Making jar files available to javac?

Sorry, this is probably just a dumb newbie question. I'm having a problem setting up a library jar file in a windows environment. I know that this is just an issue of setting up the PATH or CLASSPATH or something else correctly but I can't seem to get it right. Any hints, suggestions, nudges or outright handholding cheerfully welcomed.
Thanks,
Frank Vandenberg

Setting the classpath isn't the same in all Windowz version.
Win95/Win98/Win98 SE/Win ME:
Here you have a c:\autoexec.bat
so you can just add the following line:
path="%jdkdir%\bin\"
if there probely already is a "path-defenetion" in your autoexec.bat in that case just add it. i.e
before: path="c:\windows\"
after: path="c:\windows\;%jdkdir%\bin\"
az you probely understand is not suppose to say %jdkdir% it suppose to say where you have it installed..
I have jdk installed in "C:\jdk1.4\"
That would for me make.
path="C:\jdk1.4\bin\"
And now if you have
Windows NT 4.0/Windows NT 5.0(Win2000)/Windows NT 5.1(Windows XP)
Here you have no autoexec.bat!
So to permatetly add the path you need to use another way ->
Deskop -> My Computer -> Controlepanel -> System -> Adcanced -> EvorimentVariables.
Here you should already have some paths-defined, and you can easiely add the path you need..but be sure NOT do remove anything unless you know what/why your doing.
Notice that i translated the deskription above..becuase my Windows is in Swedish...so the names may differ but its the same button.
Good Lyck!

Similar Messages

Making jar files using API

hello!
I m trying to make a jar file using the API. When the jar tool is used for the same manifest file, it is correctly written and an executable jar is created. When I use the API, the jar file is made but it is nor executable. Here is that part of code:
jar = new File(parent, jarName);
System.out.println("Executing jar command...");
//create jar file here
Manifest mf = new Manifest(new FileInputStream(manifest));
fos = new FileOutputStream(jar);
jos = new JarOutputStream(new BufferedOutputStream(fos), mf);
BufferedInputStream reader = null;
byte[] data = new byte[BUFFER];
int byteCount = 0;
for(int i = 0; i < fileNames.length; i++)
System.out.println("Adding " + fileNames);
FileInputStream fis = new FileInputStream(fileNames);
reader = new BufferedInputStream(fis, BUFFER);
JarEntry entry = new JarEntry(new ZipEntry(fileNames));
jos.putNextEntry(entry);
while((byteCount = reader.read(data,0,BUFFER)) != -1)
jos.write(data, 0, byteCount);
}//end while
reader.close();
}//end for
jos.close();//close jar output stream
fos.close();
I m sure someone will be kind and intelligent enough to solve the problem. Thank you!
Umer

A jar file is simply a Zip file. So the two API's are quite similar. First, you create a CRC - that is a way to make sure that the file isn't corrupted and that all the bytes are there (it's a polynominal algorithm that returns a number which is written in the Zip(Jar) file as well). Each entry is a file or directory. Before writing any data to the Zip(Jar) file, you have to write info on the actual file (like name, path, length etc). That's why you use putNextEntry(). Each file can have it's storing methods, although the most common is to use the "default" methods (built in the ZipOutputStream/JarOutputStream)
But in this example you set the storing methods on the entry. You use the setMethod() method to switch between either storing or compression (don't know why, but in this example you don't compress the file - you should). I myself don't use the setSize() and setCompressedSize() methods at all (and it works fine). Don't know exactly what are the implications of using them. Neither do I use the setCrc() method (If I think a little, I've only used this API once - I don't use it that much), but it's quite straightforward. It's used to check that the data is ok, and since the CRC number is the result of an algorithm, it needs to know the bytes of UNCOMPRESSED data: that's why you use crc.update() on the bytes. After that you actually write all the info set before to the Zip (Jar) file using putNextEntry(). But remember that a ZipEntry doesn't have any data. You'll have to write it yourself: thus the need for the write method: Here are two small programs I used for Zipping and Unzipping. Changing them to work for Jar files is extremely simple:
//Zip.java
import java.io.*;
import java.util.zip.*;
public class Zip
     public static void main(String[] arg)
          try
               if (arg.length < 3)
                    System.out.println("Usage: java Zip <0-9> <zip file> <file 1> [file 2] [file 3] ...");
                    System.exit(0);
               System.out.println("Compressing files into file " + arg[1]);
               ZipOutputStream out = null;
               try
                    out = new ZipOutputStream(new FileOutputStream(arg[1]));
               catch (FileNotFoundException ex)
                    System.out.println("Cannot create file " + arg[1]);
                    System.exit(1);
               try
                    out.setLevel(Integer.parseInt(arg[0]));
               catch (IllegalArgumentException ex)
                    System.out.println("Illegal compression level");
                    new File(arg[1]).delete();
                    System.exit(1);
               for (int i=2; i<arg.length; i++)
                    System.out.println("\tCompressing file " + (i-1));
                    FileInputStream read = null;
                    try
                         read = new FileInputStream(arg);
                    catch (FileNotFoundException ex)
                         System.out.println("\tCannot find file " + arg[i]);
                         continue;
                    out.putNextEntry(new ZipEntry(arg[i]));
                    int av = 0;
                    while ((av = read.available()) != 0)
                         byte[] b = new byte[av < 64 ? av : 64];
                         read.read(b);
                         out.write(b);
                    out.closeEntry();
                    read.close();
                    System.out.println("\tDone compressing file " + i);
               System.out.println("Done compressing");
               out.finish();
               out.close();
          catch (Exception e)
               new File(arg[1]).delete();
               e.printStackTrace(System.err);
And the unzipping app:
//UnZip.java
import java.io.*;
import java.util.*;
import java.util.zip.*;
public class UnZip
     public static void main(String[] arg)
          try
               if (arg.length < 1)
                    System.out.println("Usage: java UnZip <zip file> [to dir]");
                    System.exit(0);
               String dir = "";
               if (arg.length > 1)
                    dir = arg[1] + System.getProperty("file.separator");
                    File f = new File(arg[1]);
                    if (!f.exists()) f.mkdirs();
               System.out.println("Decompressing files from file " + arg[0]);
               ZipFile read = null;
               try
                    read = new ZipFile(arg[0]);
               catch (ZipException ex)
                    System.err.println("Zip error when reading file " + arg[0]);
                    System.exit(1);
               catch (IOException ex)
                    System.err.println("I/O Exception when reading file " + arg[0]);
                    System.exit(1);
               Enumeration en = read.entries();
               while (en.hasMoreElements())
                    ZipEntry entry = (ZipEntry) en.nextElement();
                    System.out.println("\tDecompressing file " + entry.getName());
                    FileOutputStream out;
                    try
                         out = new FileOutputStream(dir + entry.getName());
                    catch (FileNotFoundException ex)
                         System.err.println("\tCannot write down to file " + dir + entry.getName());
                         continue;
                    InputStream re = read.getInputStream(entry);
                    int nRead = 0;
                    for (byte[] buffer = new byte[1024]; (nRead = re.read(buffer)) != -1; out.write(buffer, 0, nRead));
                    out.close();
                    System.out.println("\tDone decompressing file " + entry.getName());
               read.close();
               System.out.println("Done decompressing files");
          catch (Exception e)
               new File(arg[1]).delete();
               e.printStackTrace(System.err);

Making jar file in folder other than the working directory

Hello!
I am trying to make jar files by using a program. I execute jar commands at runtime. The problem is that when a command is given like this
jar cvmf C:\HTTPS\Manifest.txt C:\HTTPS\khalid.jar C:\HTTPS\client.cer
C:\HTTPS\Client.jar C:\HTTPS\client.keys C:\HTTPS\HelloRunnableWorld.class C:\HT
TPS\HelloRunnableWorld.java C:\HTTPS\keytool interaction commands.txt C:\HTTPS\M
akeJarRunnable.class C:\HTTPS\MakeJarRunnable.java C:\HTTPS\MakeJarRunnable.zip
C:\HTTPS\Manifest.txt C:\HTTPS\myjar.jar C:\HTTPS\myjar_r.jar C:\HTTPS\Problem s
cenario.txt C:\HTTPS\Run.bat C:\HTTPS\scompressed.jar C:\HTTPS\Security.jar C:\H
TTPS\Server.jar C:\HTTPS\server.keys C:\HTTPS\serverexport.cer C:\HTTPS\Starter.
class C:\HTTPS\Starter.java C:\HTTPS\Umer Farooq - What I did.ppt C:\HTTPS\Umer.jar
The jar file is created but it has 0 bytes in it. No data is written into it, despite all the files being there.
Thank You

That is not the way to create jar files. Here is how you do it.
Under say c:\ create directory "utils".
Copy all of the *.java" files which will be part of this jar file "utils.jar". After moving files into c:\utils , make sure every single file has the first line of code as "package utils;" (without the quotes though)
Type in "javac *.java" to compile all of the files at once.
cd over to the parent directory and then type in
jar cvf utils.jar c:\utils\*.class
This will create utils.jar file under c:\

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case 2: if I include the "dependence and resources" when create jar file, then the program won't run. It gives me a message:
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Hello, I have the same problem, did you solved it?
Thanks

Making jar file for the project

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As you are new to these forums, possibly you do not know:
DO NOT duplicated your posts:
http://forum.java.sun.com/thread.jspa?threadID=5262912&messageID=10094981#10094981

How to define new classpath of libraries while making jar files with ant

I am useing eclipse and ant and trying to make a jar file. I have used some external jar files. I have managed to add external jar files to my jar. But Still I have a problem. In my project all libraries in lib folder. But when I put them into jar file. They are in the root folder.Classpath has changed and It couldn't find the class path.
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My code is like that, I think it requires a little modification
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I can see why your other "question" was likely deleted. You need to drop the editorial comments and just ask a question. I believe what you want to know is how to configure something using php on your Apache server so that it can be read. If that's your question, here's a couple of places that discuss the topic:
http://www.webmasterworld.com/linux/3345459.htm
http://forums.macrumors.com/showthread.php?t=427742
http://en.allexperts.com/q/PHP5-3508/configuring-installing-permissions.htm
For a general introduction to 'nix permissions, take a look at this:
http://www.osxfaq.com/Tutorials/LearningCenter/UnixTutorials/ManagingPermissions /index.ws
And here's a whole book on the subject of Leopard permissions:
http://my.safaribooksonline.com/9780321579331
Try doing a Google search on "leopard permissions php apache" and see if you find what you are looking for.
Francine
Francine
Schwieder

Errors in making jar file

Hi all,
The following is the simple bean I try to make
import java.awt.*;
import java.io.Serializable;
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//getter method
public Color getColor() {
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at java.util.jar.Attributes.read(Attributes.java:353)
at java.util.jar.Manifest.read(Manifest.java:159)
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at sun.tools.jar.Main.run(Main.java:124)
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check this http://java.sun.com/docs/books/tutorial/jar/basics/manifest.html

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Hi,
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Thank you

* open my computer
* select view | folder options
* select file types
* locate entry "Executable Jar File", and select edit
* select actions: open, then press edit
* browse for javaw.exe
* After you have browsed for javaw.exe, append to the end of the path the following: -jar "%1" so it looks something like this:
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* click OK, and your done.
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Thank you
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I am a sysadmin with a few Windows 7 Professional machines in our network.
The current setup for our users is folder-redirection, with offline files enabled for users with laptops (some of which are said Win7 users).
This works fine, just, any shortcut files that are in the users profile (that are made available offline) make the target of the shortcut available offline aswell. Though I see the benefit of this, it is not a behaviour we want.
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Thanks Dan

Hi,
“For example if user 'dave' has a shortcut on
\\server1\users\dave\desktop, that points to
\\server2\files\shortcut_target.doc (i.e. a
directory we don't want to have offline file behaviour with) - it will make both files available offline.”
Based on my knowledge, Windows only synchronous the shortcut in this offline folder. What do you mean “it will make both files available offline.”
For further research, I suggest you may enable sync log and try to repro this problem. Then move to Event Viewer -> Applications and Service Logs ->
Microsoft -> Windows -> OfflineFiles, right click SyncLog and chose “Save all event as…” to export these event log. Here is my e-mail:
[email protected]. You could contact me, then
I would assist you for this problem.
Please refer following to enable SyncLog:
How to enable the "SyncLog Channel"
===========================================
Offline Files defines four event channels through the Windows Event Viewer.
The channels are listed under the following hierarchy in the left-most pane of the viewer,
Event Viewer -> Applications and Service Logs -> Microsoft -> Windows -> OfflineFiles
If you see only one channel in the event viewer it will be the "Operational" channel. The remaining channels are known as Analytic or Debug
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SyncLog Channel
================
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option.
This may also be configured using the channel’s Properties page, also accessible through the Event Viewer. When you no longer want
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This may be helpful when questions arise about why a particular item is or is not being synchronized or questions about why a particular
sync operation is failing.
Please remember to click “Mark as Answer” on the post that helps you, and to click “Unmark as Answer” if a marked post does not actually answer your question. This can be beneficial to other community members reading the thread. ”

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http://www.javaworld.com/javaworld/jw-12-2000/jw-1229-traps.html

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Hi All,
I have a jar file which have classes and a package in it.
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thanks a lot for your reply friend.
I tried the same you tell me to do by using the cmd;
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Thanks

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hi,
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Created-By: 1.5.0 (Sun Microsystems Inc.)
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Making jar files available to javac?

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