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好的，所以我一直在搞砸这个。我得到一个 json 数组，其中包含如下所示的对象：[{ "id": "1", "name": "some name", "url": "some url", "active": true }, { "id": "2", "name": "some other name", "url": "some other url", "active": true } ]"

现在，我希望能够将该数组反序列化为一个 java 对象，该对象包含数组中的对象列表。我制作了一个自定义反序列化器，如下所示：

public class ListSerializer extends JsonDeserializer<List<Provider>>{

private static final long serialVersionUID = 9114152571639338391L;

@Override
public List<Provider> deserialize(JsonParser jsonParser, 
DeserializationContext arg1) throws IOException, JsonProcessingException {
// TODO Auto-generated method stub
final ObjectCodec objectCodec = jsonParser.getCodec();
final JsonNode listOrObjectNode = objectCodec.readTree(jsonParser);
final List<Provider> result = new ArrayList<Provider>();
for (JsonNode node : listOrObjectNode) {
  result.add(objectCodec.treeToValue(node, Provider.class));
}
  return result;
}



}

保存列表的类如下所示：

public class ProviderList {

@JsonDeserialize(using = ListSerializer.class)
private List<Provider> providerList;

 public List<Provider> getProviderList() {
  return providerList;
}

public void setProviderList(final List<Provider> providerList) {
  this.providerList = providerList;
 }

}

现在，我知道这可能是一种愚蠢的方法，但长话短说，这是刚需。我显然做错了什么，因为我收到了这个错误：

无法将 com.wirelesscar.trailser.v1_0.domain.ProviderList 的实例从 [Source: [{"id":"1","name":"Posttrack","url":"http:\ dev.posttrack.com","active":true},{"id":"2","name":"Trackunit","url":"http:\dev.trackunit.com","active":真的}]; 行：1，列：1]

所以我的问题。这可能吗？我在哪里错了。

先谢谢了！

4

1 回答 1

2

您可以使用 TypeReference 包装器直接反序列化为列表。

@Data
public class Provider {
    private Long id;
    private String name;
    private String url;
    private boolean active;
}

@Data
public class ProviderList {
    List<Provider> providerList;
}

public class JsonTest {

    @Test
    public void  test() {
        String json = "[{\n" +
                "        \"id\": \"1\",\n" +
                "        \"name\": \"some name\",\n" +
                "        \"url\": \"some url\",\n" +
                "        \"active\": true\n" +
                "    }, {\n" +
                "        \"id\": \"2\",\n" +
                "        \"name\": \"some other name\",\n" +
                "        \"url\": \"some other url\",\n" +
                "        \"active\": true\n" +
                "    }\n" +
                "]";

        ObjectMapper mapper = new ObjectMapper();
        try {
            List<Provider> providerList = mapper.readValue(json, new TypeReference<List<Provider>>(){});
            for (Provider provider : providerList) {
                System.out.println(provider);
            }

            ProviderList list = new ProviderList();
            list.setProviderList(providerList);

        } catch (IOException e) {
            e.printStackTrace();
        }
    }
}
于 2017-05-04T10:09:09.450 回答