detect main inside a jar using java code.

Question

I am trying to detect which class inside a jar contains main or a supplied method name (if possible).

At the moment I have the following code

public static void getFromJars(String pathToAppJar) throws IOException{
    FileInputStream jar = new FileInputStream(pathToAppJar);
    ZipInputStream zipSteam = new ZipInputStream(jar);
    ZipEntry ze;
    while ((ze = zipSteam.getNextEntry()) != null) {
        System.out.println(ze.toString());          
    }
    zipSteam.close();
}

This will allow me to get packages and classes under these packages, but I do not know if it is possible to even get methods inside classes. Further, I do not know if this approach is even good for a case of several pkgs inside the jar, since each package can have a class with main in it.

I would appreciate any ideas.

you can use a java decompiler and search the decompiled results: http://jd.benow.ca/ — Simulant, Oct 26 '15 at 17:26
That is true, and I am using it at the moment, but I want something more automatic — Quantico, Oct 26 '15 at 17:27
Duplicate of http://stackoverflow.com/questions/28776205/java-binary-class-file-format-parser ? — Marged, Oct 26 '15 at 17:28
Have you seen http://stackoverflow.com/questions/9314541/analyze-jar-file-programmatically ? Once you got hold of each individual Class calling `getMethods()` should give you the info you need. — fvu, Oct 26 '15 at 17:29
@fvu, I have not seen this post. This looks like a great solution. — Quantico, Oct 26 '15 at 17:31
@fvu, looking more closely isn't load class requires the source? — Quantico, Oct 26 '15 at 17:41
@Quantico no. See the line `if (zipEntry.getName().endsWith(".class")) {` and `Class> clazz = this.loadClass(className);`. It loads the binary class. A Java class file contains enough info to make a pretty decent reproduction of the source code. — fvu, Oct 26 '15 at 17:44
What is "this" in that code? where is it being defined. In the code that I supplied above say you do ze.getName() ... you cannot just call Class> clazz = this.loadClass(className); — Quantico, Oct 26 '15 at 17:46
Note that the proposed method is part of a custom classloader, one that extends `URLClassLoader`. There should be examples floating around but I cannot find a good one except http://examples.javacodegeeks.com/core-java/net/urlclassloader/java-net-urlclassloader-example/ right now. — fvu, Oct 26 '15 at 18:02

score 3 · Accepted Answer · edited Jul 26 '18 at 18:57

Thanks to fvu's comments, I ended up with the following code.

public static void getFromJars(String pathToAppJar) throws IOException, ClassNotFoundException
{
    FileInputStream jar = new FileInputStream(pathToAppJar);
    ZipInputStream zipSteam = new ZipInputStream(jar);
    ZipEntry ze;
    URL[] urls = { new URL("jar:file:" + pathToAppJar+"!/") };
    URLClassLoader cl = URLClassLoader.newInstance(urls);

    while ((ze = zipSteam.getNextEntry()) != null) {
        // Is this a class?
        if (ze.getName().endsWith(".class")) {
            // Relative path of file into the jar.
            String className = ze.getName();

            // Complete class name
            className = className.replace(".class", "").replace("/", ".");
            Class<?> klazz = cl.loadClass(className);
            Method[] methodsArray = klazz.getMethods();
        }
    }
    zipSteam.close();
}

I removed the code that uses the methods found, since it is not important for this answer

detect main inside a jar using java code.

1 Answers1