我在要录制视频的位置编写代码,然后在录制5秒钟后将其保存。我有以下代码可以在API <24上正常工作,但是对于API> 24,我会收到错误消息。
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$MethodAndArgsCaller.run(RuntimeInit.java:537)
at com.android.internal.os.ZygoteInit.main(ZygoteInit.java:858)
我检查了该错误是由于File对象的语法,并且API> 24不支持它。
尽管我可以找到其他有效的代码,但我对此代码是否有细微调整感到好奇,这也使其适用于API> 24。也许在同一行上。
有人可以帮我吗?
编辑:
我已经尝试过:https://inthecheesefactory.com/blog/how-to-share-access-to-file-with-fileprovider-on-android-nougat/en,但是这样做也会导致应用程序在API 22上崩溃。
您可以使用FileProvider
类授予对特定文件或文件夹的访问权限,以使其他应用程序可以访问它们。创建您自己的继承FileProvider
的类,以确保您的FileProvider
与在导入的依赖项中声明的FileProviders
不冲突,如此处所述。
用file://
URI替换content://
URI的步骤:
添加扩展FileProvider的类
public class GenericFileProvider extends FileProvider {}
在下面的
<provider>
中添加FileProviderAndroidManifest.xml
标签<application>
标签。指定唯一的权限android:authorities
属性以避免冲突,导入依赖项可能指定${applicationId}.provider
和其他常用权限。
<?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>
然后在paths.xml
文件夹中创建provider_ res/xml
文件。如果文件夹不存在,则可能需要创建该文件夹。该文件的内容如下所示。它描述了我们希望共享名为“ external_files”的根文件夹(path=".")
上对外部存储的访问。
<?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);
to
fileUri= FileProvider.getUriForFile(context, context.getApplicationContext().getPackageName() + ".my.package.name.provider", mediaFile);