Jackson异常问题和解决方案总结

小灰灰 2021-12-09 03:59 1100阅读 0赞

[https://blog.csdn.net/u010429286/article/details/78395775][https_blog.csdn.net_u010429286_article_details_78395775]

转自：[http://www.baeldung.com/jackson-exception][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:

@Test  
public 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:

@Test  
public 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：

@Test  
public 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:

@Test  
public 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:

@Test  
public 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:

@Test  
public 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。  
\---------------------   
作者：我是康小小   
来源：CSDN   
原文：https://blog.csdn.net/u010429286/article/details/78395775   
版权声明：本文为博主原创文章，转载请附上博文链接！

[https_blog.csdn.net_u010429286_article_details_78395775]: https://blog.csdn.net/u010429286/article/details/78395775
[http_www.baeldung.com_jackson-exception]: http://www.baeldung.com/jackson-exception