Question

如何将以下JSON转换为Jackson java中的类

{
  "status": 0,
  "msg": "0 out of 1 Transactions Fetched Successfully",
  "transaction_details": {
    "3690": {
      "mihpayid": "Not Found",
      "status": "Not Found"
    }
  }
}

transaction_details中的整数键导致出现问题，我们无法控制从第三方支付网关返回的格式。

Answer 1

如下所示，设计以下类，

import com.fasterxml.jackson.annotation.JsonProperty;

public class Jackson {

    private String status;
    private String msg;

    private TransactionDetails transaction_details = new TransactionDetails();

    public String getStatus() {
        return status;
    }
    public void setStatus(String status) {
        this.status = status;
    }
    public String getMsg() {
        return msg;
    }
    public void setMsg(String msg) {
        this.msg = msg;
    }
    public TransactionDetails getTransaction_details() {
        return transaction_details;
    }
    public void setTransaction_details(TransactionDetails transaction_details) {
        this.transaction_details = transaction_details;
    }

}

class TransactionDetails{

    private A3690 a3690;

    @JsonProperty("3690")
    public A3690 getA3690() {
        return a3690;
    }

    public void setA3690(A3690 a3690) {
        this.a3690 = a3690;
    }

}

class A3690{

    private String mihpayid ;
    private String status;

    public String getMihpayid() {
        return mihpayid;
    }
    public void setMihpayid(String mihpayid) {
        this.mihpayid = mihpayid;
    }
    public String getStatus() {
        return status;
    }
    public void setStatus(String status) {
        this.status = status;
    }


}

你可以同样使用它来创建Jackson Object to JSON，

String jsonString = null;

        try {

            Jackson jacksonObj = new Jackson(); // populate it as per needs...
            jsonString = objectMapper.writeValueAsString(jacksonObj);

        } catch (JsonProcessingException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        }

         System.out.println(jsonString); // will return JSON into String formate.

将String JSON转换为Jackson Object也是如此，

String str = "{"status":"status","msg":"msg","transaction_details":{"3690":{"mihpayid":"mihpayid","status":"status1"}}}"; 

        try {
            Jackson jacksonObject = objectMapper.readValue(str, Jackson.class);

            System.out.println(" Status -" + jacksonObject.getStatus() +  " : msg - " + jacksonObject.getMsg());

            System.out.println(" Mihpayid -" + jacksonObject.getTransaction_details().getA3690().getMihpayid() +  " : Status - " + jacksonObject.getTransaction_details().getA3690().getStatus());

        } catch (JsonParseException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        } catch (JsonMappingException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        } catch (IOException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        }

更新：

如果键3690未修复，我的意思是它随机出现然后need to refactoring杰克逊class likewise，（这里只有杰克逊级足够所有，不需要像我上面提到的那样创建更多的类）

公共课杰克逊{

private String status;
private String msg;

//private TransactionDetails transaction_details = new TransactionDetails();

public String getStatus() {
    return status;
}
public void setStatus(String status) {
    this.status = status;
}
public String getMsg() {
    return msg;
}
public void setMsg(String msg) {
    this.msg = msg;
}

HashMap<String, HashMap<String,String>> transaction_details = new HashMap<String, HashMap<String,String>>(); 

@JsonProperty("transaction_details")
public HashMap<String, HashMap<String,String>> getMap() {
    return transaction_details;
}

public void setMap(HashMap<String, HashMap<String,String>> transaction_details) {
    this.transaction_details = transaction_details;
}

}

享受，祝你好运...... !!!

Jackson json序列化对整数键失败

1 个答案: