java Spring RestTemplate获取特定字段

Question

我的安卓应用程序中有以下JSON，我正在使用Springs for 安卓 RestTemplate。是否可以从这个json中只获取内部列表？目前，我必须创建一个包装器对象，然后才能从中获取List<Cast> casts;，这有点不方便

{
  "id": 550,
  "cast": [
    {
      "cast_id": 4,
      "character": "The Narrator",
      "credit_id": "52fe4250c3a36847f80149f3",
      "id": 819,
      "name": "Edward Norton",
      "order": 0,
      "profile_path": "/iUiePUAQKN4GY6jorH9m23cbVli.jpg"
    }
  ]
}

Answer 1

我知道两种解决方法：

1. 您所做的：使用类包装列表
2. 写你自己的JSON反序列化程序，看看这里http://wiki.fasterxml.com/JacksonHowToCustomDeserializers

但我认为最终的解决方案是在这篇帖子中： Spring/json: Convert a typed collection like List<MyPojo>

Answer 2

如果Cast是一个POJO，那么可以尝试像这样使用Jackson ObjectMapper类

ObjectMapper mapper = new ObjectMapper();
String jsonStr = ...
List<Cast> casts = mapper.readValue(jsonStr, new TypeReference<List<Cast>>(){});

Answer 3

你可以像这样抓取、施放和转换场地：

    final String json = "{\n" +
            "  \"id\": 550,\n" +
            "  \"cast\": [\n" +
            "    {\n" +
            "      \"cast_id\": 4,\n" +
            "      \"character\": \"The Narrator\",\n" +
            "      \"credit_id\": \"52fe4250c3a36847f80149f3\",\n" +
            "      \"id\": 819,\n" +
            "      \"name\": \"Edward Norton\",\n" +
            "      \"order\": 0,\n" +
            "      \"profile_path\": \"/iUiePUAQKN4GY6jorH9m23cbVli.jpg\"\n" +
            "    }\n" +
            "  ]\n" +
            "}";

    final List<Cast> casts;
    try {
        final JsonNode cast = objectMapper.readTree(json).get("cast");
        casts = objectMapper.convertValue(cast, new TypeReference<List<Cast>>() {});
    } catch (IOException e) {
        throw Throwables.propagate(e);
    }