CODEWITHSUNDEEP
javajarexe
krisp
krisp

Reputation: 4777

Run exe which is packaged inside jar file

I am executing an exe through my java program. The path is hardcoded in Java.

I have packaged my the exe in the jar.

But am stuck as I have the path name hardcoded in the Java file, so I am not able to execute my jar as a stand alone program.

Any hints for packaging such jar i.e having an exe inside and able to run it as a stand alone program?

Upvotes: 31

Views: 36991

Answers (6)

TofuBeer
TofuBeer

Reputation: 61526

This will extract the .exe to a local file on the local disk. The file will be deleted when the Java program exists.

import java.io.Closeable;
import java.io.File;
import java.io.FileNotFoundException;
import java.io.FileOutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.OutputStream;
import java.net.URI;
import java.net.URISyntaxException;
import java.net.URL;
import java.security.CodeSource;
import java.security.ProtectionDomain;
import java.util.zip.ZipEntry;
import java.util.zip.ZipException;
import java.util.zip.ZipFile;

public class Main
{
    public static void main(final String[] args)
        throws URISyntaxException,
               ZipException,
               IOException
    {
        final URI uri;
        final URI exe;

        uri = getJarURI();
        exe = getFile(uri, "Main.class");
        System.out.println(exe);
    }

    private static URI getJarURI()
        throws URISyntaxException
    {
        final ProtectionDomain domain;
        final CodeSource       source;
        final URL              url;
        final URI              uri;

        domain = Main.class.getProtectionDomain();
        source = domain.getCodeSource();
        url    = source.getLocation();
        uri    = url.toURI();

        return (uri);
    }

    private static URI getFile(final URI    where,
                               final String fileName)
        throws ZipException,
               IOException
    {
        final File location;
        final URI  fileURI;

        location = new File(where);

        // not in a JAR, just return the path on disk
        if(location.isDirectory())
        {
            fileURI = URI.create(where.toString() + fileName);
        }
        else
        {
            final ZipFile zipFile;

            zipFile = new ZipFile(location);

            try
            {
                fileURI = extract(zipFile, fileName);
            }
            finally
            {
                zipFile.close();
            }
        }

        return (fileURI);
    }

    private static URI extract(final ZipFile zipFile,
                               final String  fileName)
        throws IOException
    {
        final File         tempFile;
        final ZipEntry     entry;
        final InputStream  zipStream;
        OutputStream       fileStream;

        tempFile = File.createTempFile(fileName, Long.toString(System.currentTimeMillis()));
        tempFile.deleteOnExit();
        entry    = zipFile.getEntry(fileName);

        if(entry == null)
        {
            throw new FileNotFoundException("cannot find file: " + fileName + " in archive: " + zipFile.getName());
        }

        zipStream  = zipFile.getInputStream(entry);
        fileStream = null;

        try
        {
            final byte[] buf;
            int          i;

            fileStream = new FileOutputStream(tempFile);
            buf        = new byte[1024];
            i          = 0;

            while((i = zipStream.read(buf)) != -1)
            {
                fileStream.write(buf, 0, i);
            }
        }
        finally
        {
            close(zipStream);
            close(fileStream);
        }

        return (tempFile.toURI());
    }

    private static void close(final Closeable stream)
    {
        if(stream != null)
        {
            try
            {
                stream.close();
            }
            catch(final IOException ex)
            {
                ex.printStackTrace();
            }
        }
    }
}

Upvotes: 37

D.Snap
D.Snap

Reputation: 1780

//gets program.exe from inside the JAR file as an input stream
InputStream is = getClass().getResource("program.exe").openStream();
//sets the output stream to a system folder
OutputStream os = new FileOutputStream("program.exe");

//2048 here is just my preference
byte[] b = new byte[2048];
int length;

while ((length = is.read(b)) != -1) {
    os.write(b, 0, length);
}

is.close();
os.close();

Upvotes: 6

Mid Bifroid
Mid Bifroid

Reputation: 31

Use

getClass().getResource(what).openStream()

and copy to another file in the disk.

Upvotes: 2

gencoreoperative
gencoreoperative

Reputation: 398

Whilst the other users have answered the question correctly, extract and run then cleanup. Another point to consider is going fully native.

You are already using a native binary to achieve a specific task. Why not also create a native installer which will install your application, and install the binary to the OS specific location (Program Files on Win32) and create suitable shortcuts.

This way your application will feel more native and means you don't need to write or manage code to get around this fact. At the moment the Jar looks like a cross platform piece of code (Jar runs anywhere right?) but packs a native binary which will not run everywhere. This feels like a contradiction.

For installers I can recommend Nullsoft Installation System (NSIS) as they have many excellent tutorials and code samples to learn from.

Upvotes: 3

Kevin
Kevin

Reputation: 30429

You could write a simple java program to launch the exe using Runtime.exec(). You could then set the "Main-Class" attribute of the jar to be that launcher class. Users could then run your jar and it would run the exe.

Upvotes: -5

Joachim Sauer
Joachim Sauer

Reputation: 308041

The operating system doesn't care or know about .jar file, so you'll have to unpack the .exe file to some temporary location before you execute it.

Upvotes: 12

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