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        JSONObject toJSONString错误的解决

        时间:2023-12-10

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                  当使用Java中的JSONObject类的toJSONString()方法将Java对象序列化为Json字符串时,可能会出现错误。本文将提供一些错误的解决方法。

                  错误1:No serializer found

                  当使用toJSONString()方法将Java对象序列化为Json字符串时,可能会出现以下错误:

                  org.codehaus.jackson.map.JsonMappingException: No serializer found for class com.example.MyObject and no properties discovered to create BeanSerializer (to avoid exception, disable SerializationConfig.Feature.FAIL_ON_EMPTY_BEANS) ) (through reference chain: com.example.MyObject["property"])
                  

                  造成这个错误的原因是JSONObject不知道如何序列化您的Java对象。解决方法是让JSONObject可以序列化您的Java对象。可以使用注解@JsonAutoDetect和@JsonInclude,例如:

                  @JsonAutoDetect(fieldVisibility = JsonAutoDetect.Visibility.ANY)
                  @JsonInclude(JsonInclude.Include.NON_NULL)
                  public class MyObject {
                      private String property;
                      // getters and setters
                  }
                  

                  其中,@JsonAutoDetect指示Jackson使用默认可见性规则来检测和序列化Java类的字段,@JsonInclude指示Jackson只序列化非空值。

                  错误2:com.fasterxml.jackson.databind.JsonMappingException

                  当使用toJSONString()方法将Java对象序列化为Json字符串时,可能会出现以下错误:

                  com.fasterxml.jackson.databind.JsonMappingException: No serializer found for class com.example.MyObject and no properties discovered to create BeanSerializer (to avoid exception, disable SerializationFeature.FAIL_ON_EMPTY_BEANS) (through reference chain: com.example.MyObject["property"])
                  

                  这个错误的解决方法与错误1相同。您可以使用Jackson JSON库注释和配置来确保JSONObject可以序列化您的Java对象。

                  @JsonAutoDetect(fieldVisibility = JsonAutoDetect.Visibility.ANY)
                  @JsonInclude(JsonInclude.Include.NON_NULL)
                  public class MyObject {
                      private String property;
                      // getters and setters
                  }
                  

                  使用这些注释和配置可以让toJSONString()方法正确地序列化您的Java对象。

                  示例1:

                  下面是一个简单的Java对象,该对象具有一个名为“name”的字符串属性:

                  public class SimpleObject {
                      private String name;
                  
                      public SimpleObject(String name) {
                          this.name = name;
                      }
                  
                      public String getName() {
                          return name;
                      }
                  
                      public void setName(String name) {
                          this.name = name;
                      }
                  }
                  

                  如果尝试将此对象序列化为Json字符串,并将Json字符串解析回Java对象,就会出现上述错误。

                  我们需要添加相关注释和Java配置:

                  @JsonAutoDetect(fieldVisibility = JsonAutoDetect.Visibility.ANY)
                  @JsonInclude(JsonInclude.Include.NON_NULL)
                  public class SimpleObject {
                      private String name;
                  
                      public SimpleObject(String name) {
                          this.name = name;
                      }
                  
                      public String getName() {
                          return name;
                      }
                  
                      public void setName(String name) {
                          this.name = name;
                      }
                  }
                  

                  示例2:

                  下面是另一个Java对象,该对象具有一个名为“salary”的整数属性:

                  public class Employee {
                      private String name;
                      private int salary;
                  
                      public Employee(String name, int salary) {
                          this.name = name;
                          this.salary = salary;
                      }
                  
                      public String getName() {
                          return name;
                      }
                  
                      public void setName(String name) {
                          this.name = name;
                      }
                  
                      public int getSalary() {
                          return salary;
                      }
                  
                      public void setSalary(int salary) {
                          this.salary = salary;
                      }
                  }
                  

                  如果尝试将此对象序列化为Json字符串,并将Json字符串解析回Java对象,就会出现上述错误。

                  我们需要添加相关注释和Java配置:

                  @JsonAutoDetect(fieldVisibility = JsonAutoDetect.Visibility.ANY)
                  @JsonInclude(JsonInclude.Include.NON_NULL)
                  public class Employee {
                      private String name;
                      private int salary;
                  
                      public Employee(String name, int salary) {
                          this.name = name;
                          this.salary = salary;
                      }
                  
                      public String getName() {
                          return name;
                      }
                  
                      public void setName(String name) {
                          this.name = name;
                      }
                  
                      public int getSalary() {
                          return salary;
                      }
                  
                      public void setSalary(int salary) {
                          this.salary = salary;
                      }
                  }
                  

                  使用这些注释和配置可以让toJSONString()方法正确地序列化您的Java对象。

                  上一篇:Java如何取掉json数据中值为null的属性字段 下一篇:SQLite教程(七):数据类型详解

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