Question

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

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)

路线POST /upload的控制器方法

   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.");
        }
    }

我该如何解决这个问题？

Answer 1

我正在使用Play 2，但是如果你只是在没有检查标题的情况下获得InputStream的内容，那么你就可以将它存储在WAV文件或数据库中而没有任何问题。

Answer 2

我最近的表现完全相同。所遵循的步骤如下：

1.获取发布到record.php的Wami的示例代码

2.将记录网址从record.php更改为Play中与我的控制器对应的网址

3.在控制器内部，只保存收到的数据而不检查标题。（可能这是我在您的代码中看到的差异）。使用firebug可以帮助您查看请求类型（无论是音频/ ..还是不是）

数据包含在请求正文中，只是将其作为输入流读取并将其写入一个足以满足我的mp3。

玩！框架[1.2.4] - 通过wami-recorder接受audio / xwav Content-Type

2 个答案: