Jackson异常问题和解决方案总结

转自：http://www.baeldung.com/jackson-exception，这是我在baeldung看到的一篇关于Jackson的异常和解决的总结，觉得写的不错，自己也遇到过里面的一些问题，就翻译过来啦，收藏一下。

1. 前言

在本教程中, 我们会回顾最常见的Jackson异常 – theJsonMappingException andUnrecognizedPropertyException.

最后，我们将会简要讨论Jackson 中的no such method错误。

2. “JsonMappingException: Can not construct instance of”

2.1. The Problem

首先我们来看JsonMappingException: Can not construct instance of.

此异常是由Jackson不能完成一个对象的构造导致的 – 如果序反序列化的类是抽象类或者接口，就会导致这个异常。这里给出一个例子 – 我们想要反序列化一个Zoo对象，Zoo中有一个属性是抽象类Animal的对象：

public class Zoo {    public Animal animal;         public Zoo() { }} abstract class Animal {    public String name;         public Animal() { }} class Cat extends Animal {    public int lives;         public Cat() { }}

当我们试图将一个JSON字符串反序列化为Zoo对象时，却抛了“JsonMappingException: Can not construct instance of” 异常。代码：

@Test(expected = JsonMappingException.class)public void givenAbstractClass_whenDeserializing_thenException()   throws IOException {    String json = "{"animal":{"name":"lacy"}}";    ObjectMapper mapper = new ObjectMapper();     mapper.reader().forType(Zoo.class).readValue(json);}

完整异常如下:

com.fasterxml.jackson.databind.JsonMappingException: Can not construct instance of org.baeldung.jackson.exception.Animal,  problem: abstract types either need to be mapped to concrete types,   have custom deserializer,   or be instantiated with additional type information  at [Source: {"animal":{"name":"lacy"}}; line: 1, column: 2] (through reference chain: org.baeldung.jackson.exception.Zoo["animal"])    at c.f.j.d.JsonMappingException.from(JsonMappingException.java:148)

2.2. The Solution

可以通过在抽象类上添加@JsonDeserialize注解解决:

@JsonDeserialize(as = Cat.class)abstract class Animal {...}

3. JsonMappingException: No suitable constructor

3.1. The Problem

现在我们来看另外一个常见的异常JsonMappingException: No suitable constructor found for type.

当Jackson无法访问构造函数的时候回抛这个异常。

这里给出一个例子 –User类没有默认的构造方法:

public class User {    public int id;    public String name;     public User(int id, String name) {        this.id = id;        this.name = name;    }}

当我们试图将一个JSON字符串反序列化为User对象时，却抛异常 “JsonMappingException: No suitable constructor found” ，代码如下:

@Test(expected = JsonMappingException.class)public void givenNoDefaultConstructor_whenDeserializing_thenException()   throws IOException {    String json = "{"id":1,"name":"John"}";    ObjectMapper mapper = new ObjectMapper();     mapper.reader().forType(User.class).readValue(json);}

完整异常信息:

com.fasterxml.jackson.databind.JsonMappingException: No suitable constructor found for type[simple type, class org.baeldung.jackson.exception.User]: can not instantiate from JSON object (need to add/enable type information?) at [Source: {"id":1,"name":"John"}; line: 1, column: 2]        at c.f.j.d.JsonMappingException.from(JsonMappingException.java:148)

3.2. The Solution

解决这个问题只需要在类中添加一个默认的构造函数即可：

public class User {    public int id;    public String name;     public User() {        super();    }     public User(int id, String name) {        this.id = id;        this.name = name;    }}

这样我们再反序列化的时候就正常的反序列化了:

@Testpublic void givenDefaultConstructor_whenDeserializing_thenCorrect()   throws IOException {      String json = "{"id":1,"name":"John"}";    ObjectMapper mapper = new ObjectMapper();     User user = mapper.reader()      .forType(User.class).readValue(json);    assertEquals("John", user.name);}

4. JsonMappingException: Root name does not match expected

4.1. The Problem

接下来我们来看JsonMappingException: Root name does not match expected.

这个异常是由于JSON字符串和Jackson找到的类不匹配导致的，例如下面的JSON字符串反序列化为User时，会抛此异常：

@Test(expected = JsonMappingException.class)public void givenWrappedJsonString_whenDeserializing_thenException()  throws IOException {    String json = "{"user":{"id":1,"name":"John"}}";     ObjectMapper mapper = new ObjectMapper();    mapper.enable(DeserializationFeature.UNWRAP_ROOT_VALUE);     mapper.reader().forType(User.class).readValue(json);}

完整异常:

com.fasterxml.jackson.databind.JsonMappingException:Root name 'user' does not match expected ('User') for type [simple type, class org.baeldung.jackson.dtos.User] at [Source: {"user":{"id":1,"name":"John"}}; line: 1, column: 2]   at c.f.j.d.JsonMappingException.from(JsonMappingException.java:148)

4.2. The Solution

我们可以通过@JsonRootName注解来解决这个问题，如下：

@JsonRootName(value = "user")public class UserWithRoot {    public int id;    public String name;}

这样我们再运行反序列化时就能正常运行了:

@Testpublic void  givenWrappedJsonStringAndConfigureClass_whenDeserializing_thenCorrect()   throws IOException {      String json = "{"user":{"id":1,"name":"John"}}";     ObjectMapper mapper = new ObjectMapper();    mapper.enable(DeserializationFeature.UNWRAP_ROOT_VALUE);     UserWithRoot user = mapper.reader()      .forType(UserWithRoot.class)      .readValue(json);    assertEquals("John", user.name);}

5. JsonMappingException: No serializer found for class

5.1. The Problem

现在我们来看异常JsonMappingException: No serializer found for class.

此异常是由于我们在序列化一个对象时，此对象的属性和getter方法都是private修饰的，也就是私有的，Jackson不能访问到。

例如– 我们想序列化一个 “UserWithPrivateFields“对象:

public class UserWithPrivateFields {    int id;    String name;}

当我们序列化 “UserWithPrivateFields”的一个实例时 – 会抛出异常 “JsonMappingException: No serializer found for class” :

@Test(expected = JsonMappingException.class)public void givenClassWithPrivateFields_whenSerializing_thenException()   throws IOException {    UserWithPrivateFields user = new UserWithPrivateFields(1, "John");     ObjectMapper mapper = new ObjectMapper();    mapper.writer().writeValueAsString(user);}

完整异常:

com.fasterxml.jackson.databind.JsonMappingException: No serializer found for class org.baeldung.jackson.exception.UserWithPrivateFields and no properties discovered to create BeanSerializer (to avoid exception, disable SerializationFeature.FAIL_ON_EMPTY_BEANS) )  at c.f.j.d.ser.impl.UnknownSerializer.failForEmpty(UnknownSerializer.java:59)

5.2. The Solution

我们可以通过配置ObjectMapper的可见性来解决这个问题，例如：

@Testpublic void givenClassWithPrivateFields_whenConfigureSerializing_thenCorrect()   throws IOException {      UserWithPrivateFields user = new UserWithPrivateFields(1, "John");     ObjectMapper mapper = new ObjectMapper();    mapper.setVisibility(PropertyAccessor.FIELD, Visibility.ANY);     String result = mapper.writer().writeValueAsString(user);    assertThat(result, containsString("John"));}

或者通过@JsonAutoDetect注解解决：

@JsonAutoDetect(fieldVisibility = Visibility.ANY)public class UserWithPrivateFields { ... }

当然，如果如果我们可以修改该实体类的源码，也可以通过添加getter方法解决，这个方法应该是首选的方法。

6. JsonMappingException: Can not deserialize instance of

6.1. The Problem

接下来看异常JsonMappingException: Can not deserialize instance of.

这个异常是由于在反序列化时类型使用不当导致的，例如我们想反序列化一个List<User>对象：

@Test(expected = JsonMappingException.class)public void givenJsonOfArray_whenDeserializing_thenException()   throws JsonProcessingException, IOException {      String json       = "[{"id":1,"name":"John"},{"id":2,"name":"Adam"}]";    ObjectMapper mapper = new ObjectMapper();    mapper.reader().forType(User.class).readValue(json);}

完整异常:

com.fasterxml.jackson.databind.JsonMappingException:Can not deserialize instance of   org.baeldung.jackson.dtos.User out of START_ARRAY token  at [Source: [{"id":1,"name":"John"},{"id":2,"name":"Adam"}]; line: 1, column: 1]  at c.f.j.d.JsonMappingException.from(JsonMappingException.java:148)

6.2. The Solution

可以通过更改User 为List<User> –例如:

@Testpublic void givenJsonOfArray_whenDeserializing_thenCorrect()   throws JsonProcessingException, IOException {      String json      = "[{"id":1,"name":"John"},{"id":2,"name":"Adam"}]";        ObjectMapper mapper = new ObjectMapper();    List<User> users = mapper.reader()      .forType(new TypeReference<List<User>>() {})      .readValue(json);     assertEquals(2, users.size());}

7. UnrecognizedPropertyException

7.1. The Problem

接下来我们来看UnrecognizedPropertyException.

此异常是由于我们在反序列化时，JSON字符串中出现了未知的属性导致的，比如我们想反序列化一个JSON，此JSON中带有一个额外的属性checked：

@Test(expected = UnrecognizedPropertyException.class)public void givenJsonStringWithExtra_whenDeserializing_thenException()   throws IOException {      String json = "{"id":1,"name":"John", "checked":true}";     ObjectMapper mapper = new ObjectMapper();    mapper.reader().forType(User.class).readValue(json);}

完整异常:

com.fasterxml.jackson.databind.exc.UnrecognizedPropertyException:Unrecognized field "checked" (class org.baeldung.jackson.dtos.User), not marked as ignorable (2 known properties: "id", "name"]) at [Source: {"id":1,"name":"John", "checked":true}; line: 1, column: 38] (through reference chain: org.baeldung.jackson.dtos.User["checked"])  at c.f.j.d.exc.UnrecognizedPropertyException.from(UnrecognizedPropertyException.java:51)

7.2. The Solution

我们可以通过下面的方法配置ObjectMapper对象来解决这个问题:

@Testpublic void givenJsonStringWithExtra_whenConfigureDeserializing_thenCorrect()   throws IOException {      String json = "{"id":1,"name":"John", "checked":true}";     ObjectMapper mapper = new ObjectMapper();    mapper.disable(DeserializationFeature.FAIL_ON_UNKNOWN_PROPERTIES);     User user = mapper.reader().forType(User.class).readValue(json);    assertEquals("John", user.name);}

或者是通过@JsonIgnoreProperties注解来解决，可以忽略未知的属性:

@JsonIgnoreProperties(ignoreUnknown = true)public class User {...}

8. JsonParseException: Unexpected character (”’ (code 39))

8.1. The Problem

接下来我们讨论JsonParseException: Unexpected character (”’ (code 39)).

当我们反序列化的JSON中包含了单引号而不是双引号的时候，会抛此异常，比如：

@Test(expected = JsonParseException.class)public void givenStringWithSingleQuotes_whenDeserializing_thenException()   throws JsonProcessingException, IOException {      String json = "{'id':1,'name':'John'}";    ObjectMapper mapper = new ObjectMapper();     mapper.reader()      .forType(User.class).readValue(json);}

完整异常:

com.fasterxml.jackson.core.JsonParseException:Unexpected character (''' (code 39)):   was expecting double-quote to start field name  at [Source: {'id':1,'name':'John'}; line: 1, column: 3]  at c.f.j.core.JsonParser._constructError(JsonParser.java:1419)

8.2. The Solution

我们也可以通过配置ObjectMapper 来实现对单引号的兼容:

@Testpublic void  givenStringWithSingleQuotes_whenConfigureDeserializing_thenCorrect()   throws JsonProcessingException, IOException {      String json = "{'id':1,'name':'John'}";     JsonFactory factory = new JsonFactory();    factory.enable(JsonParser.Feature.ALLOW_SINGLE_QUOTES);    ObjectMapper mapper = new ObjectMapper(factory);     User user = mapper.reader().forType(User.class)      .readValue(json);      assertEquals("John", user.name);}

9. Jackson NoSuchMethodError

接下来我们快速的讨论一下 “No such method” errors.

当抛了java.lang.NoSuchMethodError 异常时, 通常是因为你又多个Jackson的jar包导致的。不只是Jackson，其他的时候此异常也有可能是因为jar包的冲突导致。

完整的异常:

java.lang.NoSuchMethodError:com.fasterxml.jackson.core.JsonParser.getValueAsString()Ljava/lang/String; at c.f.j.d.deser.std.StringDeserializer.deserialize(StringDeserializer.java:24)

10. Conclusion

在这篇文章中，我们深一层解析了常见的Jackson中的异常和错误，并且对每一个异常和错误分析了原因和解决方案。以上所有的代码都可以在Github上找到，这是一个基于Maven的工程，很容易就可以导入并且运行拉。