CODEWITHSUNDEEP
javadeserializationfastjson
Vineel
Vineel

Reputation: 1788

Deserialization fails when there is no default constructor

I 'm trying to use Fastjson library for JSON serialization. When I try to deserialize , it fails showing no default constructor error. Note: My class here is a toy example. I realty, it contains so many references to other classes which are in other maven projects and its practically not possible to modify every class.

Here is the code.

Student s = new Student("vineel", "20");
String hell = JSON.toJSONString(s);
Student model2 = JSON.parseObject(hell, Student.class);
System.out.println(model2);

public class Student {

private String name;
private String age;
Student(String name,String age){
this.name = name;
this.age = age;
}
@override
public String toString() {
return "Student [name=" + name + ", age=" + age + "]";
}
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
public String getAge() {
return age;
}
public void setAge(String age) {
this.age = age;
}

}

Here is the error:

Exception in thread "main" com.alibaba.fastjson.JSONException: default constructor not found. class com.alibaba.fastjson.Student
at com.alibaba.fastjson.util.JavaBeanInfo.build(JavaBeanInfo.java:467)
at com.alibaba.fastjson.util.JavaBeanInfo.build(JavaBeanInfo.java:213)
at com.alibaba.fastjson.parser.ParserConfig.createJavaBeanDeserializer(ParserConfig.java:656)
at com.alibaba.fastjson.parser.ParserConfig.getDeserializer(ParserConfig.java:573)
at com.alibaba.fastjson.parser.ParserConfig.getDeserializer(ParserConfig.java:386)
at com.alibaba.fastjson.parser.DefaultJSONParser.parseObject(DefaultJSONParser.java:658)
at com.alibaba.fastjson.JSON.parseObject(JSON.java:365)
at com.alibaba.fastjson.JSON.parseObject(JSON.java:269)
at com.alibaba.fastjson.JSON.parseObject(JSON.java:488)
at com.alibaba.fastjson.JSON.main(JSON.java:1068)

Upvotes: 0

Views: 1964

Answers (2)

huang jerome
huang jerome

Reputation: 11

Change constructor to.

@JsonCreator
public Student(@JsonProperty("name") String name, @JsonProperty("age") String age){
    this.name = name;
    this.age = age;
}

Upvotes: 0

daniu
daniu

Reputation: 15008

So create a TO class.

Student model2 = JSON.parseObject(hell, StudentTO.class).asStudent();
System.out.println(model2);

public class StudentTO {

    private String name;
    private String age;
    public String getName() {
        return name;
    }
    public void setName(String name) {
        this.name = name;
    }
    public String getAge() {
        return age;
    }
    public void setAge(String age) {
        this.age = age;
    }
    public Student asStudent() {
        return new Student(name, age);
    }
}

Upvotes: -1

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