Java IO.pdf - Nguyen Dang Binh

}
}
}
catch (IOException e) {System.err.println(e);}
Java I/O
Here are the first few lines that result from running this program on the classes.zip file from
JDK 1.1.4:
% java ZipLister /usr/local/java/lib/classes.zip
sun/net/www/protocol/systemresource/ParseSystemURL.class
java/io/ObjectInputValidation.class
sun/awt/motif/MTextFieldPeer.class
sun/tools/javac/BatchParser.class
sun/rmi/transport/proxy/HttpOutputStream.class
To get a single entry in the zip file rather than a list of the entire contents, you pass the name
of the entry to the getEntry() method:
public ZipEntry getEntry(String name)
Of course, this requires you to know the name of the entry in advance. The name is simply the
path and filename, like java/io/ObjectInputValidation.class. For example, to retrieve the zip
entry for java/io/ObjectInputValidation.class from the ZipFile zf, you might write:
ZipEntry ze = zf.getEntry("java/io/ObjectInputValidation.class");
You can also get the name with the getName() method of the ZipEntry class, discussed later
in this chapter. This method, however, requires you to have a ZipEntry object already, so
there's a little chicken-and-egg problem here.
Most of the time, you'll want more than the names of the files in the archive. You can get the
actual contents of the zip entry using getInputStream():
public InputStream getInputStream(ZipEntry ze) throws IOException
This returns an input stream from which you can read the uncompressed contents of the zip
entry (file). Example 9.8 is a simple unzip program that uses this input stream to unpack zip
archives named on the command line.
Example 9.8. Unzipper
import java.util.*;
import java.util.zip.*;
import java.io.*;
import com.macfaq.io.*;
public class Unzipper {
public static void main(String[] args) {
for (int i = 0; i < args.length; i++) {
try {
ZipFile zf = new ZipFile(args[i]);
Enumeration e = zf.entries();
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