我有以下课程:
class ServiceSnapshot {
List<ExchangeSnapshot> exchangeSnapshots = ...
...
}
class ExchangeSnapshot{
Map<String, String> properties = ...
...
}
SayI有一组ServiceSnapshots,如下所示:
Collection<ServiceSnapshot> serviceSnapshots = ...
我想过滤集合,以便生成的ServiceSnapshots集合仅包含ServiceSnapshots,其中包含ExchangeSnapshots,其中ExchangeSnapshots上的属性与给定的String匹配。
我有以下未经测试的代码,只是想知道是否有更清晰/更易读的方法来执行此操作,使用Java 7,如果需要可能还有Google Guava?
更新:另请注意,我在下面提供的代码示例不适合我的目的,因为我使用iterator.remove()来过滤集合。事实证明我不能这样做,因为它正在修改底层集合,这意味着后面对我的方法的调用导致越来越少的snashots由于之前的调用从集合中删除它们 - 这不是我想要的。
public Collection<ServiceSnapshot> getServiceSnapshotsForComponent(final String serviceId, final String componentInstanceId) {
final Collection<ServiceSnapshot> serviceSnapshots = getServiceSnapshots(serviceId);
final Iterator<ServiceSnapshot> serviceSnapshotIterator = serviceSnapshots.iterator();
while (serviceSnapshotIterator.hasNext()) {
final ServiceSnapshot serviceSnapshot = (ServiceSnapshot) serviceSnapshotIterator.next();
final Iterator<ExchangeSnapshot> exchangeSnapshotIterator = serviceSnapshot.getExchangeSnapshots().iterator();
while (exchangeSnapshotIterator.hasNext()) {
final ExchangeSnapshot exchangeSnapshot = (ExchangeSnapshot) exchangeSnapshotIterator.next();
final String foundComponentInstanceId = exchangeSnapshot.getProperties().get("ComponentInstanceId");
if (foundComponentInstanceId == null || !foundComponentInstanceId.equals(componentInstanceId)) {
exchangeSnapshotIterator.remove();
}
}
if (serviceSnapshot.getExchangeSnapshots().isEmpty()) {
serviceSnapshotIterator.remove();
}
}
return serviceSnapshots;
}
答案 0 :(得分:0)
使用番石榴:
Iterables.removeIf(serviceSnapshots, new Predicate<ServiceSnapshot>() {
@Override
public boolean apply(ServiceSnapshot serviceSnapshot) {
return !Iterables.any(serviceSnapshot.getExchangeSnapshots(), new Predicate<ExchangeSnapshot>() {
@Override
public boolean apply(ExchangeSnapshot exchangeSnapshot) {
String foundComponentInstanceId = exchangeSnapshot.getProperties().get("ComponentInstanceId");
return foundComponentInstanceId != null && foundComponentInstanceId.equals(componentInstanceId);
}
});
}
});
我的某个地方可能遗漏了某个!
,但基本策略是删除任何没有ID匹配的ServiceSnapshot
的{{1}}个对象。