Develop Reference

Setting permissions for created directory to copy files into it

During the execution of my program it creates a directory which contains two sub-directories/two folders. Into one of these folders I need to copy a Jar-File. My programm resembles an installation routine. The copying of the Jar file is not the problem here, but the permissions of the created directories.
I tried to set the permissions of the directories (before actually creating them with the mkdirs() method) with File.setWritable(true, false) and also with the .setExecutable and .setReadable methods, but the access to the sub-directories is still denied.
Here's an excerpt of my code for the creation of one of the two sub-directories:
folderfile = new File("my/path/to/directory");
folderfile.setExecutable(true, false);
folderfile.setReadable(true, false);
folderfile.setWritable(true, false);
result = folderfile.mkdirs();
if (result) {
System.out.println("Folder created.");
}else {
JOptionPane.showMessageDialog(chooser, "Error");
}
File source = new File("src/config/TheJar.jar");
File destination = folderfile;
copyJar(source, destination);
And my "copyJar" method:
private void copyJar(File source, File dest) throws IOException {
InputStream is = null;
OutputStream os = null;
try {
is = new FileInputStream(source);
os = new FileOutputStream(dest);
byte[] buffer = new byte[1024];
int length;
while ((length = is.read(buffer))>0) {
os.write(buffer, 0, length);
}
} catch (Exception e) {
e.printStackTrace();
}
is.close();
os.close();
}
At os = new FileOutputStream(dest); the debugger throws a FileNotFoundException with the description that the access to the directory has been denied.
Does anyone have an idea what I am doing wrong or have a better solution for setting the permissions via Java? Thanks in advance!

A similar question was asked there are several years.
A possible solution for Java 7 and Unix system is available here : How do i programmatically change file permissions?
Or, below the best response, a example with JNA.
I hope that that will help you !

I solved the problem. In the end it was much easier to solve than expected.
The main problem was not the permission issue but the FileNotFoundException. The file that is assigned to the OutputStream is not really a file, but just a directory so that the Stream can't find it. You have to create the file before initializing the OutputStream and after that you copy your source file into the newly created file. The code:
private void copyJar(File source, File dest) throws IOException {
InputStream is = null;
File dest2 = new File(dest+"/TheJar.jar");
dest2.createNewFile();
OutputStream os = null;
try {
is = new FileInputStream(source);
os = new FileOutputStream(dest2);
byte[] buffer = new byte[1024];
int length;
while ((length = is.read(buffer))>0) {
os.write(buffer, 0, length);
}
} catch (Exception e) {
e.printStackTrace();
}
is.close();
os.close();
}

ZipException: error in opening zip file

I'm working on a method which will take a zipped file, unzip it, and return a new file/directory containing all the unzipped files. The goal is to then take that directory and extract an excel document from it and then convert it into a Workbook class I built (which is fully unit tested and works fine). The problem is that I'm getting the following exception:
java.util.zip.ZipException: error in opening zip file
at java.util.zip.ZipFile.open(Native Method)
at java.util.zip.ZipFile.<init>(ZipFile.java:215)
at java.util.zip.ZipFile.<init>(ZipFile.java:145)
at java.util.zip.ZipFile.<init>(ZipFile.java:159)
at com.atd.core.datamigrator.BulkImageUpload.createWorkbook(BulkImageUpload.java:54)
at com.atd.core.datamigrator.BulkImageUpload.importImages(BulkImageUpload.java:38)
at com.atd.core.datamigrator.BulkImageUpload.main(BulkImageUpload.java:236)
Here is my code
private Workbook createWorkbook(File file) {
File unZipedFile = unZip(file);
File[] files = unZipedFile.listFiles();
Workbook wBook = null;
for (int i = 0; i < files.length; i++) {
if (files[i].getName().contains(".xls")) {
try {
File f = files[i];
ZipFile zip = new ZipFile(f);
wBook = new Workbook(zip);
} catch (IOException e) {
e.printStackTrace();
}
break;
}
}
return wBook;
}
private File unZip(File input) {
File output = new File("unzippedFile");
OutputStream out = null;
try {
ZipFile zipFile = new ZipFile(input);
Enumeration<? extends ZipEntry> entries = zipFile.entries();
while (entries.hasMoreElements()) {
ZipEntry entry = entries.nextElement();
File entryDestination = new File(output, entry.getName());
entryDestination.getParentFile().mkdirs();
InputStream in = zipFile.getInputStream(entry);
ZipInputStream zis = new ZipInputStream(in);
out = new FileOutputStream(entryDestination);
out.write(zis.read());
out.flush();
out.close();
}
} catch (FileNotFoundException e) {
// TODO Auto-generated catch block
e.printStackTrace();
} catch (IOException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
return output;
}
I know this is a problem with the unzip method because when I use File f = new File("some path") instead of using the unzipped file, it works fine.
Also, File I/O was never my strong point, so be nice :)

Okay, I now believe that this is the problem:
ZipInputStream zis = new ZipInputStream(in);
out = new FileOutputStream(entryDestination);
out.write(zis.read());
out.flush();
out.close();
You're creating a new file, and writing a single byte to it. That's not going to be a valid Excel file of any description. You're also failing to close streams using finally blocks, but that's a different matter. To copy the contents of one stream to another, you want something like:
byte[] buffer = new byte[8192];
int bytes;
while ((bytes = input.read(buffer)) > 0) {
output.write(buffer, 0, bytes);
}
That said, you'd be better off using a 3rd party library to hide all of this detail - look at Guava and its ByteStreams and Files classes for example.
It's worth taking a step back and working out why you didn't spot this problem for yourself, by the way. For example, the first thing I'd have done would be to look at the directory where the files were unzipped, and try to open those files. Just seeing a bunch of 1-byte files would be a bit of a giveaway. When trying to diagnose an issue, it's vital that you can split a big problem into small ones, and work out which small one is at fault.

regarding cross checking of files in an directory

I have developed a java program that copies the file from source folder to destination folder
there are 10 serialized files that it copies from source folder to destination folder
but one thing is missing in it is that let say if the files are already exists in the destination folder then in that case it should not copy
so basically a look is done within in one second that will check the destination folder contain those 10 serialized files or not
if not then in that case only it should copy and after copying it should again check within in second whether file exists or not , Please advise how to achieve this
//Create a class extends with TimerTask
class ScheduledTask extends TimerTask {
public void run() {
InputStream inStream = null;
OutputStream outStream = null;
try {
File source = new File("C:\\cache\\");
File target = new File("C:\\Authclient\\cache\\");
// Already exists. do not copy
/*if (target.exists()) {
return;
}*/
File[] files = source.listFiles();
for (File file : files) {
inStream = new FileInputStream(file);
outStream = new FileOutputStream(target + "/" + file.getName());
byte[] buffer = new byte[1024];
int length;
// copy the file content in bytes
while ((length = inStream.read(buffer)) > 0) {
outStream.write(buffer, 0, length);
}
inStream.close();
outStream.close();
}
System.out.println("File is copied successful!");
} catch (IOException e) {
e.printStackTrace();
}
}
}
public class Copycache {
public static void main(String args[]) throws InterruptedException {
Timer time = new Timer();
ScheduledTask task = new ScheduledTask();
time.schedule(task, new Date(), TimeUnit.SECONDS.toMillis(1));
}
}
the above exists implementation which is commented is not working correct rite now please advise

I am curious about your exact requirements. Consider this small example:
File file = new File("test.txt");
if (!file.exists())
{
FileOutputStream fis = new FileOutputStream(file);
fis.write("blabla".getBytes());
fis.close();
}
Now put a breakpoint on the line FileOutputStream fis...
Run it and wait at the breakpoint, then create the test.txt manually and put some data in it.
Then continue running the program.
Your program will overwrite the contents of test.txt without warning.
If timing is so crucial here you will need to figure out a different solution.
Edit: I got curious and did some more testing. It seems it won't even throw an exception if you add the line file.createNewFile();, break there, create the file and then continue the application. I wonder why..

ZipInputStream check whether zip file valid before going to extract it

How we check whether zip file corrupted or valid Zip file before going to extract it
my code`
import java.io.IOException;
import java.io.OutputStream;
import java.util.zip.ZipEntry;
import java.util.zip.ZipInputStream;
public void unzip() {
FileInputStream fin = null;
ZipInputStream zin = null;
OutputStream fout = null;
File outputDir = new File(_location);
File tmp = null;
try {
fin = new FileInputStream(_zipFile);
zin = new ZipInputStream(fin);
ZipEntry ze = null;
while ((ze = zin.getNextEntry()) != null) {
Log.d("Decompress", "Unzipping " + ze.getName());
if (ze.isDirectory()) {
dirChecker(ze.getName());
} else {
tmp = File.createTempFile( "decomp", ".tmp", outputDir );
fout = new BufferedOutputStream(new FileOutputStream(tmp));
DownloadFile.copyStream( zin, fout, _buffer, BUFFER_SIZE );
zin.closeEntry();
fout.close();
fout = null;
tmp.renameTo( new File(_location + ze.getName()) );
tmp = null;
}
}
zin.close();
zin = null;
} catch (IOException e) {
throw new RuntimeException(e);
} finally {
if ( tmp != null ) { try { tmp.delete(); } catch (Exception ignore) {;} }
if ( fout != null ) { try { fout.close(); } catch (Exception ignore) {;} }
if ( zin != null ) { try { zin.closeEntry(); } catch (Exception ignore) {;} }
if ( fin != null ) { try { fin.close(); } catch (Exception ignore) {;} }
}
}
`
this work fine with valid zipfile, but invalid zipfile it doesen't throw any exception not produce anything, but i need to confirm the validity of zip file before going to unzip it

I think it's pretty much useless for checking if the zip file is corrupted for two reasons:
Some zip files contain more bytes than just the zip part. For example, self-extracting archives have an executable part yet they're still valid zip.
The file can be corrupted without changing its size.
So, I suggest calculating the CRC for a guaranteed method of checking for corruption.

A Zip file will be valid as long as it has its zip entry catalog present. If you take the zip command, it will allow you to browse so long as the catalog is present. A parameter used for testing actually performs an extraction and CRC check.
What you can do is extract and do a CRC check on a temporary folder using the temp dir creation facility of Java. Then if it is all successful, commit the extract by copying the files from the temp dir to the final destination.

Problem in Zipping a File

When I run my code and use the files that are in the resource folder of my project itself, I face no problems. It zips the file successfully and I can extract it using WINZIP. The problem comes when I try to zip a file that is not in the project folder.
When I do the same, I am passing the Absolute Path of both the src and the dest files. My program doesn't give any exceptions, but when I try to open that zip file, I get an error saying, File is Invalid.
Can anyone tell me why this may be happening.
public static void compress(String srcPath, String destPath) {
srcFile = new File(srcPath);
destFile = new File(destPath);
try {
fileInputStream = new FileInputStream(srcFile);
fileOutputStream = new FileOutputStream(destFile);
zipEntry = new ZipEntry(srcPath);
zipOutputStream = new ZipOutputStream(fileOutputStream);
zipOutputStream.putNextEntry(zipEntry);
byte[] data = new byte[12];
while ((fileInputStream.read(data)) != -1) {
zipOutputStream.write(data);
}
} catch (Exception e) {
e.printStackTrace();
} finally {
try{
fileInputStream.close();
zipOutputStream.close();}catch (Exception e) {
e.printStackTrace();
}
}
}

You should not store paths with drive letters in your zip file because when you try to extract your zip, it will try to create a directory with the name of the drive and fail.
You will need to change your code so that it removes the drive letter from the path before creating the ZipEntry.