Java + JSON: list of anonymous objects (key + value) with different key names

Question

Maybe it's simple, but I'm stuck... I have an app using Spring Boot and springframework.web.client.RestTemplate, so there's Jackson on backseat doing JSON job automatically.

I want to deserialize REST input with JSON like this:

{
    "serial_no": 12345678,
    "last_dt": [
        {
            "lp": "2022-04-22T15:00:00"
        },
        {
            "bp": null
        },
        {
            "dp": "2022-04-22T00:00:00"
        },
        {
            "iv": "2022-04-22T00:05:11"
        }
    ]
}

Please, help me to create POJO for this model... If I created following classes (simplified with lombok):

@Data
public class LastDt {

    @JsonProperty("lp")
    private String lp;
    @JsonProperty("bp")
    private String bp;
    @JsonProperty("dp")
    private String dp;
    @JsonProperty("iv")
    private String iv;

}

@Data
public class Device {

    @JsonProperty("serial_no")
    private Long serialNo;
    @JsonProperty("last_dt")
    private List<LastDt> lastDt;

}

deserialization is technically ok - I got an object with serialNo Long value and unfortunatelly list of 4 instances of class LastDt: first instance (lastDt[0]) had field "lp" with value and "bp", "dp", "iv" null; second (lastDt[1]): "bp" with value and "lp", "dp", "iv" null and so on - only one value was set and others are not. That's far from what I want to get. As you can see, there's four pair-like anonymous objects with different key names, so my model is bad, and I know this, but got stuck trying to create other ones... Can anyone help me?

Maybe the JSON representation is badly chosen? Why do you have four objects in the array that differ in their structure? Maybe you wanted to have last_dt to be an object instead of an array? — Seelenvirtuose
– Seelenvirtuose, Commented Apr 22, 2022 at 14:54
Have you try to call the field from Device like in your json : last_dt instead of lastDt ? — Zabon
– Zabon, Commented Apr 22, 2022 at 15:38
Do you have control over the json you need to deserialize? Can you change its structure? — Chaosfire
– Chaosfire, Commented Apr 22, 2022 at 16:09

Chaosfire · Accepted Answer · 2022-05-01 14:28:11Z

2

You can use @JsonAlias. The idea is to have a single field in LastDt, with all possible names of the property mapped to that field.

public class LastDt {

  @JsonAlias({"lp", "bp", "dp", "iv"})
  private String value;

  //getter and setters

  @Override
  public String toString() {
    return "LastDt{" +
            "value='" + value + '\'' +
            '}';
  }
}

No changes are required to Device.

And a simple test:

public class Temp {

  public static void main(String[] args) throws Exception {
    ObjectMapper mapper = new ObjectMapper();
    InputStream dataStream = getTheStream();
    Device device = mapper.readValue(dataStream, Device.class);
    System.out.println(device);
  }
}

answered May 1, 2022 at 14:28

Chaosfire

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Comments

Maryam Zng · Accepted Answer · 2022-04-22 15:04:31Z

1

You should use @JsonInclude(JsonInclude.Include.NON_NULL) in your class:

@Data
@JsonInclude(JsonInclude.Include.NON_NULL)
public class LastDt {
    @JsonProperty("lp")
    private String lp;
    @JsonProperty("bp")
    private String bp;
    @JsonProperty("dp")
    private String dp;
    @JsonProperty("iv")
    private String iv;
}

edited Apr 22, 2022 at 15:04

answered Apr 22, 2022 at 14:54

Maryam Zng

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1 Comment

Seelenvirtuose Over a year ago

OP knows that obviously.

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