Jackson: Serialize only 10 Properties out of 100 - jackson

Question

Jackson by default includes all public getters. But if I have hundreds of getters and I want to serialize just few of them, how can I achieve this?
I don't want to use #JsonIgnore annotation to specify all others hundred properties.

Answer 1

You can use Jackson's views or filters to select the serialization form of your object. Here is an example using the filter which picks up the properties with the given names:
public class JacksonFilter {
#JsonFilter("filter")
public static class Bean {
private final String field1;
private final String field12;
private final String field10;
private final String field100;
public Bean(String field1, String field12, String field10, String field100) {
this.field1 = field1;
this.field12 = field12;
this.field10 = field10;
this.field100 = field100;
}
public String getField1() {
return field1;
}
public String getField12() {
return field12;
}
public String getField10() {
return field10;
}
public String getField100() {
return field100;
}
}
public static void main(String[] args) throws JsonProcessingException {
ObjectMapper mapper = new ObjectMapper();
SimpleFilterProvider filters = new SimpleFilterProvider();
filters.addFilter("filter",
SimpleBeanPropertyFilter.serializeAllExcept("field12", "field100"));
mapper.setFilters(filters);
Bean bean = new Bean("A", "B", "C", "D");
System.out.println(mapper.writerWithDefaultPrettyPrinter().writeValueAsString(bean));
}
}
Output:
{
"field1" : "A",
"field10" : "C"
}