How to use JarOutputStream to create a JAR file?
JavaJarJava Problem Overview
How does one create a JAR file programmatically using java.util.jar.JarOutputStream? The JAR file produced by my program looks correct (it extracts fine) but when I try loading a library from it Java complains that it cannot find files which are clearly stored inside it. If I extract the JAR file and use Sun's jar command-line tool to re-compress it the resulting library works fine. In short, something is wrong with my JAR file.
Please explain how to create a JAR file programmatically, complete with a manifest file.
Java Solutions
Solution 1 - Java
It turns out that JarOutputStream has three undocumented quirks:
- Directory names must end with a '/' slash.
- Paths must use '/' slashes, not '\'
- Entries may not begin with a '/' slash.
Here is the correct way to create a Jar file:
public void run() throws IOException {
Manifest manifest = new Manifest();
manifest.getMainAttributes().put(Attributes.Name.MANIFEST_VERSION, "1.0");
JarOutputStream target = new JarOutputStream(new FileOutputStream("output.jar"), manifest);
add(new File("inputDirectory"), target);
target.close();
}
private void add(File source, JarOutputStream target) throws IOException {
String name = source.getPath().replace("\\", "/");
if (source.isDirectory()) {
if (!name.endsWith("/")) {
name += "/";
}
JarEntry entry = new JarEntry(name);
entry.setTime(source.lastModified());
target.putNextEntry(entry);
target.closeEntry();
for (File nestedFile : source.listFiles()) {
add(nestedFile, target);
}
} else {
JarEntry entry = new JarEntry(name);
entry.setTime(source.lastModified());
target.putNextEntry(entry);
try (BufferedInputStream in = new BufferedInputStream(new FileInputStream(source))) {
byte[] buffer = new byte[1024];
while (true) {
int count = in.read(buffer);
if (count == -1)
break;
target.write(buffer, 0, count);
}
target.closeEntry();
}
}
}
Solution 2 - Java
There's another "quirk" to pay attention: All JarEntry's names should NOT begin with "/".
For example: The jar entry name for the manifest file is "META-INF/MANIFEST.MF" and not "/META-INF/MANIFEST.MF".
The same rule should be followed for all jar entries.
Solution 3 - Java
You can do it with this code:
public void write(File[] files, String comment) throws IOException {
FileOutputStream fos = new FileOutputStream(PATH + FILE);
JarOutputStream jos = new JarOutputStream(fos, manifest);
BufferedOutputStream bos = new BufferedOutputStream(jos);
jos.setComment(comment);
for (File f : files) {
print("Writing file: " + f.toString());
BufferedReader br = new BufferedReader(new FileReader(f));
jos.putNextEntry(new JarEntry(f.getName()));
int c;
while ((c = br.read()) != -1) {
bos.write(c);
}
br.close();
bos.flush();
}
bos.close();
// JarOutputStream jor = new JarOutputStream(new FileOutputStream(PATH + FILE), manifest);
}
PATH variable: path to JAR file
FILE variable: name and format
Solution 4 - Java
Ok, so by request, here's Gili's code, modified to use relative paths rather than absolute ones. (Replace "inputDirectory" with the directory of your choice.) I just tested it, but if it doesn't work, lemme know.
public void run() throws IOException
{
Manifest manifest = new Manifest();
manifest.getMainAttributes().put(Attributes.Name.MANIFEST_VERSION, "1.0");
JarOutputStream target = new JarOutputStream(new FileOutputStream("output.jar"), manifest);
File inputDirectory = new File("inputDirectory");
for (File nestedFile : inputDirectory.listFiles())
add("", nestedFile, target);
target.close();
}
private void add(String parents, File source, JarOutputStream target) throws IOException
{
BufferedInputStream in = null;
try
{
String name = (parents + source.getName()).replace("\\", "/");
if (source.isDirectory())
{
if (!name.isEmpty())
{
if (!name.endsWith("/"))
name += "/";
JarEntry entry = new JarEntry(name);
entry.setTime(source.lastModified());
target.putNextEntry(entry);
target.closeEntry();
}
for (File nestedFile : source.listFiles())
add(name, nestedFile, target);
return;
}
JarEntry entry = new JarEntry(name);
entry.setTime(source.lastModified());
target.putNextEntry(entry);
in = new BufferedInputStream(new FileInputStream(source));
byte[] buffer = new byte[1024];
while (true)
{
int count = in.read(buffer);
if (count == -1)
break;
target.write(buffer, 0, count);
}
target.closeEntry();
}
finally
{
if (in != null)
in.close();
}
}
Solution 5 - Java
Here's some sample code for creating a JAR file using the JarOutputStream:
Solution 6 - Java
This answer would solve the relative path problem.
private static void createJar(File source, JarOutputStream target) {
createJar(source, source, target);
}
private static void createJar(File source, File baseDir, JarOutputStream target) {
BufferedInputStream in = null;
try {
if (!source.exists()){
throw new IOException("Source directory is empty");
}
if (source.isDirectory()) {
// For Jar entries, all path separates should be '/'(OS independent)
String name = source.getPath().replace("\\", "/");
if (!name.isEmpty()) {
if (!name.endsWith("/")) {
name += "/";
}
JarEntry entry = new JarEntry(name);
entry.setTime(source.lastModified());
target.putNextEntry(entry);
target.closeEntry();
}
for (File nestedFile : source.listFiles()) {
createJar(nestedFile, baseDir, target);
}
return;
}
String entryName = baseDir.toPath().relativize(source.toPath()).toFile().getPath().replace("\\", "/");
JarEntry entry = new JarEntry(entryName);
entry.setTime(source.lastModified());
target.putNextEntry(entry);
in = new BufferedInputStream(new FileInputStream(source));
byte[] buffer = new byte[1024];
while (true) {
int count = in.read(buffer);
if (count == -1)
break;
target.write(buffer, 0, count);
}
target.closeEntry();
} catch (Exception ignored) {
} finally {
if (in != null) {
try {
in.close();
} catch (Exception ignored) {
throw new RuntimeException(ignored);
}
}
}
}