Mapping Java objects using FasterXML JsonProperty with multiple unknown keys of the same structure [duplicate]

Question

I have a JSON object structured like this:

{
    "John": { ...Some JSON object... },
    "Mary": { ...Some JSON object... },
    "Sam": { ...Some JSON object... }
}

The values for the John, Mary, and Sam keys are all structured the same, let's say it maps to a PersonInfo Java class. How do I get a List<PersonInfo> by just getting the top level JSON objects keys as a List? I can't do something like this

import com.fasterxml.jackson.annotation.JsonProperty;

@Data
public class PersonRecord {
    @JsonProperty("John")
    private PersonInfo john;
}

Because I don't know the top level keys, so I can't hardcode John. I don't care about the name of the key, I just want to select the values from each of the top level keys and put them in a list.

import com.fasterxml.jackson.annotation.JsonProperty; @Data public class PersonRecord { @JsonProperty("John") private List<PersonInfo> listofperson; } — abhinavsinghvirsen
– abhinavsinghvirsen, Commented Jul 23, 2024 at 5:20
I can't hardcode John since I don't know the top level keys — Kevin2566
– Kevin2566, Commented Jul 23, 2024 at 5:25
import com.fasterxml.jackson.annotation.JsonProperty; @Data public class PersonRecord { @JsonProperty("John") private Map<String,PersonInfo> rec; } try this — abhinavsinghvirsen
– abhinavsinghvirsen, Commented Jul 23, 2024 at 5:30

talex · Accepted Answer · 2024-07-23 06:33:37Z

-1

Use Map<String, PersonInfo> instead of PersonRecord.

answered Jul 23, 2024 at 6:33

talex

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