java - 为什么 Jackson 反序列化 json 无法按顺序打印对象数组？

Question

pom.xml

<dependencies>

        <dependency>
            <groupId>com.google.code.gson</groupId>
            <artifactId>gson</artifactId>
            <version>2.8.5</version>
        </dependency>

        <dependency>
            <groupId>com.fasterxml.jackson.core</groupId>
            <artifactId>jackson-databind</artifactId>
            <version>2.8.3</version>
        </dependency>

</dependencies>

pojo类:Function.java

public class Function{

private Integer id;

private Set<MenuToFunction> menuToFunctions;

public Integer getId(){
return this.id;
}

public void setId(Integer id){
this.id=id;
}

public Set<MenuToFunction> getMenuToFunctions(){
return this.menuToFunctions;
}

    public void setMenuToFunctions(Set<MenuToFunction> menuToFunctions){
    this.menuToFunctions=menuToFunctions;
    }
}

pojo类:MenuToFunction.java

public class MenuToFunction{

    private Integer id;

    public Integer getId(){
    return this.id;
    }

    public void setId(Integer id){
    this.id=id;
    }
}

我的 json 文件，请注意这些 id 是从 1 到 6。

{
    "id": 1,
    "menuToFunctions": [
        {
            "id": 1
        },
        {
            "id": 2
        },
        {
            "id": 3
        },
        {
            "id": 4
        },
        {
            "id": 5
        },
        {
            "id": 6
        }
    ]
}

最后是我的测试代码。我使用 jackson 和 gson 来反序列化 json。

public class ReadJsonFile{

    public static void main(String[] args){

    try{
        String fileAbsolutePath="T:/GetFunctionDetailResponse.json";

        Function function=new ObjectMapper().readValue(new File(fileAbsolutePath),Function.class);

        for(MenuToFunction menuToFunction:function.getMenuToFunctions()){

        System.out.println("menuToFunction.getMenu().getId() with jackson:"+menuToFunction.getId());
        }

        function=new GsonBuilder().create().fromJson(new FileReader(fileAbsolutePath),Function.class);

        for(MenuToFunction menuToFunction:function.getMenuToFunctions()){

        System.out.println("menuToFunction.getMenu().getId() with gson:"+menuToFunction.getId());
        }

    }catch(Exception e){

        e.printStackTrace();
    }
    }
}

控制台输出

menuToFunction.getMenu().getId() with jackson:5
menuToFunction.getMenu().getId() with jackson:3
menuToFunction.getMenu().getId() with jackson:6
menuToFunction.getMenu().getId() with jackson:2
menuToFunction.getMenu().getId() with jackson:4
menuToFunction.getMenu().getId() with jackson:1
menuToFunction.getMenu().getId() with gson:1
menuToFunction.getMenu().getId() with gson:2
menuToFunction.getMenu().getId() with gson:3
menuToFunction.getMenu().getId() with gson:4
menuToFunction.getMenu().getId() with gson:5
menuToFunction.getMenu().getId() with gson:6

那么为什么 Jackson 反序列化 json 无法按顺序打印对象数组，但 gson 可以？

Jackson 有配置吗？

我也尝试过最新版本jackson-2.10.3，但仍然有同样的问题。

感谢您的帮助。

Answer 1

有一个属性可以提供帮助:(来自 jackson 的documentation.)

@JsonPropertyOrder({ "id", "menuToFunctions" })
public class Function{

private Integer id;

private Set<MenuToFunction> menuToFunctions;

public Integer getId(){
return this.id;
}

public void setId(Integer id){
this.id=id;
}

public Set<MenuToFunction> getMenuToFunctions(){
return this.menuToFunctions;
}

    public void setMenuToFunctions(Set<MenuToFunction> menuToFunctions){
    this.menuToFunctions=menuToFunctions;
    }
}

还包括在 ObjectMapper 中排序:

public class ReadJsonFile{

    public static void main(String[] args){

    try{
        String fileAbsolutePath="T:/GetFunctionDetailResponse.json";

        Function function=new ObjectMapper().configure(SerializationFeature.ORDER_MAP_ENTRIES_BY_KEYS, true)
            .configure(MapperFeature.SORT_PROPERTIES_ALPHABETICALLY, true)
.readValue(new File(fileAbsolutePath),Function.class);

        for(MenuToFunction menuToFunction:function.getMenuToFunctions()){

        System.out.println("menuToFunction.getMenu().getId() with jackson:"+menuToFunction.getId());
        }

        function=new GsonBuilder().create().fromJson(new FileReader(fileAbsolutePath),Function.class);

        for(MenuToFunction menuToFunction:function.getMenuToFunctions()){

        System.out.println("menuToFunction.getMenu().getId() with gson:"+menuToFunction.getId());
        }

    }catch(Exception e){

        e.printStackTrace();
    }
    }
}