忽略來自json body的使用jackson的空元素

Question

我正在用java中的jackson構造一個json體。它會像下面

{ 
    "subject": "math", 
    "marks": "100", 
    "student":{ 
     "name": "x", 
     "class": "8" 
    } 
} 

基於不同REST URI的JSON的身體已經忽略某些字段或元素。我如何使用jackson忽略上述json主體中的「學生」部分？ 當我不理它，我應該能夠得到的只有

{ "subject": "math", "marks": "100"} 

，但我得到它下面是引線不

{ "subject": "math", "marks": "100","student":{}} 

我有兩個班，getter和setter，一個爲主體另一個是學生。我嘗試使用@JsonIgnore，但它忽略了所有的URI，我不想要。我也試過@JsonInclude（Include.NON_EMPTY）。我如何實現這一目標？

我在這裏添加我的代碼。基於不同的REST URI，json主體必須忽略一些字段或元素。例如，對於一個URI，它應該包含所有字段，對於另一個URI則應該忽略學生。 分數=新分數（）;

score.setSubject("math"); 
    score.setMarks("100"); 

    Score.Student student =score.new Student(); 
    score.setStudent(student); 

    switch (type) { 
    case StudentAdd: 
     score.setSubject("math"); 
     score.setMarks("100"); 
     break; 
    case StudentDelete: 
     score.setSubject("math"); 
     score.setMarks("100"); 
     break; 
    case StudentComplete: 
     score.setChangeReason("C"); 

    default: 
     break; 
     score.setSubject("math"); 
     score.setMarks("100"); 

    score.setStudent(student); 

    ObjectMapper objectMapper = new ObjectMapper(); 
    objectMapper.configure(SerializationFeature.INDENT_OUTPUT, true); 
    //objectMapper.setSerializationInclusion(Include.NON_NULL); 
    objectMapper.setSerializationInclusion(Include.NON_EMPTY); 

    StringWriter jsonBody = new StringWriter(); 
    objectMapper.writeValue(jsonBody, score); 

    return jsonBody.toString(); 
} 

    //@JsonInclude(Include.NON_EMPTY) 
    class Score { 

     private String subject; 
     private String marks; 
     private Student student; 

     public String getSubject() { 
      return subject; 
     } 

     public void setSubject(String subject) { 
      this.subject = subject; 
     } 

     public String getMarks() { 
      return marks; 
     } 

     public void setMarks(String marks) { 
      this.marks = marks; 
     } 

     public Student getStudent() { 
      return student; 
     } 

     public void setStudent(Student student) { 
      this.student = student; 
     } 

    //@JsonInclude(Include.NON_NULL) 
    class Student { 

     private String name; 

     @JsonProperty("class") 
     private String clazz; 

     public String getName() { 
      return name; 
     } 

     public void setName(String name) { 
      this.name = name; 
     } 

     public String getClazz() { 
      return clazz; 
     } 

     public void setClazz(String clazz) { 
      this.clazz = clazz; 
     } 

} 
    } 
    } 
    } 

Answer 1

EDIT

此帖子似乎是一個duplicate。

編輯2

使用@JsonView代替@JsonFilter

下面是使用@JsonView打印相同JSON對象的不同視圖的示例代碼：

{「受試者「：」math「，」marks「：」100「}

and

{ 「主題」： 「數學」， 「標誌」： 「100」， 「學生」：{ 「名」： 「X」， 「階級」： 「8」}}

public class Views { 
    public static class Filtered {} 
    public static class All extends Filtered {} 
} 

public class Student { 
    private String name; 

    @JsonProperty("class") 
    private String clazz; 

    public String getName() { 
     return name; 
    } 

    public void setName(String name) { 
     this.name = name; 
    } 

    public String getClazz() { 
     return clazz; 
    } 

    public void setClazz(String clazz) { 
     this.clazz = clazz; 
    } 
} 

public class Score { 
    @JsonView(Views.Filtered.class) 
    private String subject; 

    @JsonView(Views.Filtered.class) 
    private String marks; 

    @JsonView(Views.All.class) 
    private Student student; 

    public String getSubject() { 
     return subject; 
    } 

    public void setSubject(String subject) { 
     this.subject = subject; 
    } 

    public String getMarks() { 
     return marks; 
    } 

    public void setMarks(String marks) { 
     this.marks = marks; 
    } 

    public Student getStudent() { 
     return student; 
    } 

    public void setStudent(Student student) { 
     this.student = student; 
    } 
} 

public class JsonTest { 
    public static void main(String[] args) throws JsonProcessingException { 
     Student student = new Student(); 
     student.setName("x"); 
     student.setClazz("8"); 

     Score score = new Score(); 
     score.setSubject("math"); 
     score.setMarks("100"); 
     score.setStudent(student); 

     ObjectMapper mapper = new ObjectMapper(); 

     //do not serialize student property 
     System.out.println(mapper.writerWithView(Views.Filtered.class).writeValueAsString(score)); 

     //also serialize student property 
     System.out.println(mapper.writeValueAsString(score)); 
    } 
} 

Answer 2

在您的ObjectMapper中將WRITE_NULL_MAP_VALUES設置爲false。

代碼：

ObjectMapper mapper = new ObjectMapper(); 
    mapper.configure(SerializationConfig.Feature.WRITE_NULL_MAP_VALUES, false); 

Answer 3

這可以使用與JsonInclude的NON_NULL而非NON_EMPTY值註釋進行配置。

NON_EMPTY與NON_NULL的Javadoc不太清晰，但我的實驗顯示，使用NON_NULL的效果與您預期的那樣使用Jackson 2.6.2。

@JsonProperty 
@JsonInclude(Include.NON_NULL) 
private Student student; 

例

ObjectMapper mapper = new ObjectMapper(); 
System.out.println(mapper.writeValueAsString(new Foo(100, new Student("bob"), "math"))); 
System.out.println(mapper.writeValueAsString(new Foo(100, null, "math"))); 

輸出

{"marks":100,"subject":"math","student":{"name":"bob"}} 
{"marks":100,"subject":"math"} 

注意這也可以使用全局設置

mapper.setSerializationInclusion(JsonInclude.Include.NON_NULL);