使用jackson忽略json body中的空元素

Question

我正在用java中的jackson构建一个json体。它会像下面这样

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

基于不同的 REST URI，json 主体必须忽略某些字段或元素。如何使用杰克逊忽略上述 json 正文中的“学生”部分？当我忽略它时，我应该只能得到

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

但我得到它如下，这是不正确的-

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

我有两个带有getters和setters的类，一个作为主语，另一个作为学生。我尝试使用@JsonIgnore，但是它忽略了所有的URI，这是我不想要的。我也试过@JsonInclude(包含。非空)。我如何实现这一点？

我在这里添加我的代码。基于不同的REST URI，json主体必须忽略一些字段或元素。例如，对于一个URI，它应该包括所有字段，对于另一个URI，它应该忽略学生。分数=新分数（）；

    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

在ObjectMapper中将WRITE_NULL_MAP_VALUES设置为false。

代码：

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

Answer 2

编辑

这些帖子似乎是重复的。

编辑2

使用@JsonView而不是@JsonFilter

下面是使用 @JsonView 打印同一 JSON 对象的不同视图的示例代码：

{ "科目":"数学"，"分数":" 100"}

和

{“主题”：“数学”，“标记”：“100”，“学生”：{“名称”：“x”，“类”：“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 3

这可以在注释上使用< code>JsonInclude进行配置，其值为< code>NON_NULL而不是< code>NON_EMPTY。

对于NON_EMPTY和NON_NULL的Javadoc还不太清楚，但是我的实验表明，使用NON_NULL可以像您期望的那样使用Jackson 2.6.2。

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

示例

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")));

输出

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

注意这也可以使用全局设置

mapper.setSerializationInclusion(JsonInclude.Include.NON_NULL);