; } }
这可能是您正在寻找的:
yourStream .filter(/* your criteria */) .findFirst() .get();
更好的是,如果有可能不匹配任何元素,在这种情况下get()会抛出 NPE。所以使用:
get()
yourStream .filter(/* your criteria */) .findFirst() .orElse(null); /* You could also create a default object here */
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() .orElse(null); System.out.printf("At the first stop at Station1 there were %d passengers in the train.", firstStopAtStation1.passengerCount); }
输出是:
At the first stop at Station1 there were 250 passengers in the train.