Java 8如何操作一个列表中的对象并将其收集到另一个列表中？

Question

我是Java8的新手。我需要按照特定条件将POJO列表转换为组，并在UI上显示。

• 将属于Children类别的对象列表转换为Person 类别。

• 返回列表。

  Children.java
  private String firstName;
  private String lastName;
  private String school;
  private String personId;
  // Setters and getters.
  
  Person.java
  private String fullName;
  private String address;
  // Setters and Getters.
  
  private Person convertChildToPerson(Children child) {
           Person person = new Person();
           person.setFullName(child.getLastName() + ", " + child.getFirstName());
            ..
            return person;
   } 
  

  ..     主要代码..    ..

  List<Person> personList;  // Is populated by other functions.
  
  // Connect to DB and gets the output
  List<Children> childrenList = (Children) criteria.list();  
  
  for(Children children: childrenList) {
       personList.add(convertChildToPerson(children));
  }
  return personList; 
  

  上面的 FOR-LOOP 是否可以被Java 8 Streams取代？

Answer 1

@property int idCell;

- (void)tableView:(UITableView *)tableView didSelectRowAtIndexPath:(NSIndexPath *)indexPath {
    self.idCell = indexPath.row
    [self performSegueWithIdentifier:@"detailViewController" sender:sender];
}

- (void)prepareForSegue:(UIStoryboardSegue *)segue sender:(id)sender {
    if([segue.identifier isEqualToString:@"detailViewController"]) {
        SecondVC *vc = [segue destinationViewController];
        vc.idDetail = idCell;
    }
}

如Holger所建议的那样更短（更好）：

 List<Person> newList = childrenList.stream()
             .map(this::convertChildToPerson)
             .collect(Collectors.toList());

 personList.addAll(newList);