Article View: borland.public.jbuilder.java.api
Article #1365How to zip up directory?
From: Alex Colic
Date: Thu, 31 Jan 2002 16:39
Date: Thu, 31 Jan 2002 16:39
73 lines
2646 bytes
2646 bytes
Hi,
I am trying to zip up a directory. So far I have the below code but I get an
error when trying to open the input stream. Could someone suggest how I can
get this to work please. CRLCA is the directory.
Thanks
Alex
private boolean zipUpFile(File backupDir)
{
boolean zipped=false;
ZipOutputStream targetStream = null;
BufferedInputStream sourceStream=null;
try
{
targetStream = new ZipOutputStream(new
FileOutputStream(backupdirectory + seperator + "CRLCA.zip" ));
String filename = backupdirectory + seperator + "CRLCA";
File f = new File(filename);
FileInputStream fin=new FileInputStream(f);
ZipEntry theEntry = new ZipEntry(filename);
if (f.isDirectory())
{
targetStream.setComment("Backup of Database on: " + new
java.util.Date(System.currentTimeMillis()));
targetStream.setMethod(ZipOutputStream.DEFLATED);
targetStream.setLevel(9);
targetStream.putNextEntry(theEntry);
StreamCopier(fin, targetStream);
targetStream.closeEntry();
fin.close();
}
}
catch (Exception ex)
{
System.out.println(ex);
}
finally
{
try
{
if (targetStream != null) targetStream.close();
}
catch (Exception ex) {}
}
return zipped;
}
private static void StreamCopier(InputStream in, OutputStream out) throws
IOException
{
synchronized(in)
{
synchronized(out)
{
byte [] buffer=new byte[1024];
while(true)
{
int bytesRead=in.read(buffer);
if(bytesRead==-1)break;
out.write(buffer,0,bytesRead);
}
}
}
}
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