杰克逊提供程序同一类的多个ObjectMapper

我想序列化与两个映射器在不同的资源方法相同的类别类。
我写了两种不同的方式 CategorySerialized和CategoryTreeSerialized杰克逊提供程序同一类的多个ObjectMapper

public class MyJacksonJsonProvider implements ContextResolver<ObjectMapper> 
{ 
    private static final ObjectMapper MAPPER = new ObjectMapper(); 

    static { 
     MAPPER.enable(SerializationFeature.INDENT_OUTPUT);   
     MAPPER.registerModule(new SimpleModule() 
       .addSerializer(Category.class, new CategorySerializer(Category.class))); 
     } 

    public MyJacksonJsonProvider() { 
     System.out.println("Instantiate MyJacksonJsonProvider"); 
    } 

    @Override 
    public ObjectMapper getContext(Class<?> type) { 
     System.out.println("MyJacksonProvider.getContext() called with type: "+type); 
     return MAPPER; 
    }

系列化分类两大类，这是简单的实体分类

@Entity 
    public class Category { 

     @Id 
     @GeneratedValue(strategy = GenerationType.IDENTITY) 

     @Type(type = "objectid") 
     private String id; 
     private String name; 

     @ManyToOne 
     @JsonManagedReference 
     private Category parent; 

     @JsonBackReference 
     @OneToMany(mappedBy = "parent", fetch = FetchType.EAGER) 
     @Column(insertable = false) 
     private List<Category> children; 

     ....getter and setter .... 
    }

这是CategoryResource

@Path(value = "resource") 
public class CategoryResource { 

    @Inject 
    CategoryService categoryService; 

    @Context 
    Providers providers; 

    @GET 
    @Produces(value = MediaType.APPLICATION_JSON+";charset="+ CharEncoding.UTF_8) 
    @Path("/categories") 
    public List getCategories(){ 
     List<Category> categories = categoryService.findAll(); 
     return categories; 
    } 

    @GET 
    @Produces(value = MediaType.APPLICATION_JSON+";charset="+ CharEncoding.UTF_8) 
    @Path("/categoriestree") 
    public List getCategoriesTree(){ 
     List<Category> categories = categoryService.findAll(); 

     ContextResolver<ObjectMapper> cr = providers 
       .getContextResolver(ObjectMapper.class, MediaType.APPLICATION_JSON_TYPE); 
     ObjectMapper c = cr.getContext(ObjectMapper.class); 
     c.registerModule(new SimpleModule() 
       .addSerializer(Category.class, new CategoryTreeSerializer(Category.class))); 

     return categories; 
    }

CategorySerialized延伸StdSerializer登记与提供

MAPPER.registerModule(new SimpleModule() 
        .addSerializer(Category.class, new CategorySerializer(Category.class)));

CategoryTreeSerialized延伸StdSerializer注册的资源

ContextResolver<ObjectMapper> cr = providers 
        .getContextResolver(ObjectMapper.class, MediaType.APPLICATION_JSON_TYPE); 
      ObjectMapper c = cr.getContext(ObjectMapper.class); 
      c.registerModule(new SimpleModule() 
        .addSerializer(Category.class, new CategoryTreeSerializer(Category.class)));

内不幸的是，这并不工作，因为映射是静态的决赛。
所谓的第一资源，注册模块，然后不改变

例如，如果我称之为/categoriestree资源第一，我得到CategoryTreeSerialized系列化。
但是，如果以后我叫/类别资源总是与CategoryTreeSerialized类序列化，而不是与CategorySerialized

（反之亦然）

来源

2017-07-23 Pako

有道理。你有什么问题？ – shmosel

我想，当我打电话** getCategories **（/类别）我得到序列化** CategorySerialized **，当我打电话** getCategoryTree **（/ categoriestree）我得到序列化** CategoryTreeSerialized ** – Pako

所以使用不同的映射器。 – shmosel

不知道这可能是Spring MVC的，我的例子适用于JAX-RS，但使用谷歌搜索你应该找到一个类似的解决方案。

@GET 
@Produces(value = MediaType.APPLICATION_JSON+";charset="+ CharEncoding.UTF_8) 
@Path("/categories") 
public Response getCategories(){ 
    List<Category> categories = categoryService.findAll(); 
    ResponseBuilder response = ResponseBuilder.ok() 
      .entity(new MyCategoriesMapper() 
       .build(categories)) 
      .type(MediaType.APPLICATION_JSON)); 

    return response.build(); 
} 

@GET 
@Produces(value = MediaType.APPLICATION_JSON+";charset="+ CharEncoding.UTF_8) 
@Path("/categoriestree") 
public Response getCategoriesTree(){ 
    List<Category> categories = categoryService.findAll(); 
    ResponseBuilder response = ResponseBuilder.ok() 
      .entity(new MyCategoriesTreeMapper() 
       .build(categories)) 
      .type(MediaType.APPLICATION_JSON)); 

    return response.build(); 
}

来源

2017-07-23 11:04:25 sschrass

我发现的唯一的解决方案是创建一个包装类，使范畴类的副本和：你可以为每个Request，在身体与相应的串行连载中一样返回Response注册

我已经创建了一个名为CategoryTree类别实体的副本是实体的精确副本，我注册了CategoryTreeSerialized系列化CategoryTree模块该模块

@Provider 
@Produces(MediaType.APPLICATION_JSON) 
@Consumes(MediaType.APPLICATION_JSON) 
public class MyJacksonJsonProvider implements ContextResolver<ObjectMapper> 
{ 
    private static final ObjectMapper MAPPER_DEFAULT = new ObjectMapper(); 


    static { 
     MAPPER_DEFAULT.enable(SerializationFeature.INDENT_OUTPUT); 
     MAPPER_DEFAULT.registerModule(new SimpleModule() 
      .addSerializer(Category.class, new CategorySerializer(Category.class))); 
     MAPPER_DEFAULT.registerModule(new SimpleModule() 
      .addSerializer(CategoryTree.class, new CategoryTreeSerializer(CategoryTree.class))); 

    public MyJacksonJsonProvider() { 
     System.out.println("Instantiate MyJacksonJsonProvider"); 
    } 

    @Override 
    public ObjectMapper getContext(Class<?> type) { 
     System.out.println("MyJacksonProvider.getContext() called with 
     return MAPPER_DEFAULT; 
    } 
}

在CategoryResource类，我得到转化为列出清单副本

@GET 
@Produces(value = MediaType.APPLICATION_JSON+";charset="+ CharEncoding.UTF_8) 
@Path("/categoriestree") 
public List getCategoriesTree(){ 
    List<Category> categories = categoryService.findAll(); 

    List<CategoryTree> categoryTrees = 
      new CategoryTree().createList(categories); 
    return categoryTrees; 
}

然后CategoryTree是
类别副本这个解决方案不是很表演和优雅，但它是唯一一个作品，因为它无法注销模块

如果有人有不同，表演和优雅的解决方案写下来

来源

2017-07-23 16:36:48 Pako

杰克逊提供程序同一类的多个ObjectMapper

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