我不明白在 springboot 应用程序中如何正确编写 websocket 的测试用例。我有一个实现WebSocketHandler
的类,我在WebSocketConfigurer
中添加了这个处理程序:
@Configuration
@EnableWebSocket
public class WebSocketConfig implements WebSocketConfigurer {
@Override
public void registerWebSocketHandlers(WebSocketHandlerRegistry registry) {
registry.addHandler(logWebSocketHandler() , "/test")
.addInterceptors(new HttpSessionHandshakeInterceptor())
.setAllowedOrigins("*");
}
@Bean
public LogWebSocketHandler logWebSocketHandler(){
return new LogWebSocketHandler();
}
}
但是当我在测试用例下面写这个时,我得到一个例外:
@RunWith(SpringJUnit4ClassRunner.class)
@SpringApplicationConfiguration({App.class})
@WebAppConfiguration
public class LogsControllerTest {
private WebSocketContainer container;
@Before
public void setup() {
container = ContainerProvider.getWebSocketContainer();
}
@Test
public void testGetLog() throws Exception {
Session session = container.connectToServer(new TestEndpoint(),
URI.create("ws://127.0.0.1:8080/test"));
}
}
异常:
javax.websocket.DeploymentException
:启动WebSocket连接的HTTP请求失败
我读到如果我设置ws://127.0.0.1:8080/test/
(结尾为斜线)它会起作用,但它不起作用。
我做错了什么?
答案 0 :(得分:1)
我找到了解决方案:
添加注释@WebIntegrationTest(value = 8080)
或您可以指定randomPort = true
之后添加:
@ClientEndpoint
public class TestWebSocketClient {
Session session;
@OnOpen
public void onOpen(final Session session){
this.session = session;
}
}
并且在测试的主体上:
@Autowired
private LogWebSocketHandler socketHandler;
private WebSocketContainer container;
private TestWebSocketClient client;
@Before
public void setup() {
container = ContainerProvider.getWebSocketContainer();
client = new TestWebSocketClient();
}
@Test
public void createSessionAfterOpenLogWebSocketHandler() throws Exception {
container.connectToServer(client , URI.create("ws://localhost:8080/path"));
while( !socketHandler.isOpen()){
// sometime it is doesn't work, but I dont know solution of this problem
// wait until socket is open
}
Assert.assertTrue( socketHandler.isOpen() );
}