Converted json (from ArrayList) 有更多的元素

Question

我 post 我做了什么，因为我没有得到结果..这里我有一个方法 returns 一个 ArrayList:

public ArrayList<Label> getLabels() 
         throws ClassNotFoundException, SQLException{

    ArrayList<Label> labels = new ArrayList<>();
    sq = "SELECT * from LABELS";

    try {       
          Class.forName(typeDB);
          c = DriverManager.getConnection(path);            
          stm = c.prepareStatement(sq);  
          ResultSet rs = stm.executeQuery();

          while(rs.next()) {
             Label label = new Label(rs.getString("type"), rs.getString("description"),rs.getString("product")+"-"+rs.getString("version"), rs.getString("cutter"));
             labels.add(label); 
           }

        } catch (SQLException e) {
            System.out.println(e.getMessage());
        } finally {
            if (stm != null)
                stm.close();
            if (c != null)
        c.close();
        }

        System.out.println("Label "+ labels.size());
        return labels;
    }

然后我想将这个 ArrayList 转换为 JSON 格式。所以我执行 labelsToJSON(action.getLabels()); where:

public void labelsToJSON(ArrayList<Label> list){

      ObjectMapper mapper = new ObjectMapper();
      try{
           mapper.writeValue(new File("C:\temp\labels.json"), list);
          }catch(JsonGenerationException e){
               e.printStackTrace();
          }catch(JsonMappingException e){
               e.printStackTrace();
          }catch (IOException e){
               e.printStackTrace();
          }

     }
}

Labelclass的定义是：

public class Label {
    private String barcode;     
    private String labelCode;
    private String productCode; 
    private String type;   
    //and many others..

    public Label(){

    }

    //This is the costructor I use above in the method   
    public Label(String type, String description, String productCode, String cutter) {
        this.type = type;
        this.description = description;
        this.productCode = productCode;
        this.cutter = cutter;    
    }

    //and then some other constructors (I post 2 for example)
    public Label(String type, String description, String product, String version, String cutter) {
        this.type = type;
        this.description = description;
        this.product = product;
        this.version = version;
        this.cutter = cutter;    
    }

    public Label(String barcode, String product, String version, String dateProduction, String order , int quantity, String packetNumber, String type, String description, String cutter) {
        this.barcode = barcode;
        this.product = product;
        this.version = version;
        this.dateProduction = dateProduction;
        this.order = order;
        this.packetNumber = packetNumber;
        this.quantity = quantity;
        this.type = type;
        this.description = description;
        this.cutter = cutter;
    }

   //setters, getters etc

因此，我使用参数 String type, String description, String productCode, String cutter 从构造函数创建了一个对象。但是 labels.json 包含这些数据

[{ 
    "barcode":null,
    "labelCode":null,
    "productCode":"111123123-1123",        //<- 
    "type":"Container",                    //<-
    "description":"this is a description", //<- all these I was expected.
    "cutter":"1031",                       //<-
    "date":null,
    "time":null,
    "dateProduction":null,
    "order":null,
    "product":null,
    "version":null,
    "packetNumber":null,
    "quantity":0
  }, //and so on

我不明白为什么json文件有这么多属性？？我的对象应该只有 4 --> String type, String description, String productCode, String cutter

Answer 1

ObjectMapper 将默认序列化 class 上的所有字段值，无论它们是否为空，因此您可以从 Label class 中获取所有内容。

要仅序列化您可以配置 ObjectMapper 的非空值，请参阅 setSerializationInclusion and Include

的 JavaDoc

mapper.setSerializationInclusion(Include.NON_NULL);

编辑： 正如 Maraboc 指出的那样，您遇到了在使用 Include.NON_NULL 时 quantity 仍然被序列化的问题。为了更好地控制哪些字段被序列化，您可以使用 @JsonIgnore 注释来防止 class 中的其他字段被序列化。

或者您可以将 @JsonIgnoreProperties({"quantity"}) 添加到您的 class

Answer 2

您可以使用 JsonSerialize 注释定义 Label class 并将数量类型从原始 int 更改为 Integer Object。如果类型为 int，则默认值零将分配给变量。

@JsonSerialize(
include=JsonSerialize.Inclusion.NON_NULL)
public class Label {

    // ...other properties

    private Integer quantity;

    public Integer getQuantity() {
       return quantity;
    }

    public void setQuantity(Integer quantity) {
       this.quantity = quantity;
    }
}

Converted json (from ArrayList) 有更多的元素

Converted json (from ArrayList) has more elements

java

json

to-json