使用File对象在Android（API>24）中捕获和保存视频？

Question

我想录制一段视频，录制5秒后保存。我有以下在API上运行良好的代码

代码：

public void startRecording()
{
    File mediaFile = new
            File(Environment.getExternalStorageState().equals(Environment.MEDIA_MOUNTED)
            + "/myvideo.mp4");


    Intent intent = new Intent(MediaStore.ACTION_VIDEO_CAPTURE);
    intent.putExtra(MediaStore.EXTRA_DURATION_LIMIT,5);
    fileUri = Uri.fromFile(mediaFile);

    intent.putExtra(MediaStore.EXTRA_OUTPUT, fileUri);
    startActivityForResult(intent, VIDEO_CAPTURE);
}

protected void onActivityResult(int requestCode,
                                int resultCode, Intent data) {

    if (requestCode == VIDEO_CAPTURE) {
        if (resultCode == RESULT_OK) {
            Toast.makeText(this, "Video has been saved to:\n" +
                    data.getData(), Toast.LENGTH_LONG).show();
        } else if (resultCode == RESULT_CANCELED) {
            Toast.makeText(this, "Video recording cancelled.",
                    Toast.LENGTH_LONG).show();
        } else {
            Toast.makeText(this, "Failed to record video",
                    Toast.LENGTH_LONG).show();
        }
    }
}

错误：

2019-10-08 01:15:43.483 21573-21573/com.mobilecomputing.learn2sign E/AndroidRuntime: FATAL EXCEPTION: main
    Process: com.mobilecomputing.learn2sign, PID: 21573
    android.os.FileUriExposedException: file:///storage/emulated/0/myvideo.mp4 exposed beyond app through ClipData.Item.getUri()
        at android.os.StrictMode.onFileUriExposed(StrictMode.java:1978)
        at android.net.Uri.checkFileUriExposed(Uri.java:2371)
        at android.content.ClipData.prepareToLeaveProcess(ClipData.java:963)
        at android.content.Intent.prepareToLeaveProcess(Intent.java:10228)
        at android.content.Intent.prepareToLeaveProcess(Intent.java:10213)
        at android.app.Instrumentation.execStartActivity(Instrumentation.java:1854)
        at android.app.Activity.startActivityForResult(Activity.java:4599)
        at androidx.fragment.app.FragmentActivity.startActivityForResult(FragmentActivity.java:676)
        at android.app.Activity.startActivityForResult(Activity.java:4557)
        at androidx.fragment.app.FragmentActivity.startActivityForResult(FragmentActivity.java:663)
        at com.mobilecomputing.learn2sign.PlayHelpVideo.startRecording(PlayHelpVideo.java:125)
        at com.mobilecomputing.learn2sign.PlayHelpVideo$1.onClick(PlayHelpVideo.java:46)
        at android.view.View.performClick(View.java:6669)
        at android.view.View.performClickInternal(View.java:6638)
        at android.view.View.access$3100(View.java:789)
        at android.view.View$PerformClick.run(View.java:26145)
        at android.os.Handler.handleCallback(Handler.java:873)
        at android.os.Handler.dispatchMessage(Handler.java:99)
        at android.os.Looper.loop(Looper.java:193)
        at android.app.ActivityThread.main(ActivityThread.java:6898)
        at java.lang.reflect.Method.invoke(Native Method)
        at com.android.internal.os.RuntimeInit$MethodAndArgsCaller.run(RuntimeInit.java:537)
        at com.android.internal.os.ZygoteInit.main(ZygoteInit.java:858)

我检查了，错误是因为文件对象的语法，API不支持它

虽然我可以找到其他工作的代码，我很好奇，如果这个代码有一个小的调整，可以使它的API工作

有人能帮我吗？

编辑：

我尝试过：https://inthecheesefactory.com/blog/how-to-share-access-to-file-with-fileprovider-on-android-nougat/en但这样做也会导致应用程序在API 22上崩溃。

更新：

根据回复的答案进行更改后，代码在API上运行

private File createImageFile() throws IOException {
    // Create an image file name
    String imageFileName = "myvideo";
    File storageDir = new File(Environment.getExternalStoragePublicDirectory(
            Environment.DIRECTORY_DCIM), "Camera");
    File image = File.createTempFile(
            imageFileName,  /* prefix */
            ".mp4",         /* suffix */
            storageDir      /* directory */
    );

    // Save a file: path for use with ACTION_VIEW intents
    Log.v("myTag","FAB create image");
    mCurrentPhotoPath = "file:" + image.getAbsolutePath();
    return image;
}

这是从下面的函数调用的：

public void startRecording()
{

    Log.v("myTag","FAB recording");
    File mediaFile = null;
    try {
        mediaFile = createImageFile();
    } catch (IOException ex) {
        Log.v("myTag","Exception");
        return;
    }


    Intent intent = new Intent(MediaStore.ACTION_VIDEO_CAPTURE);
    intent.putExtra(MediaStore.EXTRA_DURATION_LIMIT,5);
    //fileUri = Uri.fromFile(mediaFile);
    fileUri = FileProvider.getUriForFile(PlayHelpVideo.this,
            BuildConfig.APPLICATION_ID + ".provider",
            mediaFile);
    intent.addFlags(Intent.FLAG_GRANT_READ_URI_PERMISSION);
    intent.putExtra(MediaStore.EXTRA_OUTPUT, fileUri);
    startActivityForResult(intent, VIDEO_CAPTURE);
}

Answer 1

您可以使用FileProvider类来访问特定的文件或文件夹，使其他应用程序可以访问这些文件或文件夹。创建您自己的继承文件提供者的类，以确保您的文件提供者不会与此处所述的导入依赖项中声明的文件提供者冲突。

将file：//URI替换为内容：//URI的步骤：

添加扩展FileProvider的类

public class GenericFileProvider extends FileProvider {}

添加文件提供者

<?xml version="1.0" encoding="utf-8"?>
<manifest xmlns:android="http://schemas.android.com/apk/res/android"
    ...
    <application
        ...
        <provider
            android:name=".GenericFileProvider"
            android:authorities="${applicationId}.fileprovider"
            android:exported="false"
            android:grantUriPermissions="true">
            <meta-data
                android:name="android.support.FILE_PROVIDER_PATHS"
                android:resource="@xml/provider_paths"/>
        </provider>
    </application>
</manifest>

然后创建提供者路径。xml文件夹中的文件。如果文件夹不存在，则可能需要创建该文件夹。文件内容如下所示。它描述了我们希望以External_files的名称共享对根文件夹（path=“.”）中外部存储的访问。

<?xml version="1.0" encoding="utf-8"?>
<paths xmlns:android="http://schemas.android.com/apk/res/android">
    <external-path name="external_files" path="."/>
</paths>

最后一步是更改下面的代码行

fileUri = Uri.fromFile(mediaFile);

到

fileUri= FileProvider.getUriForFile(context, context.getApplicationContext().getPackageName() + ".my.package.name.provider", mediaFile);