I created a Jar file from my java code :
public void create() throws IOException{
FileOutputStream stream = new FileOutputStream(this.packagePath);
JarOutputStream out = new JarOutputStream(stream, new Manifest());
out.close();
//jarFile = new JarFile(new File(this.packagePath));
}
I get a META-INF directory, with a MANIFEST.MF file inside.
Now, when I want to add a file to the jar file :
public void addFile(File file) throws IOException{
//first, make sure the package already exists
if(!file.exists()){
throw new IOException("Make" +
" sure the package file already exists.you might need to call the Package.create() " +
"method first.");
}
FileOutputStream stream = new FileOutputStream(this.packagePath);
JarOutputStream out = new JarOutputStream(stream);
/*if(jarFile.getManifest()!=null){
out = new JarOutputStream(stream,jarFile.getManifest());
}else{
out=new JarOutputStream(stream);
}*/
byte buffer[] = new byte[BUFFER_SIZE];
JarEntry jarEntry = new JarEntry(file.getName());
jarEntry.setTime(file.lastModified());
out.putNextEntry(jarEntry);
//Write file to archive
FileInputStream in = new FileInputStream(file);
while (true) {
int nRead = in.read(buffer, 0, buffer.length);
if (nRead <= 0)
break;
out.write(buffer, 0, nRead);
}
in.close();
out.close();
}
when adding a file to the JAR archive using the above code, the META-INF directory with its MANIFEST.MF disappears and I get the newly added file.
I want to be able to add the file to the jar, and still get the manifest file. inside the manifest file, I want a line with the name of the newly added jar.
Thanks.
include the ant.jar in your project classpath
codeJar jar = new Jar();
//configure the jar object
jar.execute();
Look here for info about the parameters you can set.
I think you cannot add a new entry, because you cannot open jar package for "append". I think you must create a new jar file and copying entries from old, and add your entries.
FileOutputStream stream = new FileOutputStream(this.packagePath);
This line will create a new, empty file.
To edit a jar file, you will need to back up the old version and copy its entries into your new jar file.
Related
While Uncompressing .7z file, Empty folders are ignored, I want to consider Empty folders as well after uncompressing any .7z file.
My Code is as below
SevenZFile sevenZFile = new SevenZFile(new File(filename));
SevenZArchiveEntry entry;
while ((entry = sevenZFile.getNextEntry()) != null){
if (entry.isDirectory()){
continue;
}
File curfile = new File(DestinationPath,entry.getName());
File parent = curfile.getParentFile();
if (!parent.exists()) {
parent.mkdirs();
}
FileOutputStream out = new FileOutputStream(curfile);
byte[] content = new byte[(int) entry.getSize()];
sevenZFile.read(content, 0, content.length);
out.write(content);
out.close();
Your code seems working.
Probably the folder aren't in the "yourfile.7zip" from the beginning.
This is a common issue of 7zip and you have to update your 7zip version.
If the 7Zip contains proper arguments just use:
if (entry.isDirectory()){
new File(DestinationPath,entry.getName()).mkdir();
continue;
}
Since:
A file output stream is an output stream for writing data to a File or
to a FileDescriptor.
That is the proper method to accomplish the task because there's no folder implementation by native library vendor.
I have the following situation:
I am able to zip my files with the following method:
public boolean generateZip(){
byte[] application = new byte[100000];
ByteArrayOutputStream baos = new ByteArrayOutputStream();
// These are the files to include in the ZIP file
String[] filenames = new String[]{"/subdirectory/index.html", "/subdirectory/webindex.html"};
// Create a buffer for reading the files
try {
// Create the ZIP file
ZipOutputStream out = new ZipOutputStream(baos);
// Compress the files
for (int i=0; i<filenames.length; i++) {
byte[] filedata = VirtualFile.fromRelativePath(filenames[i]).content();
ByteArrayInputStream in = new ByteArrayInputStream(filedata);
// Add ZIP entry to output stream.
out.putNextEntry(new ZipEntry(filenames[i]));
// Transfer bytes from the file to the ZIP file
int len;
while ((len = in.read(application)) > 0) {
out.write(application, 0, len);
}
// Complete the entry
out.closeEntry();
in.close();
}
// Complete the ZIP file
out.close();
} catch (IOException e) {
System.out.println("There was an error generating ZIP.");
e.printStackTrace();
}
downloadzip(baos.toByteArray());
}
This works perfectly and I can download the xy.zip which contains the following directory and file structure:
subdirectory/
----index.html
----webindex.html
My aim is to completely leave out the subdirectory and the zip should only contain the two files. Is there any way to achieve this?
(I am using Java on Google App Engine).
Thanks in advance
If you are sure the files contained in the filenames array are unique if you leave out the directory, change your line for constructing ZipEntrys:
String zipEntryName = new File(filenames[i]).getName();
out.putNextEntry(new ZipEntry(zipEntryName));
This uses java.io.File#getName()
You can use Apache Commons io to list all your files, then read them to an InputStream
Replace the line below
String[] filenames = new String[]{"/subdirectory/index.html", "/subdirectory/webindex.html"}
with the following
Collection<File> files = FileUtils.listFiles(new File("/subdirectory"), new String[]{"html"}, true);
for (File file : files)
{
FileInputStream fileStream = new FileInputStream(file);
byte[] filedata = IOUtils.toByteArray(fileStream);
//From here you can proceed with your zipping.
}
Let me know if you have issues.
You could use the isDirectory() method on VirtualFile
I am having an issue putting a folder in a zip file I am trying to create. While the path is valid, when I run the code it gives me a File Not Found Exception.
Here is my code
String outFilename = "outfile.zip";
ZipOutputStream out = new ZipOutputStream(new FileOutputStream(outFilename));
byte[] buf = new byte[1024];
File file = new File("workspace");
System.out.println(file.isDirectory());
System.out.println(file.getAbsolutePath());
FileInputStream in = new FileInputStream(file.getAbsolutePath());
out.putNextEntry(new ZipEntry(file.getAbsolutePath()));
int len;
while ((len = in.read(buf)) > 0) {
out.write(buf, 0, len);
}
out.closeEntry();
in.close();
You're trying to read bytes from a directory; it doesn't work like that. The exception says as much, too.
You need to add the directory, then add each file within the directory. If you use the file path you don't need to add the directory explicitly.
I'd be very wary of using the absolute path as the zip entry; better to use a relative path so you can unzip it anywhere and not risk overwriting something you want.
I have a java program as below for zipping a folder as a whole.
public static void zipDir(String dir2zip, ZipOutputStream zos)
{
try
{
File zipDir= new File(dir2zip);
String[] dirList = zipDir.list();
byte[] readBuffer = new byte[2156];
int bytesIn = 0;
for(int i=0; i<dirList.length; i++)
{
File f = new File(zipDir, dirList[i]);
if(f.isDirectory())
{
String filePath = f.getPath();
zipDir(filePath, zos);
continue;
}
FileInputStream fis = new FileInputStream(f);
ZipEntry anEntry = new ZipEntry(f.getPath());
zos.putNextEntry(anEntry);
while((bytesIn = fis.read(readBuffer)) != -1)
{
zos.write(readBuffer, 0, bytesIn);
}
fis.close();
}
}
catch(Exception e)
{
e.printStackTrace();
}
}
public static void main(){
String date=new java.text.SimpleDateFormat("MM-dd-yyyy").format(new java.util.Date());
ZipOutputStream zos = new ZipOutputStream(new FileOutputStream("Output/" + date + "_RB" + ".zip"));
zipDir("Output/" + date + "_RB", zos);
zos.close();
}
My query here is. The target folder(+date+_RB) to be zipped is present inside the folder named Output. After successful zipping, when I extract the zipped file, I find a folder Output inside which the (+date+_RB) required folder is present. I need not want that Output folder after the extraction of the zipped file, rather it should directly extract the required folder alone. Please advise on the same.
UPDATE:
I tried Isaac's answer. While extracting the resultant zip file, no folders are getting extracted. Only the files inside all the folders are getting extracted. I just dont need the folder "Output" alone in the resultant zip file. But what the program does is, it doesnt extracts all other folders inside the Output folder, rather it just extracts the files inside those folders. Kindly advise on how to proceed...
It happens because of this:
ZipEntry anEntry = new ZipEntry(f.getPath());
f.getPath() will return Output/ at the beginning of the string. This is due to the flow of your program and how it (mis)uses File objects.
I suggest you construct a File object called, say, tmp:
File tmp = new File(dirList[i]);
The change the construction of f:
File f = new File(zipDir, tmp.getPath());
Then, change this:
ZipEntry anEntry = new ZipEntry(f.getPath());
To this:
ZipEntry anEntry = new ZipEntry(tmp.getPath());
I didn't have time to actually test it, but in a nutshell, your problem is due to how the File object is constructed.
What is the Java equivalent to this jar command:
C:\>jar cvf myjar.jar directory
I'd like to create this jar file programmatically as I can't be assured that the jar command will be on the system path where I could just run the external process.
Edit: All I want is to archive (and compress) a directory. Doesn't have to follow any java standard. Ie: a standard zip is fine.
// These are the files to include in the ZIP file
String[] source = new String[]{"source1", "source2"};
// Create a buffer for reading the files
byte[] buf = new byte[1024];
try {
// Create the ZIP file
String target = "target.zip";
ZipOutputStream out = new ZipOutputStream(new FileOutputStream(target));
// Compress the files
for (int i=0; i<source.length; i++) {
FileInputStream in = new FileInputStream(source[i]);
// Add ZIP entry to output stream.
out.putNextEntry(new ZipEntry(source[i]));
// Transfer bytes from the file to the ZIP file
int len;
while ((len = in.read(buf)) > 0) {
out.write(buf, 0, len);
}
// Complete the entry
out.closeEntry();
in.close();
}
// Complete the ZIP file
out.close();
} catch (IOException e) {
}
You can also use the answer from this post How to use JarOutputStream to create a JAR file?
Everything you'll want is in the java.util.jar package:
http://java.sun.com/javase/6/docs/api/java/util/jar/package-summary.html