假设我们有简单的json字符串json={“key1”：“value1”，“key2”：“value2”}和java类
class Foo {
private String field1;
private Integer field2;
//setter & getter
}
此外，我们不想改变< code>Foo类。注意，json键与Foo的字段名不匹配。有没有简单的方法可以让我们用Jackson或者其他库将json字符串解复用到< code>Foo类？
您可以使用以下json库并构建自定义反序列化器，如下所示。
杰克逊注释-2.10.4， 杰克逊核心-2.10.4
，
杰克逊.数据绑定-2.10.4
import com.fasterxml.jackson.core.JsonParser;
import com.fasterxml.jackson.core.JsonProcessingException;
import com.fasterxml.jackson.databind.DeserializationContext;
import com.fasterxml.jackson.databind.JsonNode;
import com.fasterxml.jackson.databind.ObjectMapper;
import com.fasterxml.jackson.databind.deser.std.StdDeserializer;
import com.fasterxml.jackson.databind.module.SimpleModule;
import com.fasterxml.jackson.databind.node.IntNode;
import java.io.IOException;
public class FooDeserializer extends StdDeserializer<Foo> {
public static void main (String [] args) throws JsonProcessingException {
String json = "{\"key1\":\"value1\", \"key2\":100}";
ObjectMapper mapper = new ObjectMapper();
SimpleModule module = new SimpleModule();
module.addDeserializer(Foo.class, new FooDeserializer());
mapper.registerModule(module);
Foo foo = mapper.readValue(json, Foo.class);
System.out.println(foo);
}
public FooDeserializer() {
this(null);
}
public FooDeserializer(Class<?> vc) {
super(vc);
}
@Override
public Foo deserialize(JsonParser jp, DeserializationContext ctx)
throws IOException, JsonProcessingException {
JsonNode node = jp.getCodec().readTree(jp);
String field1 = node.get("key1").asText();
int field2 = (Integer) ((IntNode) node.get("key2")).numberValue();
return new Foo(field1,field2);
}
}
