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Java - Differentiate ZIP file from CSV file

I'm using a webservice that is always sending me a plain/text file. However, that file can either be a zip or a csv but I'm not being informed of its type beforehand.
Is there a way to know the file type by looking through its content programmatically wise of course. As one is in byte code and the other one an actually readeable text.
I've already thought of looking for lots of commas in the file content but that seems inaccurate.

You can use java.util.zip.ZipFile, if the constructor throws a ZipException, it's not a zip file...
try(ZipFile zip = new ZipFile(filename)) {
// It's a zip file
}
catch(ZipException e) {
// Not a valid zip
}

You could make use of the ZIP file structure.
As per the file header, each file should start with the bytes: 0x04 0x03 0x4b 0x50.
You could also use a MIME detection library such as Apache Tika import org.apache.tika.Tika;
import org.apache.tika.mime.MediaType;
import java.io.IOException;
import java.io.InputStream;
import java.nio.file.Files;
import java.nio.file.Path;
import java.nio.file.Paths;
public class Detect {
/**
* Resolves the MediaType using Tika and prints it to the standard output.
* #param file the path of the file to probe.
* #throws IOException whenever an I/O exception occurs.
*/
private void detect(Path file) throws IOException {
Tika tika = new Tika();
try(InputStream is = Files.newInputStream(file)){
MediaType mediaType = MediaType.parse(tika.detect(is));
System.out.println(mediaType);
}
}
public static void main(String[] args) throws IOException {
Detect d = new Detect();
d.detect(Paths.get("zip_file"));
d.detect(Paths.get("csv_file"));
}
}

Excel 2007 file corrupted after being created in Apache POI

I use NetBeans 8. I got problem after compiling this simple code:
package file;
import java.io.FileNotFoundException;
import java.io.FileOutputStream;
import java.io.IOException;
import javax.swing.JOptionPane;
import org.apache.poi.xssf.usermodel.XSSFWorkbook;
import org.apache.poi.ss.usermodel.Workbook;
public class File {
public static void main(String[] args) throws FileNotFoundException, IOException
{ Workbook wb = new XSSFWorkbook();
String name = "charlie.xlsx";
FileOutputStream fileOut = new FileOutputStream(name);
wb.write(fileOut);
fileOut.close();
I'm total rookie in Java so basically I rewrote that code from Apache POI documentation, trying to understand how it works. Well - it works fine until I try to open the output file in MS Excel - because then I get a message that file cannot be open cause it's corrupt.
What went wrong?

You need to create a Sheet. Add this to your code and it will work.
wb.createSheet("Test1");

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.

Error With creating an excel file from java POI

Hi I would like to create an excel file from a java code, I put this code on eclipse but nothing happen
import org.apache.poi.ss.usermodel.Workbook;
import org.apache.poi.hssf.usermodel.HSSFWorkbook;
import java.io.FileOutputStream;
public class TestPOI1 {
public static void main(String[] args) {
//create the new workbook
Workbook workbook = new HSSFWorkbook();
try {
//create the output stream to save the document on the hard drive
FileOutputStream output = new FileOutputStream("Test1.xls");
//write the file onto the hard drive
workbook.write(output);
//finish it up by closing the document
output.close();
} catch(Exception e) {
e.printStackTrace();
}
}
}
in the console, this message is written
Usage: BiffDrawingToXml [options] inputWorkbook Options:
-exclude-workbook exclude workbook-level records
-sheet-indexes output sheets with specified indexes
-sheet-namek output sheets with specified name
and I can't found my excel file in the hard drive or in the file project. thanks for help.

Actually this error you will get while the Jar is still building. Just wait for few seconds.
Right Click and then Run : It will work..!

Java UnZipping a zipped .app back to .app (OS X)

In my auto updater application i am downloading a zipped file that contains the new MyApp.app application file. So i am downloading MyApp.zip.. Then i use this following class to try and unzip it:
package update;
import java.io.BufferedOutputStream;
import java.io.File;
import java.io.FileInputStream;
import java.io.FileOutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.OutputStream;
import java.util.Enumeration;
import java.util.List;
import java.util.zip.ZipEntry;
import java.util.zip.ZipFile;
import java.util.zip.ZipInputStream;
public class UnZip {
public static final void copyInputStream(InputStream in, OutputStream out)
throws IOException
{
byte[] buffer = new byte[1024];
int len;
while((len = in.read(buffer)) >= 0)
out.write(buffer, 0, len);
in.close();
out.close();
}
public static final void unZipIt(String F1, String F2) {
Enumeration entries;
ZipFile zipFile;
try {
zipFile = new ZipFile(F1);
entries = zipFile.entries();
while(entries.hasMoreElements()) {
ZipEntry entry = (ZipEntry)entries.nextElement();
if(entry.isDirectory()) {
// Assume directories are stored parents first then children.
System.err.println("Extracting directory: " + entry.getName());
// This is not robust, just for demonstration purposes.
(new File(entry.getName())).mkdirs();
continue;
}
System.err.println("Extracting file: " + entry.getName());
copyInputStream(zipFile.getInputStream(entry),
new BufferedOutputStream(new FileOutputStream(entry.getName())));
}
zipFile.close();
} catch (IOException ioe) {
System.err.println("Unhandled exception:");
ioe.printStackTrace();
return;
}
}
}
However after the unzip the application wont launch.. any ideas?

Your executable file is most likely not flagged as executable. The trick is that .app "files" are in fact directories, so making them executable serves no practical purpose, you need to find the actual binary.
To do that, you need to open ./myApp.app/Contents/Info.plit and look for the CFBundleExecutable key: the associated string is the path of the executable file, relative to ./myApp.app/Contents/MacOS, I believe.
Once you've found that file, chmod +x it, and check whether your application still fails to start.
If it doesn't, problem solved.
If it does, try and open your application from the terminal through the open ./myApp.app command. If anything odd is printed, update your question with it and let us know what that was.
If all else fails, look into the Console application for interesting log entries - you can search for your application's name, see if anything comes up.