我们在代码中使用Jackson JSON映射器来反序列化某些配置对象。当特定字段缺失或空白时,我们希望杰克逊在反序列化时失败
杰克逊支持这种行为的唯一特征是基元:
final DeserializationConfig.Feature failOnPremitives = DeserializationConfig.Feature.FAIL_ON_NULL_FOR_PRIMITIVES;
事情是有问题的字段主要是字符串
非常感谢任何帮助
答案 0 :(得分:1)
有一个名为FAIL_ON_NULL_FOR_CREATOR_PARAMETERS
的选项。
所以我认为它可以在DeserializationConfig.Feature.FAIL_ON_NULL_FOR_CREATOR_PARAMETERS;
或yml
:
jackson:
serialization:
indent-output: false
deserialization:
fail-on-unknown-properties: true
fail-on-missing-creator-properties: true
fail-on-null-creator-properties: true
这适用于所有类型,字符串,整数,双打等。
答案 1 :(得分:0)
您考虑过Bean Validation吗?
虽然杰克逊专注于JSON解析,但Bean验证就是在bean上声明并执行验证。
您可以使用来自@NotNull
的@NotBlank
或Hibernate Validator,即Bean验证参考实现。
或者您可以使用JSON Schema。
答案 2 :(得分:0)
必须创建特定于对象的自定义反序列化器。
import com.fasterxml.jackson.core.JsonParser;
import com.fasterxml.jackson.core.JsonProcessingException;
import com.fasterxml.jackson.core.ObjectCodec;
import com.fasterxml.jackson.core.Version;
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 java.io.IOException;
class JacksonDeserializerTest {
public static void main(String[] args) throws Exception {
ObjectMapper mapper = new ObjectMapper();
SimpleModule module = new SimpleModule("CustomPersonDeserializer", new Version(1, 0, 0, null, null, null));
module.addDeserializer(Person.class, new CustomPersonDeserializer());
mapper.registerModule(module);
String jsonString = "{ \"id\": 1, \"name\": \"User 1 \"}";
Person user = mapper.readValue(jsonString, Person.class);
System.out.println("User: " + user.toString());
jsonString = "{ \"id\": 1}";
user = mapper.readValue(jsonString, Person.class);
}
static class CustomPersonDeserializer extends StdDeserializer<Person> {
private static final long serialVersionUID = -4100181951833318756L;
public CustomPersonDeserializer() {
this(null);
}
public CustomPersonDeserializer(Class<?> vc) {
super(vc);
}
@Override
public Person deserialize(JsonParser parser, DeserializationContext deserializer) throws IOException, JsonProcessingException {
Person person = new Person();
ObjectCodec codec = parser.getCodec();
JsonNode node = codec.readTree(parser);
JsonNode idNode = node.get("id");
int id = idNode.asInt();
person.setId(id);
JsonNode nameNode = node.get("name");
if(nameNode == null){
throw new IOException("name must be provided");
}
String name = nameNode.asText();
if (name.trim().length() < 1){
throw new IOException("name can not be empty");
}
person.setName(name);
return person;
}
}
static class Person {
private int id;
private String name;
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
public int getId() {
return id;
}
public void setId(int id) {
this.id = id;
}
@Override
public String toString() {
return "Person{" +
"name='" + name + '\'' +
", id=" + id +
'}';
}
}
}
答案 3 :(得分:0)
替代方法:
这同样适用于对象
ObjectMapper mapper= new ObjectMapper();
mapper.configure(DeserializationFeature.FAIL_ON_UNKNOWN_PROPERTIES, false);