根据对象id更新列表中的对象

Question

我想知道是否可以直接通过其id更新列表中的对象。下面是我的对象结构:

public class MyObject {
public UUID uuid;
public String string1;
public int int1;
...
}

在另一个对象中，我有一个MyObject的列表，我想通过对象id更新其中一个。除了下面的代码，还有其他的解决方案吗?

List<MyObject> list = new ArrayList<>();
// Add many object into my list
MyObject newObject = new MyObject(...);
list.removeIf(o -> o.getUuid().equals(newObject.getUuid()));
list.add(newObject);

Answer 1

紧凑，但粗糙…

List<MyObject> list2 =
list.stream().map(item->item.getUuid().equals(newObject.getUuid()) ? newObject : item)
.collect(Collectors.toList();

Answer 2

如果您真的想坚持使用List，您可以遍历列表并每次手动检查UUID。

List<MyObject> list = new ArrayList<>();
for (int i = 0; i < list.size(), i++) {
if (list.get(i).getUuid().equals(newObject.getUuid())) {
list.set(i, newObject());
// Probably want to stop the loop here.
}
}

使用UUID作为Map的键使它更简单。

List<MyObject> list = new ArrayList<>();
Map<UUID, MyObject> map = list.stream().collect(Collecotrs.toMap(o -> o.getUuid(), o -> o));
map.put(newObject.getUuid(), newObject);
list = new ArrayList<>(map.values());

Answer 3

我认为最简单的方法是实现hashcode/equals，它将以UUID为目标。当调用add()和remove()时，匹配UUID的对象将被添加和从列表中删除。

下面是一个例子:

import java.util.*;
public class ListExample {
public static void main(String[] args) {
System.out.println("Implementation using a LinkedList.");
List<MyObject> list = new ArrayList<>();
list.add(new MyObject(UUID.fromString("299971c6-31b8-11ec-8d3d-0242ac130003"), "A", 0));
list.add(new MyObject(UUID.fromString("31e77904-31b8-11ec-8d3d-0242ac130003"), "B", 1));
list.add(new MyObject(UUID.fromString("368b0214-31b8-11ec-8d3d-0242ac130003"), "C", 2));
System.out.println(list);
System.out.println("Remove an entry.");
list.remove(new MyObject(UUID.fromString("368b0214-31b8-11ec-8d3d-0242ac130003")));
System.out.println(list);
list.add(new MyObject(UUID.fromString("9b21edea-31b9-11ec-8d3d-0242ac130003"), "D", 3));
System.out.println(list);
System.out.println("Adding a duplicate (creates a new entry).");
list.add(new MyObject(UUID.fromString("299971c6-31b8-11ec-8d3d-0242ac130003"), "A", 0));
System.out.println(list);
System.out.println("Remove duplicate (removes the first entry but not the last).");
list.remove(new MyObject(UUID.fromString("299971c6-31b8-11ec-8d3d-0242ac130003"), "A", 0));
System.out.println(list);
}
public static class MyObject {
public UUID uuid;
public String string1;
public int int1;
public MyObject(UUID uuid) {
this.uuid = uuid;
}
public MyObject(UUID uuid, String string1, int int1) {
this.uuid = uuid;
this.string1 = string1;
this.int1 = int1;
}
@Override
public boolean equals(Object o) {
if (this == o) return true;
if (o == null || getClass() != o.getClass()) return false;
MyObject myObject = (MyObject) o;
return Objects.equals(uuid, myObject.uuid);
}
@Override
public int hashCode() {
return Objects.hash(uuid);
}
@Override
public String toString() {
return "MyObject{" +
"uuid=" + uuid +
", string1='" + string1 + ''' +
", int1=" + int1 +
'}';
}
}
}

输出:

Implementation using a LinkedList.
[MyObject{uuid=299971c6-31b8-11ec-8d3d-0242ac130003, string1='A', int1=0}, MyObject{uuid=31e77904-31b8-11ec-8d3d-0242ac130003, string1='B', int1=1}, MyObject{uuid=368b0214-31b8-11ec-8d3d-0242ac130003, string1='C', int1=2}]
Remove an entry.
[MyObject{uuid=299971c6-31b8-11ec-8d3d-0242ac130003, string1='A', int1=0}, MyObject{uuid=31e77904-31b8-11ec-8d3d-0242ac130003, string1='B', int1=1}]
[MyObject{uuid=299971c6-31b8-11ec-8d3d-0242ac130003, string1='A', int1=0}, MyObject{uuid=31e77904-31b8-11ec-8d3d-0242ac130003, string1='B', int1=1}, MyObject{uuid=9b21edea-31b9-11ec-8d3d-0242ac130003, string1='D', int1=3}]
Adding a duplicate (creates a new entry).
[MyObject{uuid=299971c6-31b8-11ec-8d3d-0242ac130003, string1='A', int1=0}, MyObject{uuid=31e77904-31b8-11ec-8d3d-0242ac130003, string1='B', int1=1}, MyObject{uuid=9b21edea-31b9-11ec-8d3d-0242ac130003, string1='D', int1=3}, MyObject{uuid=299971c6-31b8-11ec-8d3d-0242ac130003, string1='A', int1=0}]
Remove duplicate (removes the first entry but not the last).
[MyObject{uuid=31e77904-31b8-11ec-8d3d-0242ac130003, string1='B', int1=1}, MyObject{uuid=9b21edea-31b9-11ec-8d3d-0242ac130003, string1='D', int1=3}, MyObject{uuid=299971c6-31b8-11ec-8d3d-0242ac130003, string1='A', int1=0}]

这里的主要问题是可以创建重复项(因为我们使用列表)

如果您需要一种保留顺序并排除重复项的方法，请使用LinkedHashSet

import java.util.*;
public class SetExample {
public static void main(String[] args) {
System.out.println("Implementation using a LinkedHashSet");
Set<MyObject> set = new LinkedHashSet<>();
set.add(new MyObject(UUID.fromString("299971c6-31b8-11ec-8d3d-0242ac130003"), "A", 0));
set.add(new MyObject(UUID.fromString("31e77904-31b8-11ec-8d3d-0242ac130003"), "B", 1));
set.add(new MyObject(UUID.fromString("368b0214-31b8-11ec-8d3d-0242ac130003"), "C", 2));
System.out.println(set);
System.out.println("Remove an entry.");
set.remove(new MyObject(UUID.fromString("368b0214-31b8-11ec-8d3d-0242ac130003")));
System.out.println(set);
set.add(new MyObject(UUID.fromString("9b21edea-31b9-11ec-8d3d-0242ac130003"), "D", 3));
System.out.println(set);
System.out.println("Adding a duplicate (does not create a new entry).");
set.add(new MyObject(UUID.fromString("299971c6-31b8-11ec-8d3d-0242ac130003"), "A", 0));
System.out.println(set);
}
public static class MyObject {
public UUID uuid;
public String string1;
public int int1;
public MyObject(UUID uuid) {
this.uuid = uuid;
}
public MyObject(UUID uuid, String string1, int int1) {
this.uuid = uuid;
this.string1 = string1;
this.int1 = int1;
}
@Override
public boolean equals(Object o) {
if (this == o) return true;
if (o == null || getClass() != o.getClass()) return false;
MyObject myObject = (MyObject) o;
return Objects.equals(uuid, myObject.uuid);
}
@Override
public int hashCode() {
return Objects.hash(uuid);
}
@Override
public String toString() {
return "MyObject{" +
"uuid=" + uuid +
", string1='" + string1 + ''' +
", int1=" + int1 +
'}';
}
}

输出:

Implementation using a LinkedHashSet
[MyObject{uuid=299971c6-31b8-11ec-8d3d-0242ac130003, string1='A', int1=0}, MyObject{uuid=31e77904-31b8-11ec-8d3d-0242ac130003, string1='B', int1=1}, MyObject{uuid=368b0214-31b8-11ec-8d3d-0242ac130003, string1='C', int1=2}]
Remove an entry.
[MyObject{uuid=299971c6-31b8-11ec-8d3d-0242ac130003, string1='A', int1=0}, MyObject{uuid=31e77904-31b8-11ec-8d3d-0242ac130003, string1='B', int1=1}]
[MyObject{uuid=299971c6-31b8-11ec-8d3d-0242ac130003, string1='A', int1=0}, MyObject{uuid=31e77904-31b8-11ec-8d3d-0242ac130003, string1='B', int1=1}, MyObject{uuid=9b21edea-31b9-11ec-8d3d-0242ac130003, string1='D', int1=3}]
Adding a duplicate (does not create a new entry).
[MyObject{uuid=299971c6-31b8-11ec-8d3d-0242ac130003, string1='A', int1=0}, MyObject{uuid=31e77904-31b8-11ec-8d3d-0242ac130003, string1='B', int1=1}, MyObject{uuid=9b21edea-31b9-11ec-8d3d-0242ac130003, string1='D', int1=3}]

Answer 4

使用如下代码:

List<MyObject> list = new ArrayList<MyObject>();

int id = 4; //the UUID of the Object you are searching for

MyObject found = list.stream()
.filter(MyObject -> (MyObject.getUUID()==id))
.findAny()
.orElse(null);

int i = list.indexOf(found);
int newUUID = 3;
list.get(i).setUUID(newUUID);