我正在通过我的java程序执行一个exe。路径是用Java硬编码的。
我已经把我的可执行文件打包在罐子里了。
但是,由于在Java文件中硬编码了路径名,因此我无法将jar作为独立程序执行。
对于打包这样的jar有什么建议吗?也就是说,里面有一个exe文件,可以作为一个独立的程序运行。
发布于 2009-03-01 18:05:14
这会将.exe
解压到本地磁盘上的本地文件中。当Java程序存在时,该文件将被删除。
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();
}
}
}
}
发布于 2009-03-01 17:33:51
操作系统并不关心或不知道.jar文件,因此在执行之前,您必须将.exe
文件解压缩到某个临时位置。
发布于 2016-04-04 20:13:25
//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();
https://stackoverflow.com/questions/600146
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