Name: jk109818			Date: 03/12/2002


FULL PRODUCT VERSION :
JDK1.1.8_008
JDK1.2.2_008

FULL OPERATING SYSTEM VERSION :Windows NT 4.0 SP5


A DESCRIPTION OF THE PROBLEM :
There is no compatibility in the Zip file created between
JDK1.1.8 and JDK 1.2.X, JDK 1.3.X.

Reappearance procedure
1: Compress the file which used Japanese for the file name
by java.util.zip.ZipFile of JDK1.1.8.
2:The Zip file created by 1.1.8 cannot be thawed by JDK 1.2.X.
("ZipFile.getInputStream (ZipEntry)" in the following source
code will return null.)

I know that the file name encoding method in a Zip file was
changed by JDK 1.2.X.
However, it is amusing that InputStream cannot be obtained
in spite of passing ZipEntry acquired from ZipFile to
ZipFile as it is.

It deals with the problem of the file name using umlaut as a bug.
(Bug Id :4244499,4415733)
Reference is made only till these problems not having the file name
of "ZipEntry" correctly.
Since my problem cannot even perform opening a file,
it is still more serious.


Source code (extract)

	ZipFile zf = null;
	try {
	  zf = new ZipFile(src);

	  Enumeration enum = zf.entries();
	  if (enum == null) return;
	  while(enum.hasMoreElements()) {
		ze = (ZipEntry)enum.nextElement();
		int off = 0;
		int len = 0;
		long size = ze.getSize();

		if(ze.isDirectory()) {
		  // Directory
		} else {
		  // file
		  InputStream is = zf.getInputStream(ze);
//This result is null on Java2
		  bis = new BufferedInputStream(is, RLEN);
		  fos = new FileOutputStream(file);
		  while((size-=len) > 0) {
			len = bis.read(b, off, RLEN);
			//My Processing
		  }
		  bis.close();
		  fos.close();
		}

	    } // end while

	    zf.close();

	} catch (ZipException e) {
	    // ZIP Formal error:ZipFile()
	} catch (IOException e) {
	    // I/O
error:ZipFile(),FileOutputStream(),read(),write().close()
	}



STEPS TO FOLLOW TO REPRODUCE THE PROBLEM :
1.Compress the file which used Japanese for the file name by
java.util.zip.ZipFile of JDK1.1.8.
2.UnCompress The Zip file(created by 1) on JDK 1.2.X.
Then "ZipFile.getInputStream (ZipEntry)" returns null.
3.

EXPECTED VERSUS ACTUAL BEHAVIOR :
"ZipFile.getInputStream (ZipEntry)" returns some
InputStream.

ERROR MESSAGES/STACK TRACES THAT OCCUR :
No error messages or Exception is occur.

This bug can be reproduced always.

---------- BEGIN SOURCE ----------
	ZipFile zf = null;
	try {
	  zf = new ZipFile(src);

	  Enumeration enum = zf.entries();
	  if (enum == null) return;
	  while(enum.hasMoreElements()) {
		ze = (ZipEntry)enum.nextElement();
		int off = 0;
		int len = 0;
		long size = ze.getSize();

		if(ze.isDirectory()) {
		  // Directory
		} else {
		  // file
		  InputStream is = zf.getInputStream(ze);  //This result is null
on Java2
		  bis = new BufferedInputStream(is, RLEN);
		  fos = new FileOutputStream(file);
		  while((size-=len) > 0) {
			len = bis.read(b, off, RLEN);
			//My Processing
		  }
		  bis.close();
		  fos.close();
		}

	    } // end while

	    zf.close();

	} catch (ZipException e) {
	    // ZIP Formal error:ZipFile()
	} catch (IOException e) {
	    // I/O error:ZipFile(),FileOutputStream(),read(),write().close()
	}


---------- END SOURCE ----------

CUSTOMER WORKAROUND :
Do not use Java.
(Review ID: 143349) 
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