使用Java 11 HttpClient反序列化JSON，使用JSON停止自定义BodyHandler，并且不会继续
使用Java 11 HttpClient反序列化JSON，使用JSON停止自定义BodyHandler，并且不会继续
提问于 2019-08-23 15:41:43
在直接使用带有自定义HttpClient::send的Java11 HttpResponse.BodyHandler将JSON反序列化为定制对象时，我遇到了问题。我在回答这个问题时遇到了这个问题。
我正在使用的版本：
- OpenJDK 11
- 杰克逊2.9.9.3
我创建了一个简单的泛型JsonBodyHandler类，它实现了HttpResponse.BodyHandler。
public class JsonBodyHandler<W> implements HttpResponse.BodyHandler<W> {
private final Class<W> wClass;
public JsonBodyHandler(Class<W> wClass) {
this.wClass = wClass;
}
@Override
public HttpResponse.BodySubscriber<W> apply(HttpResponse.ResponseInfo responseInfo) {
return asJSON(wClass);
}
}asJSON方法定义为：
public static <W> HttpResponse.BodySubscriber<W> asJSON(Class<W> 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);
}
});
}因此，它返回一个自定义HttpResponse.BodySubscriber，它将主体作为String，然后应用从JSON到给定的targetType的映射来测试它：
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);
}Model类：
public class Model {
private String userId;
private String id;
private String title;
private boolean completed;
//getters setters constructors toString
}产出与预期相符：
Model{userId='1', id='1', title='delectus aut autem', completed=false}但是，当我首先将asJSON方法更改为读取InputStream而不是String时：
public static <W> HttpResponse.BodySubscriber<W> asJSON(Class<W> targetType) {
HttpResponse.BodySubscriber<InputStream> upstream = HttpResponse.BodySubscribers.ofInputStream();
return HttpResponse.BodySubscribers.mapping(
upstream,
(InputStream is) -> {
try (InputStream stream = is) {
ObjectMapper objectMapper = new ObjectMapper();
return objectMapper.readValue(stream, targetType);
} catch (IOException e) {
throw new UncheckedIOException(e);
}
});
}它在使用ObjectMapper调用该值后挂起，并且不继续(我已经检查它是否成功地从端点获得响应，状态代码为200)，但是它只是挂起。有人知道有什么问题吗？
Stack Overflow用户
发布于 2021-09-16 15:23:49
通过使用杰克逊TypeReference，我能够单独使用泛型来实现它，而不需要冗余的Class<T>参数。
package com.company;
import com.fasterxml.jackson.core.type.TypeReference;
import com.fasterxml.jackson.databind.ObjectMapper;
import java.io.IOException;
import java.io.InputStream;
import java.io.UncheckedIOException;
import java.net.http.HttpResponse;
import java.util.function.Supplier;
public class JsonBodyHandler<W> implements HttpResponse.BodyHandler<Supplier<W>> {
public JsonBodyHandler() {
}
private static <W> Supplier<W> toSupplierOfType(InputStream inputStream) {
return () -> {
try (InputStream stream = inputStream) {
return new ObjectMapper().readValue(stream, new TypeReference<W>() {
});
} catch (IOException e) {
throw new UncheckedIOException(e);
}
};
}
private static <W> HttpResponse.BodySubscriber<Supplier<W>> asJSON() {
return HttpResponse.BodySubscribers.mapping(
HttpResponse.BodySubscribers.ofInputStream(),
JsonBodyHandler::toSupplierOfType);
}
@Override
public HttpResponse.BodySubscriber<Supplier<W>> apply(HttpResponse.ResponseInfo responseInfo) {
return JsonBodyHandler.asJSON();
}
}并在使用中
package com.company;
import java.net.URI;
import java.net.http.HttpClient;
import java.net.http.HttpRequest;
import java.time.Duration;
import java.util.List;
public class Main {
private static final HttpRequest request = HttpRequest.newBuilder()
.uri(URI.create("https://myApi"))
.timeout(Duration.ofMinutes(1))
.header("Content-Type", "application/json")
.build();
private static final HttpClient client = HttpClient.newBuilder().build();
public static void main(String[] args) throws InterruptedException {
client.sendAsync(Main.request, new JsonBodyHandler<List<MyDto>>())
.thenAccept(response -> {
List<MyDto> myDtos = response.body().get();
System.out.println(myDtos);
}).join();
}
}页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/57629401
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