Question

假设我们有以下课程：

public abstract class Investment {

   private String investmentType;

   // getters & setters
}

public class Equity extends Investment {
}

public class Bond extends Investment {
}

public class InvestmentFactory {

    public static Investment getTypeFromString(String investmentType) {
        Investment investment = null;
        if ("Bond".equals(investmentType)) {
            investment = new Bond();
        } else if ("Equity".equals(investmentType)) {
            investment = new Equity();
        } else {
            // throw exception
        }
        return investment;
    }
}

以及以下@RestController：

@RestController
public class InvestmentsRestController {

    private InvestmentRepository investmentRepository;

    @Autowired
    public InvestmentsRestController(InvestmentRepository investmentRepository) {
        this.investmentRepository = investmentRepository;
    }

    @RequestMapping(RequestMethod.POST)
    public List<Investment> update(@RequestBody List<Investment> investments) {
       return investmentRepository.update(investments);
    }

}

以及请求正文中的以下json：

[
  {"investmentType":"Bond"},
  {"investmentType":"Equity"}
]

如何在抽象类List<Investment>上不使用Jackson的@JsonSubTypes而不是使用Investment的情况下将json绑定或转换为InvestmentFactory的请求正文？ >

Answer 1

如果可以使用@JsonDeserialize：

@JsonDeserialize(using = InvestmentDeserializer.class)
public abstract class Investment {
}



public class InvestmentDeserializer extends JsonDeserializer<Investment> {
    @Override
    public Investment deserialize(JsonParser p, DeserializationContext ctxt) throws IOException, JsonProcessingException {
        ObjectMapper objectMapper = (ObjectMapper) p.getCodec();
        TreeNode node = objectMapper.readTree(p);

        TreeNode investmentNode = node.get("investmentType");
        String type = objectMapper.readValue(investmentNode.traverse(objectMapper), String.class);
        return InvestmentFactory.getTypeFromString(type);
    }
}

示例控制器：

@RestController
public class MyController {
    @RequestMapping("/")
    public List<Class<?>> update(@RequestBody List<Investment> investments) {
        return investments.stream().map(Object::getClass).collect(Collectors.toList());
    }
}

测试：

$ curl localhost:8080 -H "Content-Type: application/json" -d '[{"investmentType":"Bond"},{"investmentType":"Equity"}]'

输出：

["com.example.demo.Bond","com.example.demo.Equity"]

Answer 2

@JsonDeserialize的效果很好，但是如果您拥有的字段不仅仅是类型，那么您将必须手动设置它们。如果您要回到杰克逊，可以使用：

Investment.class

public class InvestmentResolver extends TypeIdResolverBase {

    @Override
    public JavaType typeFromId(DatabindContext context, String id) throws IOException {
        Investment investment = InvestmentFactory.getTypeFromString(type);
        return context.constructType(investment.getClass());
    }

InvestmentResolver.class

'[{"investmentType":"Bond","investmentName":"ABC"},{"investmentType":"Equity","investmentName":"APPL"}]'

这样做的好处是，如果您开始向“投资”中添加字段，则不必在Desrializer中添加它们（至少在我的情况下是这样），但杰克逊会帮您解决这个问题。所以明天您可以拥有测试用例：

{{1}}

你应该很好！

如何将@RestController中的请求正文转换为抽象值列表？

2 个答案: