jackson - 有没有办法在运行时避免父实体的序列化？ [英] jackson - Is there any way to avoid serialization of parent entity at runtime?

问题描述

假设我有两个简单的类

@JsonFilter("filter properties by name")
public class Foo
{
    private Integer id;
    private String name;
}

@JsonFilter("filter properties by name")
public class Bar
{
    private Integer id;
    private String name;
    private Foo foo;
}

我想序列化 Bar的实例其中 foo 字段只有其id。这一切都应该在运行时完成。
我尝试使用过滤器

I'd like to serialize an instance of Bar in which the foo field has only its id. It all should be done at runtime. I tried to do this using filters

FilterProvider filter = new SimpleFilterProvider().addFilter(
            "filter properties by name", SimpleBeanPropertyFilter
                .serializeAllExcept(/*name of the field to exclude*/));
objectMapper.writer(filter).writeValuAsString(bar);

但是，这样做，我必须手动排除父类的所有字段;此外，如果此字段之一与子类的字段具有相同的名称，则它们都被排除。
在我的示例中，我不能以这种方式排除字段 name ，因为它也影响字段 name Bar 类。

However, in doing so, I have to manually exclude all the fields of the parent class; also, if one of this field has the same name of a field of the child class, they are both excluded. In my example, I can't exclude the field name in this way because it impacts also the field name of the Bar class.

那么，我该如何以最简洁/优雅的方式解决？
有没有类似上面的代码，可能使用点符号或类似的东西？

So, how can I solve in a most concise/elegant way? Is there something similar to the above code, maybe using dot notations or something like that?

例如。直到现在在我的示例过滤器中，我能够写出类似 [...]。serializeAllExcept（name，foo））; ，但它很棒拥有 [...]。serializeAllExcept（foo.name））; 或类似。

E.g. until now in my example filter I'm able to write something like [...].serializeAllExcept("name", "foo"));, but it was great to have [...].serializeAllExcept("foo.name")); or similar.

推荐答案

您不需要任何过滤器。 Foo和Bar类没有变化。

You don't need any filters for it. No changes in Foo and Bar classes.

新的MixInFoo类：

New MixInFoo class:

public class MixInFoo {
    @JsonProperty("mixinFooId")
    private Integer id;
    @JsonIgnore
    private String name;

}

我更改了'id'属性名称只是为了说明您可以完全更改响应而无需修改原始Foo类。

I've changed 'id' property name just to illustrate that you can change response completely without modifying original Foo class.

ObjectMapper objectMapper = new ObjectMapper();
objectMapper.getSerializationConfig().addMixInAnnotations(Foo.class, MixInFoo.class);
objectMapper.getDeserializationConfig().addMixInAnnotations(Foo.class, MixInFoo.class);

String result = objectMapper.writeValueAsString(json);

您需要注册MixIn类，如图所示。

You will need to register MixIn classes as shown.

******* 基于过滤器的实施 ***********

******* Filter based implementation ***********

是的，你可以使用Filter来实现相同的结果。您需要将@JsonFilter添加到Foo类并将其命名为FooFilter。然后你可以添加仅适用于Foo类的Filter：

Yes, you can use Filter to achieve the same result. You need to add @JsonFilter to the Foo class and name it like "FooFilter". Then you can add Filter that will apply only to the Foo class:

@JsonFilter("FooFilter")
public class Foo {
    private Integer id;
    private String name;
}

public class Bar {
    private Integer id;
    private String name;
    private Foo foo;
}

public static void main(String []args) throws IOException {
    ObjectMapper objectMapper = new ObjectMapper();

    String []fooIgnore = {"name"};
    SimpleBeanPropertyFilter propertyFilter = SimpleBeanPropertyFilter.serializeAllExcept(fooIgnore);
    FilterProvider filterProvider = new SimpleFilterProvider().addFilter("FooFilter", propertyFilter);

    ObjectWriter objectWriter = objectMapper.writer(filterProvider);
    String result = objectWriter.writeValueAsString(json);

    System.out.println(result);
}

在该实现中，您不需要将自定义SimpleBeanProvider扩展和实现为仅将过滤器应用于Foo类。

In that implementation you don't need to extend and implement you custom SimpleBeanProvider to apply the filter only to the Foo class.

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