Spring&Jackson-JSON列表自定义XML

Question

{
    "name": "Edward",
    "address": [
        {
            "streetName": "value1",
            "city": "value2"
        },
        {
            "streetName": "value3",
            "city": "value4"
        }
    ]
}

<root>
    <name>Edward</name>
    <streetName1>value1</streetName1>
    <city1>value2</city1>
    <streetName2>value3</streetName2>
    <city2>value4</city2>
</root>

@JacksonXmlElementWrapper(useWrapping = false)
private List<Address> address = new ArrayList<>();

<root>
    <name>Edward</name>
    <address>
        <streetName>value1</streetName>
        <city>value2</city>
    </address>
    <address>
        <streetName>value3</streetName>
        <city>value4</city>
    </address>
</root>

Answer 1

使用自定义序列化程序可以解决您的问题，因此，如果您有一个Person类，如下所示：

@JacksonXmlRootElement(localName = "root")
@JsonSerialize(using = PersonSerializer.class)
public class Person {

    private String name;

    private List<Address> address = new ArrayList<>();

}

您可以构建一个自定义的PersonSerializerserializer类，扩展JSONSerializer类，如下所示：

public class PersonSerializer extends JsonSerializer<Person> {

    @Override
    public void serialize(Person t, JsonGenerator jg, SerializerProvider sp) throws IOException {
        String streetName = "streetName";
        String city = "city";
        int nStreet = 1;
        int nCity = 1;
        jg.writeStartObject();
        jg.writeStringField("name", t.getName());
        for (Address address : t.getAddress()) {

            jg.writeStringField(streetName + String.valueOf(nStreet++), address.getStreetName());
            jg.writeStringField(city + String.valueOf(nCity++), address.getCity());

        }
        jg.writeEndObject();
    }

}

然后，您可以像下面这样序列化Person对象，获得预期的xml结果：

XmlMapper xmlMapper = new XmlMapper();
xmlMapper.enable(SerializationFeature.INDENT_OUTPUT);
//you already have your person object to serialize
String result = xmlMapper.writeValueAsString(person);