将输入流传递给视频视图答案

【问题标题】：Passing input stream to video view将输入流传递给视频视图
【发布时间】：2013-07-17 12:23:32
【问题描述】：

我有输入流，我需要将其传递给视频视图：

但它说 OutOfMemoryError

这是我的代码：

//to convert input stream to string
public static String convertStreamToString(InputStream is) throws Exception {
        BufferedReader reader = new BufferedReader(new InputStreamReader(is));
        StringBuilder sb = new StringBuilder();
        String line = null;

        while ((line = reader.readLine()) != null) {
            sb.append(line);
        }

        is.close();

        return sb.toString();
    }

//passing that string to video view
String total = convertStreamToString(inputStream);
Uri uri = Uri.parse(total);
videoHolder.setVideoURI(uri);

如何做到这一点？

编辑

这是堆栈跟踪

07-17 17:46:34.242: E/AndroidRuntime(5384): FATAL EXCEPTION: main
07-17 17:46:34.242: E/AndroidRuntime(5384): java.lang.OutOfMemoryError
07-17 17:46:34.242: E/AndroidRuntime(5384):     at java.lang.AbstractStringBuilder.enlargeBuffer(AbstractStringBuilder.java:94)
07-17 17:46:34.242: E/AndroidRuntime(5384):     at java.lang.AbstractStringBuilder.append0(AbstractStringBuilder.java:145)
07-17 17:46:34.242: E/AndroidRuntime(5384):     at java.lang.StringBuilder.append(StringBuilder.java:216)
07-17 17:46:34.242: E/AndroidRuntime(5384):     at com.abc.Utility.Utils.convertStreamToString(Utils.java:196)
07-17 17:46:34.242: E/AndroidRuntime(5384):     at com.abc.xyz.VideoPlayer.onCreate(VideoPlayer.java:257)
07-17 17:46:34.242: E/AndroidRuntime(5384):     at android.app.Activity.performCreate(Activity.java:4635)
07-17 17:46:34.242: E/AndroidRuntime(5384):     at android.app.Instrumentation.callActivityOnCreate(Instrumentation.java:1049)
07-17 17:46:34.242: E/AndroidRuntime(5384):     at android.app.ActivityThread.performLaunchActivity(ActivityThread.java:2031)
07-17 17:46:34.242: E/AndroidRuntime(5384):     at android.app.ActivityThread.handleLaunchActivity(ActivityThread.java:2092)
07-17 17:46:34.242: E/AndroidRuntime(5384):     at android.app.ActivityThread.access$600(ActivityThread.java:126)
07-17 17:46:34.242: E/AndroidRuntime(5384):     at android.app.ActivityThread$H.handleMessage(ActivityThread.java:1172)
07-17 17:46:34.242: E/AndroidRuntime(5384):     at android.os.Handler.dispatchMessage(Handler.java:99)
07-17 17:46:34.242: E/AndroidRuntime(5384):     at android.os.Looper.loop(Looper.java:137)
07-17 17:46:34.242: E/AndroidRuntime(5384):     at android.app.ActivityThread.main(ActivityThread.java:4586)
07-17 17:46:34.242: E/AndroidRuntime(5384):     at java.lang.reflect.Method.invokeNative(Native Method)
07-17 17:46:34.242: E/AndroidRuntime(5384):     at java.lang.reflect.Method.invoke(Method.java:511)
07-17 17:46:34.242: E/AndroidRuntime(5384):     at com.android.internal.os.ZygoteInit$MethodAndArgsCaller.run(ZygoteInit.java:784)
07-17 17:46:34.242: E/AndroidRuntime(5384):     at com.android.internal.os.ZygoteInit.main(ZygoteInit.java:551)
07-17 17:46:34.242: E/AndroidRuntime(5384):     at dalvik.system.NativeStart.main(Native Method)

【问题讨论】：

请发布整个堆栈跟踪。
您是否将视频文件投射到字符串对象？ :D 或者换种说法 - 在字符串对象中加载视频文件？
@CommonsWare 我已经发布了堆栈跟踪
你从哪里得到这个流？请记住，setVideoURI() 采用 URI，例如 file:// 或 http:// 路径。
输入流是一个classLoader.getResourceAsStream，

标签： java android video inputstream android-videoview

【解决方案1】：

好的，我自己解决了，但是在视频文件开始之前有 3 到 4 秒的延迟

希望对某人有所帮助：

public static String getDataSource(String path) throws IOException {
      if (!URLUtil.isNetworkUrl(path)) {
            return path;
        } else {
           URL url = new URL(path);
           URLConnection cn = url.openConnection();
           cn.connect();
            InputStream stream = cn.getInputStream();
            if (stream == null)
                throw new RuntimeException("stream is null");
            File temp = File.createTempFile("mediaplayertmp", "dat");
            temp.deleteOnExit();
            String tempPath = temp.getAbsolutePath();
            FileOutputStream out = new FileOutputStream(temp);
            byte buf[] = new byte[128];
            do {
                int numread = stream.read(buf);
                if (numread <= 0)
                    break;
                out.write(buf, 0, numread);
            } while (true);
            try {
                stream.close();
                out.close();
            } catch (IOException ex) {
              //  Log.e(TAG, "error: " + ex.getMessage(), ex);
            }
            return tempPath;
        }
    }

【讨论】：