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java zip filesystem reading zip entries

SO, from what I've gathered, one is supposed to be able to create a filesystem from a zip from java 7 and beyond. I'm trying this, the ultimate goal is to use the File object and access these files, just as if I accessed an unzipped file.
import java.io.File;
import java.io.IOException;
import java.net.URI;
import java.net.URISyntaxException;
import java.nio.file.*;
import java.util.*;
import java.util.zip.ZipEntry;
import java.util.zip.ZipFile;
public class MainZipTest {
public static void main(String[] args) throws IOException, URISyntaxException {
Map<String, String> env = new HashMap<>();
env.put("read", "true");
File file = new File("C:/pathtoazip/data.zip");
URI uri = file.toURI();
String path = "jar:" + uri;
FileSystem fs = FileSystems.newFileSystem(URI.create(path), env);
for (Path p : fs.getRootDirectories()) {
System.out.println("root" + p);
//says "/"
System.out.println(new File(p.toString()).exists());
for (File f : new File(p.toString()).listFiles())
System.out.println(f.getAbsolutePath());
//lists the contents of my c drive!
}
System.out.println(new File("somefile.txt").exists());
System.out.println(fs.getPath("somefile.txt").toFile().exists());
System.out.println(new File("/somefile.txt").exists());
System.out.println(fs.getPath("/somefile.txt").toFile().exists());
}
}
it all prints "false". What am I doing wrong here? Or am I wrong in my assumption that I can access these files through the File object? If so, how does one access them?

Path was introduced as generalization of File (disk file). A Path can be inside a zip file, an URL, and more.
You can use Files with Path for similar File functionality.
for (Path p : fs.getRootDirectories()) {
System.out.println("root: " + p);
System.out.println(Files.exists(p));
Files.list(p).forEach(f -> System.out.println(f.toAbsolutePath()));
}
Note that a Path, like from a zip will maintain its actual file system view (fs, the zip).
So avoid File.

Create a new file in .zip folder using Scala

I have a requirement where I have to create 1 xml file inside my .zip folder.
Can you please let me know how can I achieve this using java/scala?
Also let me know can I directly update .zip folder Array[Byte] and add extra Array[Byte] for my new xml file ?

If you are using Java 7 and later, then you should use Zip File System
The following code sample shows how to create a zip file system and
copy a file to the new zip file system.
import java.util.*;
import java.net.URI;
import java.nio.file.Path;
import java.nio.file.*;
public class ZipFSPUser {
public static void main(String [] args) throws Throwable {
Map<String, String> env = new HashMap<>();
env.put("create", "true");
// locate file system by using the syntax
// defined in java.net.JarURLConnection
URI uri = URI.create("jar:file:/codeSamples/zipfs/zipfstest.zip");
try (FileSystem zipfs = FileSystems.newFileSystem(uri, env)) {
Path externalTxtFile = Paths.get("/codeSamples/zipfs/SomeTextFile.txt");
Path pathInZipfile = zipfs.getPath("/SomeTextFile.txt");
// copy a file into the zip file
Files.copy( externalTxtFile,pathInZipfile,
StandardCopyOption.REPLACE_EXISTING );
}
}
}
This post could help you: Appending files to a zip file with Java

Can a Jar File be updated programmatically without rewriting the whole file?

It is possible to update individual files in a JAR file using the jar command as follows:
jar uf TicTacToe.jar images/new.gif
Is there a way to do this programmatically?
I have to rewrite the entire jar file if I use JarOutputStream, so I was wondering if there was a similar "random access" way to do this. Given that it can be done using the jar tool, I had expected there to be a similar way to do it programmatically.

It is possible to update just parts of the JAR file using Zip File System Provider available in Java 7:
import java.net.URI;
import java.nio.file.FileSystem;
import java.nio.file.FileSystems;
import java.nio.file.Files;
import java.nio.file.Path;
import java.nio.file.Paths;
import java.nio.file.StandardCopyOption;
import java.util.HashMap;
import java.util.Map;
public class ZipFSPUser {
public static void main(String [] args) throws Throwable {
Map<String, String> env = new HashMap<>();
env.put("create", "true");
// locate file system by using the syntax
// defined in java.net.JarURLConnection
URI uri = URI.create("jar:file:/codeSamples/zipfs/zipfstest.zip");
try (FileSystem zipfs = FileSystems.newFileSystem(uri, env)) {
Path externalTxtFile = Paths.get("/codeSamples/zipfs/SomeTextFile.txt");
Path pathInZipfile = zipfs.getPath("/SomeTextFile.txt");
// copy a file into the zip file
Files.copy( externalTxtFile,pathInZipfile,
StandardCopyOption.REPLACE_EXISTING );
}
}
}

Yes, if you use this opensource library you can modify it in this way as well.
https://truevfs.java.net
public static void main(String args[]) throws IOException{
File entry = new TFile("c:/tru6413/server/lib/nxps.jar/dir/second.txt");
Writer writer = new TFileWriter(entry);
try {
writer.write(" this is writing into a file inside an archive");
} finally {
writer.close();
}
}

Moving large files in java

I have to move files from one directory to other directory.
Am using property file. So the source and destination path is stored in property file.
Am haivng property reader class also.
In my source directory am having lots of files. One file should move to other directory if its complete the operation.
File size is more than 500MB.
import java.io.File;
import java.nio.file.Files;
import java.nio.file.StandardCopyOption;
import static java.nio.file.StandardCopyOption.*;
public class Main1
{
public static String primarydir="";
public static String secondarydir="";
public static void main(String[] argv)
throws Exception
{
primarydir=PropertyReader.getProperty("primarydir");
System.out.println(primarydir);
secondarydir=PropertyReader.getProperty("secondarydir");
File dir = new File(primarydir);
secondarydir=PropertyReader.getProperty("secondarydir");
String[] children = dir.list();
if (children == null)
{
System.out.println("does not exist or is not a directory");
}
else
{
for (int i = 0; i < children.length; i++)
{
String filename = children[i];
System.out.println(filename);
try
{
File oldFile = new File(primarydir,children[i]);
System.out.println( "Before Moving"+oldFile.getName());
if (oldFile.renameTo(new File(secondarydir+oldFile.getName())))
{
System.out.println("The file was moved successfully to the new folder");
}
else
{
System.out.println("The File was not moved.");
}
}
catch (Exception e)
{
e.printStackTrace();
}
}
System.out.println("ok");
}
}
}
My code is not moving the file into the correct path.
This is my property file
primarydir=C:/Desktop/A
secondarydir=D:/B
enter code here
Files should be in B drive. How to do? Any one can help me..!!

Change this:
oldFile.renameTo(new File(secondarydir+oldFile.getName()))
To this:
oldFile.renameTo(new File(secondarydir, oldFile.getName()))
It's best not to use string concatenation to join path segments, as the proper way to do it may be platform-dependent.
Edit: If you can use JDK 1.7 APIs, you can use Files.move() instead of File.renameTo()

Code - a java method:
/**
* copy by transfer, use this for cross partition copy,
* #param sFile source file,
* #param tFile target file,
* #throws IOException
*/
public static void copyByTransfer(File sFile, File tFile) throws IOException {
FileInputStream fInput = new FileInputStream(sFile);
FileOutputStream fOutput = new FileOutputStream(tFile);
FileChannel fReadChannel = fInput.getChannel();
FileChannel fWriteChannel = fOutput.getChannel();
fReadChannel.transferTo(0, fReadChannel.size(), fWriteChannel);
fReadChannel.close();
fWriteChannel.close();
fInput.close();
fOutput.close();
}
The method use nio, it make use os underling operation to improve performance.
Here is the import code:
import java.io.File;
import java.io.FileInputStream;
import java.io.FileOutputStream;
import java.io.IOException;
import java.nio.ByteBuffer;
import java.nio.channels.FileChannel;
If you are in eclipse, just use ctrl + shift + o.

Creating a directory wherever jar file is located through java

I have already surveyed SO for an answer, and could not find an appropriate one.
When I launch my program from a jar I need to create a folder in the directory where the jar file is located. It should not matter where the user saves the jar file.
Here is the newest code I was playing with: A System.out.println will print out the correct directory but the folder will not be created. In contrast,everything is being saved to my System32 folder as of now.
public static String getProgramPath() throws IOException{
String currentdir = System.getProperty("user.dir");
currentdir = currentdir.replace( "\\", "/" );
return currentdir;
}
File dir = new File(getProgramPath() + "Comics/");//The name of the directory to create
dir.mkdir();//Creates the directory

To get a Jar's path can be a little trickier than simply getting the user.dir directory. I can't remember the details why, but user.dir does not return this path reliably in all situations. If you absolutely must get the jar's path, then you need to do a little black magic and first get the class's protectionDomain. Something like:
import java.io.File;
import java.io.IOException;
import java.io.UnsupportedEncodingException;
import java.net.URL;
import java.net.URLDecoder;
import javax.swing.JOptionPane;
public class MkDirForMe {
public static void main(String[] args) {
try {
String path = getProgramPath2();
String fileSeparator = System.getProperty("file.separator");
String newDir = path + fileSeparator + "newDir2" + fileSeparator;
JOptionPane.showMessageDialog(null, newDir);
File file = new File(newDir);
file.mkdir();
} catch (IOException e) {
e.printStackTrace();
}
}
public static String getProgramPath2() throws UnsupportedEncodingException {
URL url = MkDirForMe.class.getProtectionDomain().getCodeSource().getLocation();
String jarPath = URLDecoder.decode(url.getFile(), "UTF-8");
String parentPath = new File(jarPath).getParentFile().getPath();
return parentPath;
}
}
Even this isn't guaranteed to work, and you'll have to resign yourself to the fact that there are just some times (for instance for security reasons) when you won't be able to get a Jar's path.

With some changes (such as adding a "/" before Comics), I managed to create the directory where you expected it to. Here is the full code I used.
import java.io.*;
public class TestClass {
public static String getProgramPath() throws IOException{
String currentdir = System.getProperty("user.dir");
currentdir = currentdir.replace( "\\", "/" );
return currentdir;
}
public static void main(String[] argv) {
try {
String d = getProgramPath() + "/Comics/";
System.out.println("Making directory at " + d);
File dir = new File(d);//The name of the directory to create
dir.mkdir();//Creates the directory
}
catch (Exception e) { System.out.println("Exception occured" + e);}
}
}
In the future, please don't hard code things like "/" and such. Use built-in libraries which will ask the OS what is right in this case. This ensures the functionality doesn't break (as easily) cross platform.
Of course, catch the exception properly etc. This is just quick and dirty attempt to mold your code into something that works.