英文:
How to combine two lists from the same relation in Java-8
问题
我有以下实体 Manager 和 Colleague。
Manager 实体
@Entity@Table(name = "Manager")@Datapublic class Manager implements java.io.Serializable {@Id@Column(name = "id")private Long id;@Column(name = "name")private String name;@OneToMany@JoinColumn(name = "id")private List<Colleague> colleagues;}
Colleague 实体
@Entity@Table(name = "Colleague")@Datapublic class Colleague implements java.io.Serializable {@Id@Column(name = "id")private Long id;@Column(name = "name")private String name;}
上述关系可以表示为 JSON：
[{"id": "101","name": "manager1","colleagues": [{"id": "101","name": "colleague1"},{"id": "101","name": "colleague2"}]},{"id": "101","name": "manager2","colleagues": [{"id": "101","name": "colleague3"},{"id": "101","name": "colleague4"}]}]
我正在使用 managerReposiotry.findAll() 检索结果：
List<Manager> managerList = managerReposiotry.findAll();
我想要创建一个包含来自 Manager 和 Colleague 的所有姓名的超级列表。
我目前正在执行以下操作：
List<String> names = new ArrayList<>();managerList.stream().forEach(manager -> {List<String> nameList =manager.getColleagues().stream().map(colleague -> colleague.getName()).collect(Collectors.toList());names.addAll(nameList);});
在 Java 8 中是否有其他方法可以改进上述代码？
谢谢！
英文:
I have below entities Manager and Colleague
Manager Entity
@Entity@Table(name = "Manager")@Datapublic class Manager implements java.io.Serializable {@Id@Column(name = "id")private Long id;@Column(name = "name")private String name;@OneToMany@JoinColumn(name = "id")private List<Colleague> colleagues;}
Colleague Entity
@Entity@Table(name = "Colleague")@Datapublic class Colleague implements java.io.Serializable {@Id@Column(name = "id")private Long id;@Column(name = "name")private String name;}
Above relation can be represented in JSON as
[{"id": "101","name": "manager1","colleagues": [{"id": "101","name": "colleague1"},{"id": "101","name": "colleague2"}]},{"id": "101","name": "manager2","colleagues": [{"id": "101","name": "colleague3"},{"id": "101","name": "colleague4"}]}]
I am retrieving the result on managerReposiotry.findAll() as
List<Manager> managerList = managerReposiotry.findAll();
I want to create a super list of all names from Manager and Colleague
What I am currently doing is
List<String> names = new ArrayList<>();managerList.stream().forEach(manager -> {List<String> nameList =manager.getColleagues().stream().map(colleague -> colleague.getName()).collect(Collectors.toList());names.addAll(nameList);});
Is there any other way in Java-8 to improve the above code?
Thank you !!
答案1
得分: 3
你可以使用 flatMap 来展开所有 Manager 的 Colleague，然后仅获取 Colleague 的姓名并以列表形式收集。
List<String> names =managerList.stream() // ...Stream<Manager>.flatMap(m -> m.getColleagues().stream()) // ...Stream<Colleague>.map(c -> c.getName()) // ...Stream<String>.collect(Collectors.toList());
但更好的方式是，如果所有的同事都有经理，直接从数据库中获取。
@Query("select c.name from Colleagues c")List<String> findAllColleagueName();
英文:
You can use flatMap to flatten all Colleague of Manager then map Colleague name only and collect as list.
List<String> names =managerList.stream() // ...Stream<Manager>.flatMap(m -> m.getColleagues().stream()) // ...Stream<Colleague>.map(c-> c.getName()) // ...Stream<String>.collect(Collectors.toList());
But the better way is directly fetched from the database if all colleagues have manager.
@Query("select c.name from Colleagues c")List<String> findAllColleagueName();
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