我编写了一个 Spring RestController
,它返回一个 SseEmitter
(用于服务器发送的事件),并向每个事件添加 HATEOAS 链接。这是此 Controller 的一个简化但有效的示例:
package hello;
import static org.springframework.hateoas.mvc.ControllerLinkBuilder.linkTo;
import static org.springframework.hateoas.mvc.ControllerLinkBuilder.methodOn;
import hello.Greeting.Status;
import java.io.IOException;
import org.slf4j.Logger;
import org.slf4j.LoggerFactory;
import org.springframework.web.bind.annotation.RequestMapping;
import org.springframework.web.bind.annotation.RequestParam;
import org.springframework.web.bind.annotation.RestController;
import org.springframework.web.servlet.mvc.method.annotation.ResponseBodyEmitter;
import org.springframework.web.servlet.mvc.method.annotation.SseEmitter;
@RestController
public class GreetingController {
private static final Logger log = LoggerFactory.getLogger(GreetingController.class);
private static final String template = "Hello, %s!";
class GreetingRequestHandler implements Runnable {
private ResponseBodyEmitter emitter;
private Greeting greeting;
public GreetingRequestHandler(final ResponseBodyEmitter emitter, final Greeting greeting) {
this.emitter = emitter;
this.greeting = greeting;
}
@Override
public void run() {
try {
log.info(this.greeting.toString());
this.emitter.send(this.greeting);
Thread.sleep(5000);
if (Status.COMPLETE.equals(this.greeting.getStatus())) {
this.emitter.complete();
} else {
this.greeting.incrementStatus();
new Thread(new GreetingRequestHandler(this.emitter, this.greeting)).start();
}
} catch (IOException | InterruptedException e) {
e.printStackTrace();
}
}
}
@RequestMapping(path = "/greeting")
public SseEmitter greeting(@RequestParam(value = "name", defaultValue = "World") final String name) {
SseEmitter emitter = new SseEmitter();
Greeting greeting = new Greeting(String.format(template, name));
greeting.add(linkTo(methodOn(GreetingController.class).greeting(name)).withSelfRel());
new Thread(new GreetingRequestHandler(emitter, greeting)).start();
log.info("returning emitter");
return emitter;
}
}
Greeting
类如下:
package hello;
import java.util.concurrent.atomic.AtomicInteger;
import org.springframework.hateoas.ResourceSupport;
import com.fasterxml.jackson.annotation.JsonCreator;
import com.fasterxml.jackson.annotation.JsonIgnoreProperties;
import com.fasterxml.jackson.annotation.JsonProperty;
@JsonIgnoreProperties(ignoreUnknown = true)
public class Greeting extends ResourceSupport {
private final String content;
private final static AtomicInteger idProvider = new AtomicInteger();
private int greetingId;
private Status status;
enum Status {
ENQUEUED,
PROCESSING,
COMPLETE;
}
@JsonCreator
public Greeting(@JsonProperty("content") final String content) {
this.greetingId = idProvider.addAndGet(1);
this.status = Status.ENQUEUED;
this.content = content;
}
public Status getStatus() {
return this.status;
}
protected void setStatus(final Status status) {
this.status = status;
}
public int getGreetingId() {
return this.greetingId;
}
public String getContent() {
return this.content;
}
@Override
public String toString() {
return "Greeting{id='" + this.greetingId + "', status='" + this.status + "' content='" + this.content + "', " + super.toString() + "}";
}
public void incrementStatus() {
switch (this.status) {
case ENQUEUED:
this.status = Status.PROCESSING;
break;
case PROCESSING:
this.status = Status.COMPLETE;
break;
default:
break;
}
}
}
此代码完美运行。如果我尝试使用 Web 浏览器访问 REST 服务,我会看到出现的事件具有正确的内容和链接。
结果看起来像(每个事件在前一个事件后出现 5 秒):
data:{"content":"Hello, Kraal!","greetingId":8,"status":"ENQUEUED","_links":{"self":{"href":"http://localhost:8080/greeting?name=Kraal"}}}
data:{"content":"Hello, Kraal!","greetingId":8,"status":"PROCESSING","_links":{"self":{"href":"http://localhost:8080/greeting?name=Kraal"}}}
data:{"content":"Hello, Kraal!","greetingId":8,"status":"COMPLETE","_links":{"self":{"href":"http://localhost:8080/greeting?name=Kraal"}}}
现在我需要调用此 REST 服务并从另一个 Spring 应用程序读取这些事件...但我不知道如何使用 Spring 编写客户端代码。这不起作用,因为 RestTemplate
是为同步客户端 HTTP 访问而设计的...
ObjectMapper mapper = new ObjectMapper();
mapper.configure(DeserializationFeature.FAIL_ON_UNKNOWN_PROPERTIES, false);
mapper.registerModule(new Jackson2HalModule());
// required for HATEOAS
MappingJackson2HttpMessageConverter converter = new MappingJackson2HttpMessageConverter();
converter.setSupportedMediaTypes(MediaType.parseMediaTypes("application/hal+json"));
converter.setObjectMapper(mapper);
// required in order to be able to read serialized objects
MappingJackson2HttpMessageConverter converter2 = new MappingJackson2HttpMessageConverter();
converter2.setSupportedMediaTypes(MediaType.parseMediaTypes("application/octet-stream"));
converter2.setObjectMapper(mapper);
// required to understand SSE events
MappingJackson2HttpMessageConverter converter3 = new MappingJackson2HttpMessageConverter();
converter3.setSupportedMediaTypes(MediaType.parseMediaTypes("text/event-stream"));
List<HttpMessageConverter<?>> converters = new ArrayList<HttpMessageConverter<?>>();
converters.add(converter);
converters.add(converter2);
converters.add(converter3);
// probably wrong template
RestTemplate restTemplate = new RestTemplate();
restTemplate = new RestTemplate(converters);
// this does not work as I receive events and no a single object
Greeting greeting = restTemplate.getForObject("http://localhost:8080/greeting/?name=Kraal", Greeting.class);
log.info(greeting.toString());
我得到的错误信息是:
Caused by: com.fasterxml.jackson.core.JsonParseException: Unrecognized token 'data': was expecting ('true', 'false' or 'null')
事实上,每个事件都是一个 SSE 事件,并以“数据:”开头......
所以问题是:
- 为了能够将 SSE 映射到 Jackson,我应该注册什么 ObjectMapper 模块?
- 如何使用 Spring 订阅传入的 SSE 事件(观察者模式)?
提前致谢。
旁注:由于我正在努力使用 Spring 来完成它,所以我尝试使用 Jersey SSE 支持来完成它,如下所示。使用 Jersey 我收到了预期的事件,但是我无法将它们转换为 Greeting
类(出于与上述相同的原因,我猜这是我没有正确的转换器模块。) :
Client client = ClientBuilder.newBuilder().register(converter).register(SseFeature.class).build();
WebTarget target = client.target("http://localhost:8080/greeting/?name=Kraal");
EventInput eventInput = target.request().get(EventInput.class);
while (!eventInput.isClosed()) {
final InboundEvent inboundEvent = eventInput.read();
if (inboundEvent == null) {
// connection has been closed
break;
}
// this works fine and prints out events as they are incoming
System.out.println(inboundEvent.readData(String.class));
// but this doesn't as no proper way to deserialize the
// class with HATEOAS links can be found
// Greeting greeting = inboundEvent.readData(Greeting.class);
// System.out.println(greeting.toString());
}
最佳答案
你可以使用inboundEvent.readData(Class<T> type)
关于java - 如何使用 Spring 订阅传入的 SSE 事件,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/34394019/