Question

我正在尝试将以下jsonString结构转换为Java列表/对象数组：

[
    [1,21940000,1905386136,null,"KR","akshay","04/06/2017","03/06/2017",2017,9,40,"JPY",7478,"JPY",7478,"WHT (Residen",null,0,"03/06/2017","03/06/2017","20170604",null],
    [2,21940000,1903732187,null,"KR",null,"06/06/2017","05/06/2017",2017,9,40,"JPY",608547485,"JPY",608547485,"WHT (Non-Resi",null,0,"05/06/2017","05/06/2017","20170606",null],
    [3,21940000,2001898163, ............... ]
.
.
.
.
.
.
.
.
]

以下是Java代码：

ObjectMapper mapper = new ObjectMapper();
MyData[][] data = mapper.readValue(jsonString, MyData[][].class);

但是，我得到以下错误：

    com.fasterxml.jackson.databind.JsonMappingException: 
    Can not construct instance of com.org.model.MyData: 
     no String-argument constructor/factory method to deserialize from String value ('KR')
     at [Source: java.io.StringReader@1327cf05; line: 1, column: 30] (through reference chain: java.lang.Object[][0]->java.lang.Object[][4])

有人可以帮帮我吗？感谢

编辑：以下是我的POJO MyData.java代码：

@Entity
@Table(schema = "My_Schema", name = "My_Data_Table")
@SuppressFBWarnings(value = { "EI_EXPOSE_REP", "EI_EXPOSE_REP2" }, justification = "I prefer to suppress these FindBugs warnings")

public class MyData implements Serializable {

    /**
     * 
     */
    private static final long serialVersionUID = -6936461726389768288L;

    public MyData() {
    super();
    }

    /**
     * @param id
     */
    public MyData(Long id) {
    super();
    this.id = id;
    }

    @Id
    private Long id;

    @Column(name = "ACCOUNT")
    private long account;
    @Column(name = "DOC_NUMBER")
    private long docNumber;
    @Column(name = "TYPE")
    private String type;
    @JsonFormat(shape = JsonFormat.Shape.STRING, pattern = "dd/MM/yyyy")
    @Column(name = "DOC_DATE")
    private Date docDate;
    @JsonFormat(shape = JsonFormat.Shape.STRING, pattern = "dd/MM/yyyy")
    @Column(name = "POSTING_DATE")
    private Date postingDate;
    @Column(name = "YEAR")
    private long year;
    @Column(name = "PERIOD")
    private long period;
    @Column(name = "PK")
    private long pk;
    @Column(name = "TAX_CODE")
    private String taxCode;
    @Column(name = "CCY")
    private String ccy;
    @Column(name = "DOC_CCY_AMT")
    private long docCcyAmt;
    @Column(name = "LOCAL_CCY")
    private String localCcy;
    @Column(name = "LOCAL_CCY_AMT")
    private long localCcyAmt;
    @Column(name = "TEXT")
    private String text;
    @Column(name = "DOC_HEADER_TEXT")
    private String docHeaderText;
    @JsonFormat(shape = JsonFormat.Shape.STRING, pattern = "dd/MM/yyyy")
    @Column(name = "CLEARING_DATE")
    private Date clearingDate;
    @Column(name = "CLEARING_DOC")
    private long clearingDoc;
    @JsonFormat(shape = JsonFormat.Shape.STRING, pattern = "dd/MM/yyyy")
    @Column(name = "ENTRY_DATE")
    private Date entryDate;
    @JsonFormat(shape = JsonFormat.Shape.STRING, pattern = "dd/MM/yyyy")
    @Column(name = "VALUE_DATE")
    private Date valueDate;
    @Column(name = "ASSIGNMENT")
    private String assignment;
    @Column(name = "REMARKS")
    private String remarks;
    // Getters and setters to follow .....

所以，问题是我的输入JSON字符串是一个数组数组，我希望它是一个Java表示形式，无论是ArrayList还是普通数组......

Answer 1

您可能缺少＆＃34; KR＆＃34;在你的＆＃34; MyData＆＃34;类。在此处发布MyData类的定义之前，请先查看此代码。它一定会帮到你。

class Student {
   private String name;
   private int age;

   public Student(){}

   public String getName() {
      return name;
   }

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

   public int getAge() {
      return age;
   }

   public void setAge(int age) {
      this.age = age;
   }
   public String toString(){
      return "Student [ name: "+name+", age: "+ age+ " ]";
   }
}

并测试它

import java.io.IOException;

import org.codehaus.jackson.JsonParseException;
import org.codehaus.jackson.map.JsonMappingException;
import org.codehaus.jackson.map.ObjectMapper;
import org.codehaus.jackson.map.SerializationConfig;

public class JacksonTester {
   public static void main(String args[]){

      ObjectMapper mapper = new ObjectMapper();
      String jsonString = "{\"name\":\"Mahesh\", \"age\":21}";

      //map json to student

      try{
         Student student = mapper.readValue(jsonString, Student.class);

         System.out.println(student);

         mapper.enable(SerializationConfig.Feature.INDENT_OUTPUT);
         jsonString = mapper.writeValueAsString(student);

         System.out.println(jsonString);
      }
      catch (JsonParseException e) { e.printStackTrace();}
      catch (JsonMappingException e) { e.printStackTrace(); }
      catch (IOException e) { e.printStackTrace(); }
   }
}

使用Jackson将json数组数组转换为某些Java表示形式

1 个答案: