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Create a binary file in java

These are the related questions that might cause my question to be closed, even though I specify another question:
Java: How to write binary files? -> Doesn't really cover the point that I am talking about
create a binary file -> Absolutely doesn't cover the point
Editing a binary file in java -> They are talking about offsets and stuff, when I just need to write the data and stop
Binary files in java -> Vague.
And now to the point. I've got a file with a specific extension, to be more exact it's .nbs. I want to create a file and then write the specific data to it.
That might have sounded vague so let me show you the code I have started with.
try {
File bpdn = new File(getDataFolder() + "song.nbs");
if (!bpdn.exists()) {
bpdn.createNewFile();
}
FileOutputStream fos = new FileOutputStream(bpdn);
} catch (IOException e) {
e.printStackTrace();
}
I'll provide you even more details. So I've got a song.nbs file that I have created in the past, for myself. And now, whenever a person runs my application, I want it so there's a new song.nbs file with the exact contents of a file that I have on my PC right now. Therefore, I need to somehow get the bytes of my existing song.nbs and then copy and paste them in my Java application... or is it the way? I neither know how to get the bytes of my own file right now, nor do I know how to write them.

You need to create a resources folder. More info here.
Assuming your project structure is
ProjectName/src/main/java/Main.java
you can create a resources folder inside main/:
ProjectName/src/main/java/Main.java
ProjectName/src/main/resources/
Move your song.nbs you want to read inside resources/:
ProjectName/src/main/java/Main.java
ProjectName/src/main/resources/song.nbs
Now, get the InputStream of song.nbs stored there:
final ClassLoader classloader = Thread.currentThread().getContextClassLoader();
final InputStream is = classloader.getResourceAsStream("song.nbs");
Then, write this input stream to your new file:
final File bpdn = new File(getDataFolder() + "song.nbs");
if (!bpdn.exists()) bpdn.createNewFile();
final FileOutputStream os = new FileOutputStream(bpdn);
byte[] buffer = new byte[1024];
int read;
while ((read = is.read(buffer)) != -1) {
os.write(buffer, 0, read);
}

I think I came up with a solution, but I am not sure if this is works. I'd appreciate if you would take a look.
try {
File bpdn = new File(getDataFolder() + "badpiggiesdrip.nbs");
if (!bpdn.exists()) {
bpdn.createNewFile();
}
FileOutputStream fos = new FileOutputStream(bpdn);
fos.write(new byte[] {
Byte.parseByte(Arrays.toString(Base64.getDecoder().decode(Common.myString)))
});
} catch (IOException e) {
e.printStackTrace();
}
Common.myString is just a string, that contains data of this type:
(byte) 0x21, (byte) 0x5a, .....
and it's encoded in Base64.

Cant extract single files from Tar, 0kb size, Java

I have been created a application which shall extract single files from tar-archive. The application reads the *.tar properly, but when i try to extract files, the application just create new files with correct filename... The files is empty (0kb). So... I probably just create new files instead of extract...
I'm a totally beginner at this point...
for(TarArchiveEntry tae : tarEntries){
System.out.println(tarEntries.size());
try {
fOutput = new FileOutputStream(new File(tae.getFile(), tae.getName()));
byte[] buf = new byte[(int) tae.getSize()];
int len;
while ((len = tarFile.read(buf)) > 0) {
fOutput.write(buf, 0, len);
}
fOutput.close();
} catch (IOException e) {
e.printStackTrace();
}
}

Assuming tarFile is a TarArchiveInputStream you can only read an entry's content right after calling tarFile.getNextTarEntry().
The stream is processed sequentially, so when you invoke getNextTarEntry you skip over the content of the current entry right to the next entry. It looks as if you had read the whole archive in order to fill tarEntries in which case you've already read past the last entry and the stream is exhausted.

java Extracting Zip file

I'm looking for a way to extract Zip file. So far I have tried java.util.zip and org.apache.commons.compress, but both gave a corrupted output.
Basically, the input is a ZIP file contain one single .doc file.
java.util.zip: Output corrupted.
org.apache.commons.compress: Output blank file, but with 2 mb size.
So far only the commercial software like Winrar work perfectly. Is there a java library that make use of this?
This is my method using java.util library:
public void extractZipNative(File fileZip)
{
ZipInputStream zis;
StringBuilder sb;
try {
zis = new ZipInputStream(new FileInputStream(fileZip));
ZipEntry ze = zis.getNextEntry();
byte[] buffer = new byte[(int) ze.getSize()];
FileOutputStream fos = new FileOutputStream(this.tempFolderPath+ze.getName());
int len;
while ((len=zis.read(buffer))>0)
{
fos.write(buffer);
}
fos.flush();
fos.close();
} catch (FileNotFoundException e) {
// TODO Auto-generated catch block
e.printStackTrace();
} catch (IOException e) {
// TODO Auto-generated catch block
e.printStackTrace();
} finally
{
if (zis!=null)
{
try { zis.close();
} catch (IOException e) {
e.printStackTrace();
}
}
}
}
Many thanks,
Mike

I think your input may be compressed by some "incompatible" zip program like 7zip.
Try investigating first if it can be unpacked with a classical WinZip or such.
Javas zip handling is very well able to deal with zipped archives that come from a "compatible" zip compressor.

It is an error in my code. I need to specify the offset and len of bytes write.

it works for me
ZipFile Vanilla = new ZipFile(new File("Vanilla.zip")); //zipfile defined and needs to be in directory
Enumeration<? extends ZipEntry> entries = Vanilla.entries();// all (files)entries of zip file
while(entries.hasMoreElements()){//runs while there is files in zip
ZipEntry entry = entries.nextElement();//gets name of file in zip
File folderw =new File("tkwgter5834");//creates new directory
InputStream stream = Vanilla.getInputStream(entry);//gets input
FileInputStream inpure= new FileInputStream("Vanilla.zip");//file input stream for zip file to read bytes of file
FileOutputStream outter = new FileOutputStream(new File(folderw +"//"+ entry.toString())); //fileoutput stream creates file inside defined directory(folderw variable) by file's name
outter.write(inpure.readAllBytes());// write into files which were created
outter.close();//closes fileoutput stream
}

Have you tried jUnrar? Perhaps it might work:
https://github.com/edmund-wagner/junrar
If that doesn't work either, I guess your archive is corrupted in some way.

If you know the environment that you're going to be running this code in, I think you're much better off just making a call to the system to unzip it for you. It will be way faster than anything that you implement in java.
I wrote the code to extract a zip file with nested directories and it ran slowly and took a lot of CPU. I wound up replacing it with this:
Runtime.getRuntime().exec(String.format("unzip %s -d %s", archive.getAbsolutePath(), basePath));
That works a lot better.

Java: Error when save file in Resource after Deployment

My program has a function that read/write file from resource. This function I have tested smoothly.
For example, I write something to file, restart and loading again, I can read that data again.
But after I export to jar file, I faced problems when write file. Here is my code to write file:
URL resourceUrl = getClass().getResource("/resource/data.sav");
File file = new File(resourceUrl.toURI());
FileOutputStream output = new FileOutputStream(file);
ObjectOutputStream writer = new ObjectOutputStream( output);
When this code run, I has notice error in Command Prompt:
So, My data cannot saved. (I know it because after I restarted app, nothing changed !!!)
Please help me solve this problem.
Thanks :)

You simply can't write files into a jar file this way. The URI you get from getResource() isn't a file:/// URI, and it can't be passed to java.io.File's constructor. The only way to write a zip file is by using the classes in java.util.zip that are designed for this purpose, and those classes are designed to let you write entire jar files, not stream data to a single file inside of one. In a real installation, the user may not even have permission to write to the jar file, anyway.
You're going to need to save your data into a real file on the file system, or possibly, if it's small enough, by using the preferences API.

You need to read/write file as an input stream to read from jar file.
public static String getValue(String key)
{
String _value = null;
try
{
InputStream loadedFile = ConfigReader.class.getClassLoader().getResourceAsStream(configFileName);
if(loadedFile == null) throw new Exception("Error: Could not load the file as a stream!");
props.load(loadedFile);
}
catch(Exception ex){
try {
System.out.println(ex.getMessage());
props.load(new FileInputStream(configFileName));
} catch (FileNotFoundException e) {
ExceptionWriter.LogException(e);
} catch (IOException e) {
ExceptionWriter.LogException(e);
}
}
_value = props.getProperty(key);
if(_value == null || _value.equals("")) System.out.println("Null value supplied for key: "+key);
return _value;
}

ByteArray To XML file in Android

I am sending a file in byte[] format from web service to android device.
If that file is an XML file i.e. byte[] array then how could i convert it to original XML file.
If that file is an image i.e. byte[] array then how could i convert it to original Image.
I am using android sdk 2.2 on samsung galaxy tab.

Your Webservice should send you some identifier about the file type. whether the byte array is for image or is for general file. then only you can know about which type of file it is. after knowing file type you can convert the byte array into your desired file type. Also you can write to file. If you want to print that xml in logcat you can use
String xmlData = new String(byte[] data); System.out.println(xmlData)
create a file (whether xml or image or anything) if you know the file format
String extension = ".xml"//or ".jpg" or anything
String filename = myfile+extension;
byte[] data = //the byte array which i got from server
File f = new File(givepathOfFile+filename );
try {
f.createNewFile();
// write the bytes in file
FileOutputStream fo = new FileOutputStream(f);
fo.write(data );
fo.close();
} catch (IOException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
Thanks
Deepak

A byte array containing XML already is an XML document. If you want to make it an XML file, then just write the bytes to a file.
The same applies for an image file.
Are you really just asking how to write a byte array as a file?
Here's how to write bytes to a file in Java.
byte[] bytes = ...
FileOutputStream fos = new FileOutputStream("someFile.xml");
try {
fos.write(bytes);
} finally {
fos.close();
}
Note that you need to close an opened stream in a finally block or else you risk leaking a file descriptors. If you leak too many file descriptors, later attempts to open files or sockets could start failing.