Question

我正在迁移到Java的ASK SDK v2，在SDK v2中，每个onLaunch，onSessionEnded和ontIntent情况都被隔离到不同的处理程序中。

问题：

如何将我的输入传递到handle(HandlerInput input)的{{1}}，以及如何从我的代码中调用LaunchRequestHandler的{{1}}？

此外，在V2中，handle(HandlerInput input)类已被删除，因此如何创建LaunchRequestHandler以获取SpeechletRequestEnvelope和requestEnvelope的详细信息？

Request

呼叫地点：

Session

如何将public class LaunchRequestHandler implements RequestHandler { @Override public boolean canHandle(HandlerInput input) { return input.matches(Predicates.requestType(LaunchRequest.class)); } @Override public Optional<Response> handle(HandlerInput input) { String speechText = "Welcome to the Alexa Skills Kit, you can say hello"; return input.getResponseBuilder() .withSpeech(inputString) .withSimpleCard("HelloWorld", inputString) .withReprompt(inputString) .build(); } }传递到String body = myO ject.getAdditionalProperties().get("request").toString(); byte[] myRequest = URLDecoder.decode(body, StandardCharsets.UTF_8.name()).getBytes();的{{1}}以及如何调用相同的myRequest方法？

Answer 1

第二个与请求和会话详细信息有关的问题

public ResponseEntity<String> handleAlexaRequest(HttpServletRequest request) {

    RequestEnvelope requestEnvelope;
    ServletRequest servletRequest;

    byte[] serializedRequestEnvelope;

    try {
        serializedRequestEnvelope = IOUtils.toByteArray(request.getInputStream());
        RequestEnvelope deserializedRequestEnvelope = this.serializer.deserialize(new String(serializedRequestEnvelope, StandardCharsets.UTF_8), RequestEnvelope.class);

        servletRequest = new ServletRequest(request,serializedRequestEnvelope,deserializedRequestEnvelope);

        requestEnvelope = verifier.verify(alexaHttpRequest);

    }


    requestEnvelope.getRequest();
    requestEnvelope.getSession()

如何在Java版Alexa SDK V2中将Input传递给handle（）以及如何调用LaunchRequestHandler的handle（）？

问题：

1 个答案: