Develop Reference

Detecting Audio File Bit Rate - Processing / Java

Trying to build a little app for sorting through audio files based on some of their properties. Have managed to grab the Sample Rate and Bit Depth using Minim but can't find anything anywhere for getting the Bit Rate?
Happy to look at taking the program to Javascript if needed but just desperate to find a method for detecting bit rate of a given file.
EDIT: Attempted to try and form an equation based off file size but cannot find a method for detecting MP3 file size either.

You can use jaudiotagger
You will need to download the jar, I managed to get it from maven central
Go to Sketch -> Add File... and select the downloaded jar, it should be added in a folder named code within your sketch folder.
Assuming you have placed an mp3 file in your data folder named audio.mp3 the following code should work, printing out the bit rate in the terminal.
import org.jaudiotagger.audio.mp3.*;
import org.jaudiotagger.audio.AudioFileIO;
void setup() {
File f = new File(dataPath("audio.mp3"));
try {
MP3File mp3 = (MP3File) AudioFileIO.read(f);
MP3AudioHeader audioHeader = mp3.getMP3AudioHeader();
println("" + audioHeader.getBitRate());
}
catch(Exception e) {
e.printStackTrace();
}
}
JAudiotagger supports a variety of file formats and you can use the relevant classes and methods for each one of these.
I suggest you take a look at the javadoc. Be careful of the examples though, the one I used in order to answer your question seems to be faulty, as you can see I had to swap getAudioHeader with getMP3AudioHeader.

Can I make a Netbeans "empty file" part of compiled jar file?

I have 80,000 words for a crossword (among others) puzzle word pattern matcher. (User inputs "ba??" and gets, among other things, "ball, baby, bank, ..." or enters "ba*" and gets the aforementioned as well as "bat, basket, babboon...".)
I stuck the words in a Netbeans "empty file" and named it "dictionary". The file's contents are just (80,000) words, one per line. This code works like a charm to read the dictionary (code that filters is omitted):
static void showMatches(String pattern, String legal, String w) throws IOException
{
Path p = Paths.get("C:\\Users\\Dov\\Documents\\NetBeansProjects\\Masterwords\\src\\masterwords\\dictionary");
String word;
Scanner sc = new Scanner(p).useDelimiter("\r");
while(sc.hasNext()){
word = sc.next().substring(1);
gui.appendOutput(word);
}
sc.reset();
}
Is there a way to make the file (named "dictionary") become part of the compiled jar file so that I only need to "ship" one file to new, (largely helpless) users?
In another matter of curiosity...
Is it possible to make the argument to Paths.get(...) something like "masterwords/src/dictionary" to make the connection for the Scanner object to be able read it? I'm wondering if this might relate to an answer my first question. (If there's a way, I can't stumble onto it. Whatever similar string I use, I get no error, no output, no "build successful"--gotta click Run > Stop build/run.)

I'm not certain, based on your description, that my solution addresses your issue, but let me restate the problem as I understand it: You have a .jar file that relies on a dictionary resource. That resource is subject to change, and you'd like to be able to update it without having to ship out a whole new .jar containing a new dictionary.
If I'm reading you correctly, you want something like:
private File getInstallPath()
{
return new File(MyClass.class.getProtectionDomain().getCodeSource().getLocation().getPath());
}
This will return the install directory of your .jar file, which is where you can put your dictionary resource so that the .jar knows where to find it. Of course, now you have a bit of a training issue, because users can move, delete or misplace your dictionary file.
Part II:
Now that you've clarified your question, let me again restate: You want to be able to read an arbitrary file included in your .jar file. Fine. You're probably trying to open the file as a file, but once the file is in your .jar, you need to treat it as a resource.
Try using:
Class myClass = Class.forName("MyClass");
ClassLoader myLoader = myclass.getClassLoader();
InputStream myStream = myLoader.getResourceAsStream(myFile);
Do you really need me to explain what "myClass," "myLoader," etc. refer to? Hint: "myClass" is whatever your class is that needs to read the file.

After leaving this thread in frustration for a couple of weeks, yesterday I found a similar question at this forum, which led me to Google "java resource files" and visit ((this URL)).
Between the two I figured out how to read a file named 'dictionary' that was created as a Netbeans "empty Java file", which was located in Source Packages ... [default package] (as shown in Netbeans Projects window) and stored as C:\Users\Dov\!Docs\Documents\NetBeansProjects\WordPatternHelp\src\dictionary:
File file = new File("src/dictionary");
...
p = file.toPath();
sc = new Scanner(p).useDelimiter("\r");
Success. Hooray.
But after compiling and executing the .jar file from a DOS command line, 'dictionary' couldn't be found. So the above only works from within Netbeans IDE.
After mostly erroneous attempts caused by the above 'success', I finally got success using #Mars' second suggestion like so:
package masterwords;
public class Masterwords
...
InputStream myStream = Class.forName("masterwords.Masterwords").
getClassLoader().getResourceAsStream("dictionary");
sc = new Scanner(myStream).useDelimiter("\r"); // NULL PTR EXCEPTION HERE
So, for whatever it might be worth, a very belated thanks (and another apology) to #Mars. It was as straightforward as he indicated. Wish I'd tried it 2 weeks ago, but I'd never seen any of the methods and didn't want to take the time to learn how they work back then with other more pressing issues at hand. So I had no idea Mars had actually written the exact code I needed (except for the string arguments). Boy, do I know how the methods work now.

How to open a file without saving it to disk

My Question: How do I open a file (in the system default [external] program for the file) without saving the file to disk?
My Situation: I have files in my resources and I want to display those without saving them to disk first. For example, I have an xml file and I want to open it on the user's machine in the default program for reading xml file without saving it to the disk first.
What I have been doing: So far I have just saved the file to a temporary location, but I have no way of knowing when they no longer need the file so I don't know when/if to delete it. Here's my SSCCE code for that (well, it's mostly sscce, except for the resource... You'll have to create that on your own):
package main;
import java.io.*;
public class SOQuestion {
public static void main(String[] args) throws IOException {
new SOQuestion().showTemplate();
}
/** Opens the temporary file */
private void showTemplate() throws IOException {
String tempDir = System.getProperty("java.io.tmpdir") + "\\BONotifier\\";
File parentFile = new File(tempDir);
if (!parentFile.exists()) {
parentFile.mkdirs();
}
File outputFile = new File(parentFile, "template.xml");
InputStream inputStream = getClass().getResourceAsStream("/resources/template.xml");
int size = 4096;
try (OutputStream out = new FileOutputStream(outputFile)) {
byte[] buffer = new byte[size];
int length;
while ((length = inputStream.read(buffer)) > 0) {
out.write(buffer, 0, length);
}
inputStream.close();
}
java.awt.Desktop.getDesktop().open(outputFile);
}
}

Because of this line:
String tempDir = System.getProperty("java.io.tmpdir") + "\\BONotifier\\";
I deduce that you're working on Windows. You can easily make this code multiplatform, you know.
The answer to your question is: no. The Desktop class needs to know where the file is in order to invoke the correct program with a parameter. Note that there is no method in that class accepting an InputStream, which could be a solution.
Anyway, I don't see where the problem is: you create a temporary file, then open it in an editor or whatever. That's fine. In Linux, when the application is exited (normally) all its temporary files are deleted. In Windows, the user will need to trigger the temporary files deletion. However, provided you don't have security constraints, I can't understand where the problem is. After all, temporary files are the operating system's concern.

Depending on how portable your application needs to be, there might be no "one fits all" solution to your problem. However, you can help yourself a bit:
At least under Linux, you can use a pipe (|) to direct the output of one program to the input of another. A simple example for that (using the gedit text editor) might be:
echo "hello world" | gedit
This will (for gedit) open up a new editor window and show the contents "hello world" in a new, unsaved document.
The problem with the above is, that this might not be a platform-independent solution. It will work for Linux and probably OS X, but I don't have a Windows installation here to test it.
Also, you'd need to find out the default editor by yourself. This older question and it's linked article give some ideas on how this might work.

I don't understand your question very well. I can see only two possibilities to your question.
Open an existing file, and you wish to operate on its stream but do not want to save any modifications.
Create a file, so that you could use file i/o to operate on the file stream, but you don't wish to save the stream to file.
In either case, your main motivation is to exploit file i/o existingly available to your discretion and programming pleasure, am I correct?
I have feeling that the question is not that simple and this my answer is probably not the answer you seek. However, if my understanding of the question does coincide with your question ...
If you wish to use Stream io, instead of using FileOutputStream or FileInputStream which are consequent to your opening a File object, why not use non-File InputStream or OutputStream? Your file i/o utilities will finally boil down to manipulating i/o streams anyway.
http://docs.oracle.com/javase/7/docs/api/java/io/OutputStream.html
http://docs.oracle.com/javase/7/docs/api/java/io/InputStream.html
No need to involve temp files.

how do i write a batch file that i already have made to a specific directory in Java?

I am trying to make this program that will write a file to the users computer but am having trouble really, I want to write a file called desktop.bat that I have to the c:/ directory but it doesn't seem like it's working. this is the code:
package javawriter;
import java.io.BufferedWriter;
import java.io.File;
import java.io.FileWriter;
import java.io.IOException;
public class JavaWriterObject {
public void TryThis(){
try{
File file = new File("C:\\Desktop.bat");
if (file.exists()){
System.out.println("The file exists \ndirectory is found");
}else{
System.out.println("file is not found yet ".concat("file will be created"));
file.createNewFile();
}
FileWriter out = new FileWriter(file);
BufferedWriter writer = new BufferedWriter(out);
writer.write();
writer.close();
}catch(IOException e){
e.printStackTrace();
}
}
}
Do I have to write a String for the writer.write(); or can I do something where I can write the file I want instead?

Java 7's Files.copy method helps you copy a file from one location to another.
If you're using an older version of Java, you will need to either read the original file in yourself and copy its contents into your new file, or you'll have to use a third-party library. See the answers to this question for several good solutions, including using Apache Commons IO FileUtils or just the standard Java API.
Once you've decided how you want to copy the file, you might run into another problem. By default, Windows 7 will prevent you from writing to certain directories such as C:\. You can try writing to a different directory, such as in the temp directory or any location within the user's home directory. If you must write to C:\, the easiest solution (aside from creating the file ahead of time in Windows and overwriting it in your program, which probably defeats the purpose) is to disable UAC and make sure your user account has write permission on that directory--but this, of course, has security implications.

You have to get the content you want to write from somewhere, either a String literal in your application, or, the preferred method would be from a resource (bundled along with your application) via. YourClass.class.getResourceAsStream(name).
Copying an input stream to a file is a bit of effort, Guava can copy, but, if you're using Guava, you may as well copy directly from the resource to the file.

How to convert a Hadoop Path object into a Java File object

Is there a way to change a valid and existing Hadoop Path object into a useful Java File object. Is there a nice way of doing this or do I need to bludgeon to code into submission? The more obvious approaches don't work, and it seems like it would be a common bit of code
void func(Path p) {
if (p.isAbsolute()) {
File f = new File(p.toURI());
}
}
This doesn't work because Path::toURI() returns the "hdfs" identifier and Java's File(URI uri) constructor only recognizes the "file" identifier.
Is there a way to get Path and File to work together?
**
Ok, how about a specific limited example.
Path[] paths = DistributedCache.getLocalCacheFiles(job);
DistributedCache is supposed to provide a localized copy of a file, but it returns a Path. I assume that DistributedCache make a local copy of the file, where they are on the same disk. Given this limited example, where hdfs is hopefully not in the equation, is there a way for me to reliably convert a Path into a File?
**

I recently had this same question, and there really is a way to get a file from a path, but it requires downloading the file temporarily. Obviously, this won't be suitable for many tasks, but if time and space aren't essential for you, and you just need something to work using files from Hadoop, do something like the following:
import java.io.File;
import java.io.IOException;
import org.apache.hadoop.conf.Configuration;
import org.apache.hadoop.fs.FileSystem;
import org.apache.hadoop.fs.Path;
public final class PathToFileConverter {
public static File makeFileFromPath(Path some_path, Configuration conf) throws IOException {
FileSystem fs = FileSystem.get(some_path.toUri(), conf);
File temp_data_file = File.createTempFile(some_path.getName(), "");
temp_data_file.deleteOnExit();
fs.copyToLocalFile(some_path, new Path(temp_data_file.getAbsolutePath()));
return temp_data_file;
}
}

If you get a LocalFileSystem
final LocalFileSystem localFileSystem = FileSystem.getLocal(configuration);
You can pass your hadoop Path object to localFileSystem.pathToFile
final File localFile = localFileSystem.pathToFile(<your hadoop Path>);

Not that I'm aware of.
To my understanding, a Path in Hadoop represents an identifier for a node in their distributed filesystem. This is a different abstraction from a java.io.File, which represents a node on the local filesystem. It's unlikely that a Path could even have a File representation that would behave equivalently, because the underlying models are fundamentally different.
Hence the lack of translation. I presume by your assertion that File objects are "[more] useful", you want an object of this class in order to use existing library methods? For the reasons above, this isn't going to work very well. If it's your own library, you could rewrite it to work cleanly with Hadoop Paths and then convert any Files into Path objects (this direction works as Paths are a strict superset of Files). If it's a third party library then you're out of luck; the authors of that method didn't take into account the effects of a distributed filesystem and only wrote that method to work on plain old local files.