问题
I'm using the Flickr API. When calling the flickr.test.login method, the default JSON result is:
{
"user": {
"id": "21207597@N07",
"username": {
"_content": "jamalfanaian"
}
},
"stat": "ok"
}
I'd like to parse this response into a Java object:
public class FlickrAccount {
private String id;
private String username;
// ... getter & setter ...
}
The JSON properties should be mapped like this:
"user" -> "id" ==> FlickrAccount.id
"user" -> "username" -> "_content" ==> FlickrAccount.username
Unfortunately, I'm not able to find a nice, elegant way to do this using Annotations. My approach so far is, to read the JSON String into a Map<String, Object>
and get the values from there.
Map<String, Object> value = new ObjectMapper().readValue(response.getStream(),
new TypeReference<HashMap<String, Object>>() {
});
@SuppressWarnings( "unchecked" )
Map<String, Object> user = (Map<String, Object>) value.get("user");
String id = (String) user.get("id");
@SuppressWarnings( "unchecked" )
String username = (String) ((Map<String, Object>) user.get("username")).get("_content");
FlickrAccount account = new FlickrAccount();
account.setId(id);
account.setUsername(username);
But I think, this is the most non-elegant way, ever. Is there any simple way, either using Annotations or a custom Deserializer?
This would be very obvious for me, but of course it doesn't work:
public class FlickrAccount {
@JsonProperty( "user.id" ) private String id;
@JsonProperty( "user.username._content" ) private String username;
// ... getter and setter ...
}
回答1:
You can write custom deserializer for this class. It could look like this:
class FlickrAccountJsonDeserializer extends JsonDeserializer<FlickrAccount> {
@Override
public FlickrAccount deserialize(JsonParser jp, DeserializationContext ctxt) throws IOException, JsonProcessingException {
Root root = jp.readValueAs(Root.class);
FlickrAccount account = new FlickrAccount();
if (root != null && root.user != null) {
account.setId(root.user.id);
if (root.user.username != null) {
account.setUsername(root.user.username.content);
}
}
return account;
}
private static class Root {
public User user;
public String stat;
}
private static class User {
public String id;
public UserName username;
}
private static class UserName {
@JsonProperty("_content")
public String content;
}
}
After that, you have to define a deserializer for your class. You can do this as follows:
@JsonDeserialize(using = FlickrAccountJsonDeserializer.class)
class FlickrAccount {
...
}
回答2:
Since I don't want to implement a custom class (Username
) just to map the username, I went with a little bit more elegant, but still quite ugly approach:
ObjectMapper mapper = new ObjectMapper();
JsonNode node = mapper.readTree(in);
JsonNode user = node.get("user");
FlickrAccount account = new FlickrAccount();
account.setId(user.get("id").asText());
account.setUsername(user.get("username").get("_content").asText());
It's still not as elegant as I hoped, but at least I got rid of all the ugly casting.
Another advantage of this solution is, that my domain class (FlickrAccount
) is not polluted with any Jackson annotations.
Based on @Michał Ziober's answer, I decided to use the - in my opinion - most straight forward solution. Using a @JsonDeserialize
annotation with a custom deserializer:
@JsonDeserialize( using = FlickrAccountDeserializer.class )
public class FlickrAccount {
...
}
But the deserializer does not use any internal classes, just the JsonNode
as above:
class FlickrAccountDeserializer extends JsonDeserializer<FlickrAccount> {
@Override
public FlickrAccount deserialize(JsonParser jp, DeserializationContext ctxt) throws
IOException, JsonProcessingException {
FlickrAccount account = new FlickrAccount();
JsonNode node = jp.readValueAsTree();
JsonNode user = node.get("user");
account.setId(user.get("id").asText());
account.setUsername(user.get("username").get("_content").asText());
return account;
}
}
回答3:
You can also use SimpleModule.
SimpleModule module = new SimpleModule();
module.setDeserializerModifier(new BeanDeserializerModifier() {
@Override public JsonDeserializer<?> modifyDeserializer(
DeserializationConfig config, BeanDescription beanDesc, JsonDeserializer<?> deserializer) {
if (beanDesc.getBeanClass() == YourClass.class) {
return new YourClassDeserializer(deserializer);
}
return deserializer;
}});
ObjectMapper objectMapper = new ObjectMapper();
objectMapper.registerModule(module);
objectMapper.readValue(json, classType);
回答4:
You have to make Username a class within FlickrAccount and give it a _content field
来源:https://stackoverflow.com/questions/19158345/custom-json-deserialization-with-jackson