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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.

How to fix NotOfficeXmlFileException in java Apache POI?

I'm trying to create a new excel file with just "hello" in it.
Here's my code:
import java.io.File;
import java.io.FileInputStream;
import java.io.FileNotFoundException;
import java.io.FileOutputStream;
import java.io.IOException;
import java.util.logging.Level;
import java.util.logging.Logger;
import javax.swing.JFileChooser;
import org.apache.poi.hssf.usermodel.HSSFWorkbook;
import org.apache.poi.ss.usermodel.WorkbookFactory;
import org.apache.poi.xssf.usermodel.XSSFSheet;
import org.apache.poi.xssf.usermodel.XSSFWorkbook;
/**
*
* #author kamal
*/
public class JavaApplication4 {
private static String dir = "";
/**
* #param args the command line arguments
*/
public static void main(String[] args) {
try {
// TODO code application logic here
JFileChooser jc = new JFileChooser();
jc.setFileSelectionMode(JFileChooser.DIRECTORIES_ONLY);
int output = jc.showOpenDialog(null);
if(output == JFileChooser.APPROVE_OPTION){
File f = jc.getSelectedFile();
String directory = f.getAbsolutePath();
setDir(directory);
}
FileOutputStream out = new FileOutputStream(new File(getDir()+"\\Book2.xlsx"));
FileInputStream in = new FileInputStream(new File(getDir()+"\\Book2.xlsx"));
org.apache.poi.ss.usermodel.Workbook workbook = new XSSFWorkbook(in);
org.apache.poi.ss.usermodel.Sheet sheet = workbook.getSheetAt(0);
sheet.createRow(0).createCell(0).setCellValue("hello");
workbook.write(out);
workbook.close();
} catch (FileNotFoundException ex) {
ex.printStackTrace();
} catch (IOException ex) {
Logger.getLogger(JavaApplication4.class.getName()).log(Level.SEVERE, null, ex);
}
}
/**
* #return the dir
*/
public static String getDir() {
return dir;
}
/**
* #param dir the dir to set
*/
public static void setDir(String directory) {
dir = directory;
}
}
..And when I run it I get the following error:
Exception in thread "main" org.apache.poi.openxml4j.exceptions.NotOfficeXmlFileException: No valid entries or contents found, this is not a valid OOXML (Office Open XML) file
at org.apache.poi.openxml4j.opc.ZipPackage.getPartsImpl(ZipPackage.java:286)
at org.apache.poi.openxml4j.opc.OPCPackage.getParts(OPCPackage.java:758)
at org.apache.poi.openxml4j.opc.OPCPackage.open(OPCPackage.java:327)
at org.apache.poi.util.PackageHelper.open(PackageHelper.java:37)
at org.apache.poi.xssf.usermodel.XSSFWorkbook.<init>(XSSFWorkbook.java:291)
at javaapplication4.JavaApplication4.main(JavaApplication4.java:46)
C:\Users\kamal\AppData\Local\NetBeans\Cache\8.2\executor-snippets\run.xml:53: Java returned: 1
BUILD FAILED (total time: 7 seconds)
I looked up this code in youtube and it's same but i'm not sure why am i getting the error? Can you help me with this?

I think the most likely explanations are that either the file is corrupt, or it is an older format spreadsheet file that XSSFWorkbook does not understand.
It is unlikely anyone can give you a definite diagnosis without looking at the file itself.

Okay. Today I encountered the same problem . The server was linux and the excel file is copied from windows to linux through winscp. Winscp has options like transferring the file in binary mode , text mode etc. When we copy the excel file through text mode , I got the same error you mentioned. The error got resolved when I copy the excel file using binary mode. To summarize , this issue came because we copied excel file from windows to linux. Just make sure you are copying in binary mode if using winscp. Make sure the file is copied correctly.

I was facing the same problem. I did create the Excel file by doing right click inside the folder and then ->New->Microsoft Excel Worksheet.
As a trial I removed this file and then created the new Excel through Start Menu->Microsoft Office->Excel
It worked for me, Hopefully same will work for you too.

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.

Why does my test filesystem only work in memory?

I am working on a concept of a filesystem for a program. I am writing in Java (using JDK 7 u17).
To get started I built off of some tutorial that were showing my how to create a zip based filesystem using the FileSystemProvider class.
When I execute the code I have it do similar task to the examples which is copy a text file from the my desktop and place it in the zip file. The problem is once it copies the file it does not write it into the zip file, it seems to leave the file in memory which is destroyed when the program is terminated.
The problem is I cannot understand why, as far as I can tell everything looks to be in order but something is clearly not!
Oh yeah the same thing goes for directories too. If I tell the filesystem to make a new directory it just creates it in memory and there is nothing in the zip file.
Anyhow here is my working code;
import java.io.IOException;
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.util.HashMap;
import java.util.Map;
public class Start {
public static void main(String[] args) {
Map <String, String> env = new HashMap<>();
env.put("create", "true");
env.put("encoding", "UTF-8");
FileSystem fs = null;
try {
fs = FileSystems.newFileSystem(URI.create("jar:file:/Users/Ian/Desktop/test.zip"), env);
} catch (IOException e) {
e.printStackTrace();
}
Path externalTxtFile = Paths.get("/Users/Ian/Desktop/example.txt");
Path pathInZipFile = fs.getPath("/example.txt");
try {
Files.createDirectory(fs.getPath("/SomeDirectory"));
} catch (IOException e) {
e.printStackTrace();
}
if (Files.exists(fs.getPath("/SomeDirectory"))) {
System.out.println("Yes the directory exists in memory.");
} else {
System.out.println("What directory?");
}
// Why is the file only being copied into memory and not written out the jar/zip archive?
try {
Files.copy(externalTxtFile, pathInZipFile);
} catch (IOException e) {
e.printStackTrace();
}
// The file clearly exists just before the program ends, what is going on?
if (Files.exists(fs.getPath("/example.txt"))) {
System.out.println("Yes the file has been copied into memory.");
} else {
System.out.println("What file?");
}
}
}

I just want to add something.
Perhaps the example that you found was incomplete (I can not check since you do not references it) but in all examples I found the FileSystem instance is closed properly.
The FileSystem abstract class implements Closeable, so the close() method is called (automatically) leaving the try in the following code:
try (final FileSystem fs = FileSystems.newFileSystem(theUri, env)) {
/* ... do everything you want here ; do not need to call fs.close() ... */
}
http://docs.oracle.com/javase/tutorial/essential/exceptions/tryResourceClose.html