Develop Reference

Create a directory with ./ folder in java on Windows OS

I am working on a Java project in which the Tar file has to be created. These tar files are created on Linux OS so the folder structure always as : C:\Users\Admin\Desktop\Tar\My.tar\.\
In this path, the files are kept i.e. if I open My.tar using 7z or anything then I see files within the path : C:\Users\Admin\Desktop\Tar\My.tar\.\
Now on windows, I have written the code to generate tar file with some files. But it does not create folder having ./ I want the same structure to be followed to generate tar so that existing code wor fine.
How to create a directory with ./ - C:\Users\Admin\Desktop\Tar\My.tar\.\
The code to generate tar file is :
FileOutputStream dest = new FileOutputStream(outputTarFileName);
// Create a TarOutputStream
TarOutputStream out = new TarOutputStream(new BufferedOutputStream(dest));
File folder = new File(inputFolder);
File[] filesToTar = new File[folder.listFiles().length];
int index = 0;
for (final File fileEntry : folder.listFiles()) {
filesToTar[index++] = new File((fileEntry.getAbsolutePath()));
}
for (File f : filesToTar) {
out.putNextEntry(new TarEntry(f, f.getName()));
BufferedInputStream origin = new BufferedInputStream(new FileInputStream(f));
int count;
byte data[] = new byte[2048];
while ((count = origin.read(data)) != -1) {
out.write(data, 0, count);
}
out.flush();
origin.close();
}
out.close();
createTarFile("C:\\Users\\Admin\\Desktop\\Tar\\MyDir","C:\\Users\\Admin\\Desktop\\Tar\\\\.\\My.tar");
I tried by creating a directory having such structure :
new File("C:\Users\Admin.000\Desktop\testfolder\./\W").mkdir();
new File("C:\Users\Admin.000\Desktop\testfolder\\.\W").mkdir();
But nothing worked.

How to create ZIP files using list of Input streams?

In my case I have to download images from the resources folder in my web app. Right now I am using the following code to download images through URL.
url = new URL(properties.getOesServerURL() + "//resources//WebFiles//images//" + imgPath);
filename = url.getFile();
is = url.openStream();
os = new FileOutputStream(sClientPhysicalPath + "//resources//WebFiles//images//" + imgPath);
b = new byte[2048];
while ((length = is.read(b)) != -1) {
os.write(b, 0, length);
}
But I want a single operation to read all images at once and create a zip file for this.
I don't know so much about the use of sequence input streams and zip input streams so if it is possible through these, please let me know.

The only way I can see you being able to do this is something like the following:
try {
ZipOutputStream zip = new ZipOutputStream(new FileOutputStream("C:/archive.zip"));
//GetImgURLs() is however you get your image URLs
for(URL imgURL : GetImgURLs()) {
is = imgURL.openStream();
zip.putNextEntry(new ZipEntry(imgURL.getFile()));
int length;
byte[] b = new byte[2048];
while((length = is.read(b)) > 0) {
zip.write(b, 0, length);
}
zip.closeEntry();
is.close();
}
zip.close();
}
Ref: ZipOutputStream Example

The url should return zip file. Else you have to take one by one and create a zip using your program

Can't Retrieve Resources from External Jar File

NOTE: This is a followup to my question here.
I have a program that takes the contents of a directory and bundles everything into a JAR file. The code I use to do this is here:
try
{
FileOutputStream stream = new FileOutputStream(target);
JarOutputStream jOS = new JarOutputStream(stream);
LinkedList<File> fileList = new LinkedList<File>();
buildList(directory, fileList);
JarEntry jarAdd;
String basePath = directory.getAbsolutePath();
byte[] buffer = new byte[4096];
for(File file : fileList)
{
String path = file.getPath().substring(basePath.length() + 1);
path.replaceAll("\\\\", "/");
jarAdd = new JarEntry(path);
jarAdd.setTime(file.lastModified());
jOS.putNextEntry(jarAdd);
FileInputStream in = new FileInputStream(file);
while(true)
{
int nRead = in.read(buffer, 0, buffer.length);
if(nRead <= 0)
break;
jOS.write(buffer, 0, nRead);
}
in.close();
}
jOS.close();
stream.close();
So, all is well and good and the jar gets created, and when I explore its contents with 7-zip it has all the files I need. However, when I try to access the contents of the Jar via a URLClassLoader (the jar is not on the classpath and I don't intend it to be), I get null pointer exceptions.
The odd thing is, when I use a Jar that I've exported from Eclipse, I can access the contents of it in the way I want. This leads me to believe that I'm somehow not creating the Jar correctly, and am leaving something out. Is there anything missing from the method up above?

I figured it out based on this question - the problem was me not properly handling backslashes.
Fixed code is here:
FileOutputStream stream = new FileOutputStream(target);
JarOutputStream jOS = new JarOutputStream(stream);
LinkedList<File> fileList = new LinkedList<File>();
buildList(directory, fileList);
JarEntry entry;
String basePath = directory.getAbsolutePath();
byte[] buffer = new byte[4096];
for(File file : fileList)
{
String path = file.getPath().substring(basePath.length() + 1);
path = path.replace("\\", "/");
entry = new JarEntry(path);
entry.setTime(file.lastModified());
jOS.putNextEntry(entry);
FileInputStream in = new FileInputStream(file);
while(true)
{
int nRead = in.read(buffer, 0, buffer.length);
if(nRead <= 0)
break;
jOS.write(buffer, 0, nRead);
}
in.close();
jOS.closeEntry();
}
jOS.close();
stream.close();

separate multiple jpg files from a single file

I have merged multiple jpeg files into one single .bin file.
.....
.........
while(true){
if (q.numOfFiles() > 0) {
source = q.getNextFile();
in = new DataInputStream(new BufferedInputStream(new FileInputStream(source)));
byte[] buf = new byte[1024];
int len;
while ((len = in.read(buf)) > 0) {
out.write(buf, 0, len);
}
String s = "filename=="+source.getName()+"==filename";
out.write(s.getBytes());
out.flush();
System.out.println("merged--"+source.getName());
}
}
........
........
as you can see i am appending "filename=="+source.getName()+"==filename" after end of each file.
Now i want to separate all those jpegs with their actual file names.
How can i read the separators that I've inserted in the merged files ?

As above in the comments:
Sugestion/answer to adapt your point of view:
Using a single file zip to ship multiple files.
How to create zip in Java
And to extract:
Extract from zip in Java

Unzipping Files with a large amount of small Files inside

I have a lot of zips with a large amount of Files which i have to unzip to the SD-Card.
Currently I am using the util.zip Package with Zipentry for each File inside the zips. This works fine but is very slow.
So I wonder if there is a lib which can handle those Files faster than the normal zip of Java/Android.
Edit: The zips are archives which are about 5 to 10MB large and contain about 50 to 100 jpg-Files whiche are each part of a picture. I need to extract all zips to a specific folder.
Slow means that the same files are extracted on an IPhone in fraction of the time.
Edit 2 the code:
ZipInputStream zin = new ZipInputStream(fd);
ZipEntry ze = null;
while ((ze = zin.getNextEntry()) != null) {
System.out.println("Unzipping " + ze.getName());
if (ze.isDirectory()) {
File f = new File(targetFilePath + ze.getName());
if (!f.isDirectory()) {
f.mkdirs();
}
} else {
int size;
byte[] buffer = new byte[2048];
FileOutputStream outStream = new FileOutputStream(targetFilePath + ze.getName() + ".tile");
BufferedOutputStream bufferOut = new BufferedOutputStream(outStream, buffer.length);
while ((size = zin.read(buffer, 0, buffer.length)) != -1) {
bufferOut.write(buffer, 0, size);
}
bufferOut.flush();
bufferOut.close();
}
}
zin.close();

The best case you could do is go NDK way. AFAIK, Google even provides supported headers/API in NDK on zlib.