I've a dir (with sub dirs) template that is kept as a resource inside a jar file. During run
time I need to extract it (template) to tmp dir change some content and finally publish it as a zipped artifact.
My question is: how to extract this content easily? I was trying getResource() as well as getResourceAsStream()..
Following code works fine here: (Java7)
String s = this.getClass().getResource("").getPath();
if (s.contains("jar!")) {
// we have a jar file
// format: file:/location...jar!...path-in-the-jar
// we only want to have location :)
int excl = s.lastIndexOf("!");
s = s.substring(0, excl);
s = s.substring("file:/".length());
Path workingDirPath = workingDir = Files.createTempDirectory("demo")
try (JarFile jf = new JarFile(s);){
Enumeration<JarEntry> entries = jf.entries();
while (entries.hasMoreElements()) {
JarEntry je = entries.nextElement();
String name = je.getName();
if (je.isDirectory()) {
// directory found
Path dir = workingDirPath.resolve(name);
Files.createDirectory(dir);
} else {
Path file = workingDirPath.resolve(name);
try (InputStream is = jf.getInputStream(je);) {
Files.copy(is, file, StandardCopyOption.REPLACE_EXISTING);
}
}
}
}
} else {
// debug mode: no jar
}