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I am using Storage Access Framework(SAF) in my app. In order to scan a file(photo or document) i need send a local file path trough an API. I've managed to setup the SAF correctly, and now when a user chooses a file and i am getting a Uri as it supposed to be.

As it seems the Uri is a key for the cloud and not a local file.

The Uri value looks like that:

content://com.android.providers.media.documents/document/image:11862

How can i convert this Uri into a local file? Should i download the file from the cloud? How can i do it?

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As it seems the Uri is a key for the cloud and not a local file.

The Uri is an opaque reference to a piece of content. You have no way of knowing where the data is, nor should you care.

How can i convert this Uri into a local file?

Ideally, you don't. Ideally, you "scan a file(photo or document)" using some library that supports an InputStream. In that case, you use ContentResolver and openInputStream() to get the stream to pass to the library.

If your library does not support InputStream as a data source, you will need to use openInputStream() yourself, using Java I/O to make a copy of the content as a file in the filesystem, for you to pass to the library.

于 2016-01-14T13:16:21.030 回答
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You can read directly from the underlying file, with random access, under the following conditions:

  • The URI points to a local file, you can check this comparing the URI authority with the value com.android.externalstorage.documents;
  • You have both read and write access to the URI.
ParcelFileDescriptor parcelFileDescriptor = getContext().getContentResolver().openFileDescriptor(documentUri, "rw");
FileDescriptor fileDescriptor = parcelFileDescriptor.getFileDescriptor();
FileChannel channel = new FileInputStream(fileDescriptor).getChannel();

Store the FileDescriptor somewhere as long as you keep the FileChannel open: if that object gets garbage collected the channel will be no longer able to access the file.

于 2020-12-09T01:42:29.030 回答