I have a content of dash mpd file on string variable
String MPDString = "<?xml version="1.0"?>
" +
"<!-- MPD file Generated with GPAC version 0.5.2-DEV-revVersion: 0.5.2-426-gc5ad4e4+dfsg5-1build1 at 2018-03-07T11:10:44.851Z-->\n" +
...........
</MPD>
";
I want to use this string as dashsource but confused
I looked the link : Create MPD file from String
But still confused
Can anyone please provide me some more information
If you look at the answer you linked to in your question you will see the last example is creating the manifest from a string named 'manifestString'.
DataSource.Factory manifestDataSourceFactory = new DataSource.Factory() {
#Override
public DataSource createDataSource() {
return new ByteArrayDataSource(manifestString.getBytes());
}
};
Assunming your 'MPDString' in your example is a correctly formatted mod then you simply use it above in place of 'manifestString':
DataSource.Factory manifestDataSourceFactory = new DataSource.Factory() {
#Override
public DataSource createDataSource() {
return new ByteArrayDataSource(MPDString.getBytes());
}
};
Help needed
I use the JID3 library to edit mp3 tags. This works fine as long as the music is on an internal sdcard. However, for api 19+ this no longer works for the external sdcard.
I understand I need to implement the Storage Access Framework but have no idea how to go about this.
The way JID3 works is it reads the mp3 file, extracts the tag(s) under consideration, say artist, updates the tag(s) with the new value.
the following steps need amendment:
create a .tmp file in the actual folder where the track resides, which ends up as the actual track with amended tag(s)
delete original track
rename the tmp file with the original track name.
I posted the relevant pieces of code
public class MP3File extends MediaFile
// create temporary file to work with
try
{
oTmpFileSource = m_oFileSource.createTempFile("id3.", ".tmp");
}
catch (Exception e)
{
throw new ID3Exception("Unable to create temporary file.", e);
}
m_oFileSource.delete();
oTmpFileSource.renameTo(m_oFileSource);
public class FileSource implements IFileSource
public IFileSource createTempFile(String sPrefix, String sSuffix)
throws IOException
{
File oTmpFile = File.createTempFile("id3.", ".tmp", m_oFile.getAbsoluteFile().getParentFile());
return new FileSource(oTmpFile);
}
public boolean delete()
{
return m_oFile.delete();
}
public boolean renameTo(IFileSource oFileSource)
throws IOException
{
if ( ! (oFileSource instanceof FileSource))
{
throw new IOException("Cannot rename between different file source types.");
}
return m_oFile.renameTo(((FileSource)oFileSource).m_oFile);
}
I want to initialize MediaPlayer instances for all of the soundfiles found in res/raw:
/res/raw/test1.mp3
/res/raw/test2.mp3
/res/raw/testN.mp3
Purpose is to play different samples on a button click, without delays.
List<MediaPlayer> player = new ArrayList<>();
//TODO how to loop properly?
for (Rawfile file : rawfiles) {
pl = MediaPlayer.create(getBaseContext(), R.raw.test1);
player.add(pl);
}
Lateron, if eg button2 is clicked:
player.get(1).start();
Question: how can I get the R.raw.* files dynamically during initialization of the app?
Update: the following is quite close, but there are 2 problems:
1) If eg only one file "test.mp3" is placed in my /res/raw folder, the function shows 3 files.
2) How can I then load those files to mediaplayer?
public void listRaw(){
Field[] fields=R.raw.class.getFields();
for(int count=0; count < fields.length; count++){
Log.i("Raw Asset: ", fields[count].getName());
}
}
Result:
I/Raw Asset:: $change
I/Raw Asset:: serialVersionUID
I/Raw Asset:: test
For the moment solved as follows, but feels kinda hacky:
public static List<Integer> listRawMediaFiles() {
List<Integer> ids = new ArrayList<>();
for (Field field : R.raw.class.getFields()) {
try {
ids.add(field.getInt(field));
} catch (Exception e) {
//compiled app contains files like '$change' or 'serialVersionUID'
//which are no real media files
}
}
return ids;
}
I Really forget from where i get this, it can be duplicated any way not my code but works perfectly :
private boolean listFiles(String path) {
String [] list;
try {
list = getAssets().list(path);
if (list.length > 0) {
// folder founded
for (String file : list) {
if (!listAssetFiles(path + "/" + file))
return false;
}
} else {
//file founded
}
} catch (IOException e) {
return false;
}
return true;
}
Though undervoted, this answer from CommonsWare is comprehensive. The best you can do is iterate over the raw fields by reflection. If you find non-resource fields, you should discard them manually (I see you did it in an answer).
One point: putting files in raw directory is done during development time, and programming the iteration over raw resources is also done during development time. It's a problem you should solve before compilation, rather than finding out what files you have in run time, that is, you should list the files by name in your code.
Is there a Java equivalent for System.IO.Path.Combine() in C#/.NET? Or any code to accomplish this?
This static method combines one or more strings into a path.
Rather than keeping everything string-based, you should use a class which is designed to represent a file system path.
If you're using Java 7 or Java 8, you should strongly consider using java.nio.file.Path; Path.resolve can be used to combine one path with another, or with a string. The Paths helper class is useful too. For example:
Path path = Paths.get("foo", "bar", "baz.txt");
If you need to cater for pre-Java-7 environments, you can use java.io.File, like this:
File baseDirectory = new File("foo");
File subDirectory = new File(baseDirectory, "bar");
File fileInDirectory = new File(subDirectory, "baz.txt");
If you want it back as a string later, you can call getPath(). Indeed, if you really wanted to mimic Path.Combine, you could just write something like:
public static String combine(String path1, String path2)
{
File file1 = new File(path1);
File file2 = new File(file1, path2);
return file2.getPath();
}
In Java 7, you should use resolve:
Path newPath = path.resolve(childPath);
While the NIO2 Path class may seem a bit redundant to File with an unnecessarily different API, it is in fact subtly more elegant and robust.
Note that Paths.get() (as suggested by someone else) doesn't have an overload taking a Path, and doing Paths.get(path.toString(), childPath) is NOT the same thing as resolve(). From the Paths.get() docs:
Note that while this method is very convenient, using it will imply an assumed reference to the default FileSystem and limit the utility of the calling code. Hence it should not be used in library code intended for flexible reuse. A more flexible alternative is to use an existing Path instance as an anchor, such as:
Path dir = ...
Path path = dir.resolve("file");
The sister function to resolve is the excellent relativize:
Path childPath = path.relativize(newPath);
The main answer is to use File objects. However Commons IO does have a class FilenameUtils that can do this kind of thing, such as the concat() method.
platform independent approach (uses File.separator, ie will works depends on operation system where code is running:
java.nio.file.Paths.get(".", "path", "to", "file.txt")
// relative unix path: ./path/to/file.txt
// relative windows path: .\path\to\filee.txt
java.nio.file.Paths.get("/", "path", "to", "file.txt")
// absolute unix path: /path/to/filee.txt
// windows network drive path: \\path\to\file.txt
java.nio.file.Paths.get("C:", "path", "to", "file.txt")
// absolute windows path: C:\path\to\file.txt
I know its a long time since Jon's original answer, but I had a similar requirement to the OP.
By way of extending Jon's solution I came up with the following, which will take one or more path segments takes as many path segments that you can throw at it.
Usage
Path.combine("/Users/beardtwizzle/");
Path.combine("/", "Users", "beardtwizzle");
Path.combine(new String[] { "/", "Users", "beardtwizzle", "arrayUsage" });
Code here for others with a similar problem
public class Path {
public static String combine(String... paths)
{
File file = new File(paths[0]);
for (int i = 1; i < paths.length ; i++) {
file = new File(file, paths[i]);
}
return file.getPath();
}
}
To enhance JodaStephen's answer, Apache Commons IO has FilenameUtils which does this. Example (on Linux):
assert org.apache.commons.io.FilenameUtils.concat("/home/bob", "work\\stuff.log") == "/home/bob/work/stuff.log"
It's platform independent and will produce whatever separators your system needs.
Late to the party perhaps, but I wanted to share my take on this. I prefer not to pull in entire libraries for something like this. Instead, I'm using a Builder pattern and allow conveniently chained append(more) calls. It even allows mixing File and String, and can easily be extended to support Path as well. Furthermore, it automatically handles the different path separators correctly on both Linux, Macintosh, etc.
public class Files {
public static class PathBuilder {
private File file;
private PathBuilder ( File root ) {
file = root;
}
private PathBuilder ( String root ) {
file = new File(root);
}
public PathBuilder append ( File more ) {
file = new File(file, more.getPath()) );
return this;
}
public PathBuilder append ( String more ) {
file = new File(file, more);
return this;
}
public File buildFile () {
return file;
}
}
public static PathBuilder buildPath ( File root ) {
return new PathBuilder(root);
}
public static PathBuilder buildPath ( String root ) {
return new PathBuilder(root);
}
}
Example of usage:
File root = File.listRoots()[0];
String hello = "hello";
String world = "world";
String filename = "warez.lha";
File file = Files.buildPath(root).append(hello).append(world)
.append(filename).buildFile();
String absolute = file.getAbsolutePath();
The resulting absolute will contain something like:
/hello/world/warez.lha
or maybe even:
A:\hello\world\warez.lha
If you do not need more than strings, you can use com.google.common.io.Files
Files.simplifyPath("some/prefix/with//extra///slashes" + "file//name")
to get
"some/prefix/with/extra/slashes/file/name"
Here's a solution which handles multiple path parts and edge conditions:
public static String combinePaths(String ... paths)
{
if ( paths.length == 0)
{
return "";
}
File combined = new File(paths[0]);
int i = 1;
while ( i < paths.length)
{
combined = new File(combined, paths[i]);
++i;
}
return combined.getPath();
}
This also works in Java 8 :
Path file = Paths.get("Some path");
file = Paths.get(file + "Some other path");
This solution offers an interface for joining path fragments from a String[] array. It uses java.io.File.File(String parent, String child):
public static joinPaths(String[] fragments) {
String emptyPath = "";
return buildPath(emptyPath, fragments);
}
private static buildPath(String path, String[] fragments) {
if (path == null || path.isEmpty()) {
path = "";
}
if (fragments == null || fragments.length == 0) {
return "";
}
int pathCurrentSize = path.split("/").length;
int fragmentsLen = fragments.length;
if (pathCurrentSize <= fragmentsLen) {
String newPath = new File(path, fragments[pathCurrentSize - 1]).toString();
path = buildPath(newPath, fragments);
}
return path;
}
Then you can just do:
String[] fragments = {"dir", "anotherDir/", "/filename.txt"};
String path = joinPaths(fragments);
Returns:
"/dir/anotherDir/filename.txt"
Assuming all given paths are absolute paths. you can follow below snippets to merge these paths.
String baseURL = "\\\\host\\testdir\\";
String absoluteFilePath = "\\\\host\\testdir\\Test.txt";;
String mergedPath = Paths.get(baseURL, absoluteFilePath.replaceAll(Matcher.quoteReplacement(baseURL), "")).toString();
output path is \\host\testdir\Test.txt.
I am scanning all directories starting from "/" to find some particular directories like "MYFOLDER". However, the folder is that I get double instances of the same folder. This occurs because one folder is located in "/mnt/sdcard/MYFOLDER" and the same folder has a symbolic link in "/sdcard/MYFOLDER"..
My Question is, "Is there any way to determine whether the folder is a symbolic link or not?". Please give me some suggestions..
This is essentially how they do in Apache Commons (subject to their license):
public static boolean isSymlink(File file) throws IOException {
File canon;
if (file.getParent() == null) {
canon = file;
} else {
File canonDir = file.getParentFile().getCanonicalFile();
canon = new File(canonDir, file.getName());
}
return !canon.getCanonicalFile().equals(canon.getAbsoluteFile());
}
Edit thanks to #LarsH comment.
The above code only checks whether the children file is a symlink.
In order to answer the OP question, it's even easier:
public static boolean containsSymlink(File file) {
return !file.getCanonicalFile().equals(file.getAbsoluteFile());
}