没有注释的jackson过滤器属性

Question

public Class User {
    private String name;
    private Integer age;
    ...
}

ObjectMapper om = new ObjectMapper();
om.writeValueAsString(user);

如何在不使用@JsonIgnore等任何注释的情况下过滤属性？

Answer 1

你有两种可能的方式使用杰克逊

Mixin批注: - http://www.cowtowncoder.com/blog/archives/2009/08/entry_305.html

JSON过滤器: - http://wiki.fasterxml.com/JacksonFeatureJsonFilter

Answer 2

按名称排除属性的示例：

public Class User {
    private String name = "abc";
    private Integer age = 1;
    //getters
}

@JsonFilter("dynamicFilter")
public class DynamicMixIn {
}

User user = new User();
String[] propertiesToExclude = {"age"};
ObjectMapper mapper = new ObjectMapper()
      .addMixIn(Object.class, DynamicMixIn.class);
FilterProvider filterProvider = new SimpleFilterProvider()
                .addFilter("dynamicFilter", SimpleBeanPropertyFilter.serializeAllExcept(propertiesToExclude));
        mapper.setFilterProvider(filterProvider);

mapper.writeValueAsString(user); // {"name":"abc"}

您可以代替DynamicMixIn创建MixInByPropName

@JsonIgnoreProperties(value = {"age"})
public class MixInByPropName {
}

ObjectMapper mapper = new ObjectMapper()
      .addMixIn(Object.class, MixInByPropName.class);

mapper.writeValueAsString(user); // {"name":"abc"}

注意：如果您只想排除属性，User可以将Object.class方法的参数更改addMixIn为User.class

按您可以创建的类型排除属性MixInByType

@JsonIgnoreType
public class MixInByType {
}

ObjectMapper mapper = new ObjectMapper()
      .addMixIn(Integer.class, MixInByType.class);

mapper.writeValueAsString(user); // {"name":"abc"}