Java 在 Spring RESTful 服务中生成和使用自定义 JSON 对象
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Producing and consuming custom JSON Objects in Spring RESTful services
提问by user592748
I have some JSON Objects that are more complex than the JSON representations of the java objects I have. I have methods that build these JSON Objects and I would like to return and consume these directly. I use org.jsonlibrary to build my JSONs. I could get the GETmethod working by returning the JSON Object as a String. Is this the correct way to go about it?
我有一些 JSON 对象，它们比我拥有的 Java 对象的 JSON 表示更复杂。我有构建这些 JSON 对象的方法，我想直接返回和使用这些对象。我使用org.json库来构建我的 JSON。我可以GET通过将 JSON 对象作为String. 这是正确的方法吗？
@RequestMapping(value = "/getjson", method = RequestMethod.GET, produces="application/json")
@ResponseBody
public String getJson() {
JSONObject json = new JSONObject();
JSONObject subJson = new JSONObject();
subJson .put("key", "value");
json.put("key", subJson);
return json.toString();
}
Now I want to know how do I consume a JSON Object? As a string and convert it to a JSON Object?
现在我想知道如何使用 JSON 对象？作为字符串并将其转换为 JSON 对象？
@RequestMapping(value = "/post", method = RequestMethod.POST, produces="application/json", consumes="application/json")
@ResponseBody
public String post(@RequestBody String json) {
JSONObject obj = new JSONObject(json);
//do some things with json, put some header information in json
return obj.toString();
}
Is this the correct way to go about my problem? I am a novice, so kindly point out anything that can be done better. Please note: I do not want to return POJOs.
这是解决我的问题的正确方法吗？我是新手，所以请指出任何可以做得更好的地方。请注意：我不想返回 POJO。
采纳答案by RE350
I think using Hymanson library you can do something like below.
我认为使用 Hymanson 库，您可以执行以下操作。
@RequestMapping(value = "/getjson", method = RequestMethod.GET, produces="application/json")
@ResponseBody
public String getJson() {
//your logic
ObjectMapper mapper = new ObjectMapper();
return mapper.writeValueAsString(json);
}
@RequestMapping(value = "/post", method = RequestMethod.POST, produces="application/json", consumes="application/json")
@ResponseBody
public String post(@RequestBody String json) {
POJO pj = new POJO();
ObjectMapper mapper = new ObjectMapper();
pj = mapper.readValue(json, POJO.class);
//do some things with json, put some header information in json
return mapper.writeValueAsString(pj);
}
回答by kemenov
You can use Hymanson lib, Hymanson allows convert to/from json using spring mvc.
您可以使用 Hymanson lib，Hymanson 允许使用 spring mvc 与 json 进行转换。
回答by paul
I much rather the alternative to use Hymanson with Spring mvc since you dont have to worry of the serialization and deserialization of your objects/json-json/object. But if you still want to the process I like to use gson of google.
我更愿意将 Hymanson 与 Spring mvc 一起使用，因为您不必担心对象/json-json/object 的序列化和反序列化。但是如果你还想要这个过程我喜欢用google的gson。
回答by Hypothetical inthe Clavicle
I couldn't get the auto serialization and deserialization to work using the default Spring boot beans. In the end, this worked well for me after including Project Lombok and apache BeanUtils. Spring calls a String arg constructor if it can't automatically do the serialization.
我无法使用默认的 Spring 引导 bean 使自动序列化和反序列化工作。最后，在包含 Project Lombok 和 apache BeanUtils 后，这对我来说效果很好。如果 Spring 不能自动执行序列化，则调用 String arg 构造函数。
@PostMapping("/create")
@ResponseBody
public User createUser(HttpServletRequest request, @RequestParam("user") User u) throws IOException {
LOG.info("Got user! " + u);
return u;
}
@ToString() @Getter() @Setter() @NoArgsConstructor()
public class User {
private String email;
private String bio;
private String image;
private String displayName;
private String userId;
private long lat;
private long lng;
public User(String json) throws JsonParseException, JsonMappingException, IOException, IllegalAccessException, InvocationTargetException {
ObjectMapper om = new ObjectMapper();
User u = om.readValue(json, User.class);
BeanUtils.copyProperties(this, u);
}
}
http://commons.apache.org/proper/commons-beanutils/download_beanutils.cgihttps://projectlombok.org/
http://commons.apache.org/proper/commons-beanutils/download_beanutils.cgi https://projectlombok.org/
