java - 玩!框架 [1.2.4] - 通过 wami-recorder 接受音频/xwav 内容类型

Question

我正在尝试使用 https://code.google.com/p/wami-recorder/ 在浏览器端录制一段声音片段，然后通过 wami 示例和服务器端 Controller 方法上传该文件。出于某种原因，request.body InputStream 抛出此错误:

java.io.IOException: Bad file descriptor
    at java.io.FileInputStream.readBytes(Native Method)
    at java.io.FileInputStream.read(FileInputStream.java:198)
    at controllers.APIVoicemailController.addNewRecording(APIVoicemailController.java:56)
    at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method)
    at sun.reflect.NativeMethodAccessorImpl.invoke(NativeMethodAccessorImpl.java:39)
    at sun.reflect.DelegatingMethodAccessorImpl.invoke(DelegatingMethodAccessorImpl.java:25)
    at java.lang.reflect.Method.invoke(Method.java:597)
    at play.mvc.ActionInvoker.invokeWithContinuation(ActionInvoker.java:548)
    at play.mvc.ActionInvoker.invoke(ActionInvoker.java:502)
    at play.mvc.ActionInvoker.invokeControllerMethod(ActionInvoker.java:478)
    at play.mvc.ActionInvoker.invokeControllerMethod(ActionInvoker.java:473)
    at play.mvc.ActionInvoker.invoke(ActionInvoker.java:161)
    at play.server.PlayHandler$NettyInvocation.execute(PlayHandler.java:256)
    at play.Invoker$Invocation.run(Invoker.java:276)
    at play.server.PlayHandler$NettyInvocation.run(PlayHandler.java:234)
    at java.util.concurrent.Executors$RunnableAdapter.call(Executors.java:441)
    at java.util.concurrent.FutureTask$Sync.innerRun(FutureTask.java:303)
    at java.util.concurrent.FutureTask.run(FutureTask.java:138)
    at java.util.concurrent.ScheduledThreadPoolExecutor$ScheduledFutureTask.access$301(ScheduledThreadPoolExecutor.java:98)
    at java.util.concurrent.ScheduledThreadPoolExecutor$ScheduledFutureTask.run(ScheduledThreadPoolExecutor.java:206)
    at java.util.concurrent.ThreadPoolExecutor$Worker.runTask(ThreadPoolExecutor.java:886)
    at java.util.concurrent.ThreadPoolExecutor$Worker.run(ThreadPoolExecutor.java:908)
    at java.lang.Thread.run(Thread.java:680)

路由POST/upload的 Controller 方法

   public static void addNewRecording() {
        if (request.method.equals(Constants.METH_POST)) {
            try {
                InputStream inputStream = request.body;
                Map<String, Header> headerMap = request.headers;
                for (Entry<String, Header> entry : headerMap.entrySet()) {
                    Logger.info("Head: " + entry.getKey() + " name; "
                        + entry.getValue().name + " val: " + entry.getValue().value());
                }

                OutputStream out = new FileOutputStream(new File(
                    "/Users/franklovecchio/Desktop/file.mp3"));

                int read = 0;
                byte[] bytes = new byte[1024];

                            /* ERROR ON LINE BELOW */
                while ((read = inputStream.read(bytes)) != -1) {
                    out.write(bytes, 0, read);
                }

                inputStream.close();
                out.flush();
                out.close();
            } catch (IOException e) {
                e.printStackTrace();
            }

            for (Entry<String, String> params : request.params.allSimple().entrySet()) {
                Logger.info("k: " + params.getKey() + "v: " + params.getValue());
            }
            renderCustomMessage("Created new recording.", "Success.");
        }
    }

我该如何解决？

Answer 1

我正在使用 Play 2，但如果您只是获取 InputStream 的内容而不检查 header ，您将能够毫无问题地将其存储在 WAV 文件或 DB 中。