问题
I'm new to the Java 11 HttpClient and would like to give it a try. I have a simple GET request that return JSON and I would like to map the JSON response to a Java class called Questionnaire
.
I understand that I can turn the response out of box into a String or an input stream like this
HttpRequest request = HttpRequest.newBuilder(new URI(String.format("%s%s", this.baseURI, "/state")))
.header(ACCEPT, APPLICATION_JSON)
.PUT(noBody()).build();
HttpResponse<String> response = this.client.send(request, HttpResponse.BodyHandlers.ofString());
How can I write something that converts the JSON string to my Questionnaire class like this?
HttpResponse<Questionnaire> response = this.client.send(request, HttpResponse.BodyHandlers./* what can I do here? */);
I use Jackson to transform JSON into Java class instances. Is there Jackson support for the new Java standard HttpClient yet?
UPDATE 1 I was not precise enough, sorry about that. I am looking for a blocking get example. I was aware of http://openjdk.java.net/groups/net/httpclient/recipes.html#jsonGet
回答1:
Solution for Java 11 HttpClient::sendAsync
only
Based on this link you can do something like this :
public static void main(String[] args) throws IOException, URISyntaxException, ExecutionException, InterruptedException {
UncheckedObjectMapper uncheckedObjectMapper = new UncheckedObjectMapper();
HttpRequest request = HttpRequest.newBuilder(new URI("https://jsonplaceholder.typicode.com/todos/1"))
.header("Accept", "application/json")
.build();
Model model = HttpClient.newHttpClient()
.sendAsync(request, HttpResponse.BodyHandlers.ofString())
.thenApply(HttpResponse::body)
.thenApply(uncheckedObjectMapper::readValue)
.get();
System.out.println(model);
}
static class UncheckedObjectMapper extends com.fasterxml.jackson.databind.ObjectMapper {
/**
* Parses the given JSON string into a Map.
*/
Model readValue(String content) {
try {
return this.readValue(content, new TypeReference<Model>() {
});
} catch (IOException ioe) {
throw new CompletionException(ioe);
}
}
}
static class Model {
private String userId;
private String id;
private String title;
private boolean completed;
//getters setters constructors toString
}
I used some dummy endpoint which provides sample JSON input and sample model class to map the response directly to Model
class using Jackson.
Solution for Java 11 HttpClient::send
and HttpClient::sendAsync
I found a way by defining custom HttpResponse.BodyHandler
:
public class JsonBodyHandler<W> implements HttpResponse.BodyHandler<W> {
private Class<W> wClass;
public JsonBodyHandler(Class<W> wClass) {
this.wClass = wClass;
}
@Override
public HttpResponse.BodySubscriber<W> apply(HttpResponse.ResponseInfo responseInfo) {
return asJSON(wClass);
}
public static <T> HttpResponse.BodySubscriber<T> asJSON(Class<T> targetType) {
HttpResponse.BodySubscriber<String> upstream = HttpResponse.BodySubscribers.ofString(StandardCharsets.UTF_8);
return HttpResponse.BodySubscribers.mapping(
upstream,
(String body) -> {
try {
ObjectMapper objectMapper = new ObjectMapper();
return objectMapper.readValue(body, targetType);
} catch (IOException e) {
throw new UncheckedIOException(e);
}
});
}
}
Then I call it :
public static void main(String[] args) throws URISyntaxException, IOException, InterruptedException {
HttpRequest request = HttpRequest.newBuilder(new URI("https://jsonplaceholder.typicode.com/todos/1"))
.header("Accept", "application/json")
.build();
Model model = HttpClient.newHttpClient()
.send(request, new JsonBodyHandler<>(Model.class))
.body();
System.out.println(model);
}
The response is :
Model{userId='1', id='1', title='delectus aut autem', completed=false}
The JavaDoc of HttpResponse.BodySubscribers::mapping
was particulary useful to solve this. It can be further improved to use HttpResponse.BodySubscribers::ofInputStream
instead of HttpResponse.BodySubscribers.ofString(StandardCharsets.UTF_8)
to define the BodySubscriber
for the JsonBodyHandler
.
来源:https://stackoverflow.com/questions/57627218/how-to-map-a-json-response-to-a-java-class-using-java-11-httpclient-and-jackson