如何保存媒体文件（比如MP3播放），从现有的URI，这我从隐含意图得到些什么？
Answer 1:
使用这种方法，它的工作原理
void savefile(URI sourceuri)
{
String sourceFilename= sourceuri.getPath();
String destinationFilename = android.os.Environment.getExternalStorageDirectory().getPath()+File.separatorChar+"abc.mp3";
BufferedInputStream bis = null;
BufferedOutputStream bos = null;
try {
bis = new BufferedInputStream(new FileInputStream(sourceFilename));
bos = new BufferedOutputStream(new FileOutputStream(destinationFilename, false));
byte[] buf = new byte[1024];
bis.read(buf);
do {
bos.write(buf);
} while(bis.read(buf) != -1);
} catch (IOException e) {
e.printStackTrace();
} finally {
try {
if (bis != null) bis.close();
if (bos != null) bos.close();
} catch (IOException e) {
e.printStackTrace();
}
}
}
Answer 2:
private static String FILE_NAM = "video";
String outputfile = getFilesDir() + File.separator+FILE_NAM+"_tmp.mp4";
InputStream in = getContentResolver().openInputStream(videoFileUri);
private static File createFileFromInputStream(InputStream inputStream, String fileName) {
try{
File f = new File(fileName);
f.setWritable(true, false);
OutputStream outputStream = new FileOutputStream(f);
byte buffer[] = new byte[1024];
int length = 0;
while((length=inputStream.read(buffer)) > 0) {
outputStream.write(buffer,0,length);
}
outputStream.close();
inputStream.close();
return f;
}catch (IOException e) {
System.out.println("error in creating a file");
e.printStackTrace();
}
return null;
}
Answer 3:
如果URI是从谷歌驱动器接收到的，它可以是一个虚拟文件URI了。 勾选此文章从CommonsWare以获取更多信息。 所以，你必须要考虑的同时也从乌里保存文件的条件。
找到，如果文件URI是虚拟的或者不是你可以使用
private static boolean isVirtualFile(Context context, Uri uri) {
if (Build.VERSION.SDK_INT >= Build.VERSION_CODES.KITKAT) {
if (!DocumentsContract.isDocumentUri(context, uri)) {
return false;
}
Cursor cursor = context.getContentResolver().query(
uri,
new String[]{DocumentsContract.Document.COLUMN_FLAGS},
null, null, null);
int flags = 0;
if (cursor.moveToFirst()) {
flags = cursor.getInt(0);
}
cursor.close();
return (flags & DocumentsContract.Document.FLAG_VIRTUAL_DOCUMENT) != 0;
} else {
return false;
}
}
你可以从这样的这个虚拟文件流数据：
private static InputStream getInputStreamForVirtualFile(Context context, Uri uri, String mimeTypeFilter)
throws IOException {
ContentResolver resolver = context.getContentResolver();
String[] openableMimeTypes = resolver.getStreamTypes(uri, mimeTypeFilter);
if (openableMimeTypes == null || openableMimeTypes.length < 1) {
throw new FileNotFoundException();
}
return resolver
.openTypedAssetFileDescriptor(uri, openableMimeTypes[0], null)
.createInputStream();
}
为了找到MIME类型的尝试
private static String getMimeType(String url) {
String type = null;
String extension = MimeTypeMap.getFileExtensionFromUrl(url);
if (extension != null) {
type = MimeTypeMap.getSingleton().getMimeTypeFromExtension(extension);
}
return type;
}
总体而言，你可以使用
public static boolean saveFile(Context context, String name, Uri sourceuri, String destinationDir, String destFileName) {
BufferedInputStream bis = null;
BufferedOutputStream bos = null;
InputStream input = null;
boolean hasError = false;
try {
if (isVirtualFile(context, sourceuri)) {
input = getInputStreamForVirtualFile(context, sourceuri, getMimeType(name));
} else {
input = context.getContentResolver().openInputStream(sourceuri);
}
boolean directorySetupResult;
File destDir = new File(destinationDir);
if (!destDir.exists()) {
directorySetupResult = destDir.mkdirs();
} else if (!destDir.isDirectory()) {
directorySetupResult = replaceFileWithDir(destinationDir);
} else {
directorySetupResult = true;
}
if (!directorySetupResult) {
hasError = true;
} else {
String destination = destinationDir + File.separator + destFileName;
int originalsize = input.available();
bis = new BufferedInputStream(input);
bos = new BufferedOutputStream(new FileOutputStream(destination));
byte[] buf = new byte[originalsize];
bis.read(buf);
do {
bos.write(buf);
} while (bis.read(buf) != -1);
}
} catch (Exception e) {
e.printStackTrace();
hasError = true;
} finally {
try {
if (bos != null) {
bos.flush();
bos.close();
}
} catch (Exception ignored) {
}
}
return !hasError;
}
private static boolean replaceFileWithDir(String path) {
File file = new File(path);
if (!file.exists()) {
if (file.mkdirs()) {
return true;
}
} else if (file.delete()) {
File folder = new File(path);
if (folder.mkdirs()) {
return true;
}
}
return false;
}
呼叫从AsycTask此方法。 让我知道，如果这有助于。
Answer 4:
1.创建从URI路径作为一个文件：
File from = new File(uri.toString());
2.创建您想要的文件作为保存另一个文件：
File to = new File("target file path");
3.Rename的文件：
from.renameTo(to);
有了这个从默认的路径中的文件被自动删除，在新的路径中创建。
Answer 5:
这是最简单和最干净的：
private void saveFile(Uri sourceUri, File destination)
try {
File source = new File(sourceUri.getPath());
FileChannel src = new FileInputStream(source).getChannel();
FileChannel dst = new FileOutputStream(destination).getChannel();
dst.transferFrom(src, 0, src.size());
src.close();
dst.close();
} catch (IOException ex) {
ex.printStackTrace();
}
}
Answer 6:
对于未来的游客。 科特林代码复制从拾取意图挑文件。
val file = File("//add your destination address")
val fileContentURI = FileProvider.getUriForFile(context!!, "com.minutecodes.openote.fileprovider", file)
val out = context!!.contentResolver.openOutputStream(fileContentURI)
context!!.contentResolver.openInputStream(clip.getItemAt(0)!!.uri)!!.copyTo(out!!, DEFAULT_BUFFER_SIZE)
文章来源: Android: save a file from an existing URI
