什么是最简单的方法从android.net.Uri对象持有一个文件:类型转换为java.io.File对象在Android?
我尝试了下面的方法，但不管用:
File file = new File(Environment.getExternalStorageDirectory(), "read.me");
Uri uri = Uri.fromFile(file);
File auxFile = new File(uri.toString());
assertEquals(file.getAbsolutePath(), auxFile.getAbsolutePath());
当前回答
这些对我都没用。我发现这是可行的解决方案。但我的情况仅限于图像。
String[] filePathColumn = { MediaStore.Images.Media.DATA };
Cursor cursor = getActivity().getContentResolver().query(uri, filePathColumn, null, null, null);
cursor.moveToFirst();
int columnIndex = cursor.getColumnIndex(filePathColumn[0]);
String filePath = cursor.getString(columnIndex);
cursor.close();
其他回答
对于那些在这里寻找图像解决方案的人，特别是在这里。
private Bitmap getBitmapFromUri(Uri contentUri) {
String path = null;
String[] projection = { MediaStore.Images.Media.DATA };
Cursor cursor = getContentResolver().query(contentUri, projection, null, null, null);
if (cursor.moveToFirst()) {
int columnIndex = cursor.getColumnIndexOrThrow(MediaStore.Images.Media.DATA);
path = cursor.getString(columnIndex);
}
cursor.close();
Bitmap bitmap = BitmapFactory.decodeFile(path);
return bitmap;
}
使用这个来写入文件，它为我工作时，gif的uri是由GBoard提供的，我必须在我的应用程序数据复制该gif。
try {
String destinationFilePath = getExternalFilesDir("gifs") + "/tempFile.txt";
InputStream inputStream = getContentResolver().openInputStream(uri);
OutputStream outputStream = new FileOutputStream(destinationFilePath);
byte[] buffer = new byte[1024];
int bytesRead;
while ((bytesRead = inputStream.read(buffer)) != -1) {
outputStream.write(buffer, 0, bytesRead);
}
inputStream.close();
outputStream.close();
}
catch (Exception e) {
e.printStackTrace();
}
通过下面的代码，我能够获得adobe应用程序共享pdf文件作为流，并保存到android应用程序路径
Android.Net.Uri fileuri =
(Android.Net.Uri)Intent.GetParcelableExtra(Intent.ExtraStream);
fileuri i am getting as {content://com.adobe.reader.fileprovider/root_external/
data/data/com.adobe.reader/files/Downloads/sample.pdf}
string filePath = fileuri.Path;
filePath I am gettings as root_external/data/data/com.adobe.reader/files/Download/sample.pdf
using (var stream = ContentResolver.OpenInputStream(fileuri))
{
byte[] fileByteArray = ToByteArray(stream); //only once you can read bytes from stream second time onwards it has zero bytes
string fileDestinationPath ="<path of your destination> "
convertByteArrayToPDF(fileByteArray, fileDestinationPath);//here pdf copied to your destination path
}
public static byte[] ToByteArray(Stream stream)
{
var bytes = new List<byte>();
int b;
while ((b = stream.ReadByte()) != -1)
bytes.Add((byte)b);
return bytes.ToArray();
}
public static string convertByteArrayToPDF(byte[] pdfByteArray, string filePath)
{
try
{
Java.IO.File data = new Java.IO.File(filePath);
Java.IO.OutputStream outPut = new Java.IO.FileOutputStream(data);
outPut.Write(pdfByteArray);
return data.AbsolutePath;
}
catch (System.Exception ex)
{
return string.Empty;
}
}
编辑:对不起，我之前应该测试得更好。这应该可以工作:
new File(new URI(androidURI.toString()));
URI是java.net.URI。
科特林 2022
suspend fun Context.createFileFromAsset(assetName: String, fileName: String): File? {
return withContext(Dispatchers.IO) {
runCatching {
val stream = assets.open(assetName)
val file = File(cacheDir.absolutePath, fileName)
org.apache.commons.io.FileUtils.copyInputStreamToFile(stream, file)
file
}.onFailure { Timber.e(it) }.getOrNull()
}
}
处理完文件后，请确保对其调用.delete()。向@Mohsent致敬
