I have a problem in code:
private static String compress(String str)
{
String str1 = null;
ByteArrayOutputStream bos = null;
try
{
bos = new ByteArrayOutputStream();
BufferedOutputStream dest = null;
byte b[] = str.getBytes();
GZIPOutputStream gz = new GZIPOutputStream(bos,b.length);
gz.write(b,0,b.length);
bos.close();
gz.close();
}
catch(Exception e) {
System.out.println(e);
e.printStackTrace();
}
byte b1[] = bos.toByteArray();
return new String(b1);
}
private static String deCompress(String str)
{
String s1 = null;
try
{
byte b[] = str.getBytes();
InputStream bais = new ByteArrayInputStream(b);
GZIPInputStream gs = new GZIPInputStream(bais);
ByteArrayOutputStream baos = new ByteArrayOutputStream();
int numBytesRead = 0;
byte [] tempBytes = new byte[6000];
try
{
while ((numBytesRead = gs.read(tempBytes, 0, tempBytes.length)) != -1)
{
baos.write(tempBytes, 0, numBytesRead);
}
s1 = new String(baos.toByteArray());
s1= baos.toString();
}
catch(ZipException e)
{
e.printStackTrace();
}
}
catch(Exception e) {
e.printStackTrace();
}
return s1;
}
public String test() throws Exception
{
String str = "teststring";
String cmpr = compress(str);
String dcmpr = deCompress(cmpr);
}
This code throw java.io.IOException: unknown format (magic number ef1f)
GZIPInputStream gs = new GZIPInputStream(bais);
It turns out that when converting byte new String (b1)
and the byte b [] = str.getBytes ()
bytes are "spoiled." At the output of the line we have already more bytes. If you avoid the conversion to a string and work on the line with bytes - everything works. Sorry for my English.
public String unZip(String zipped) throws DataFormatException, IOException {
byte[] bytes = zipped.getBytes("WINDOWS-1251");
Inflater decompressed = new Inflater();
decompressed.setInput(bytes);
byte[] result = new byte[100];
ByteArrayOutputStream buffer = new ByteArrayOutputStream();
while (decompressed.inflate(result) != 0)
buffer.write(result);
decompressed.end();
return new String(buffer.toByteArray(), charset);
}
I'm use this function to decompress server responce. Thanks for help.