[xsl] [java] transforming a file to itself - buffered?

Subject: [xsl] [java] transforming a file to itself - buffered?
From: Robert Koberg <rob@xxxxxxxxxx>
Date: Tue, 04 Jun 2002 10:34:44 -0700
Hi,

What is the best way to handle transforming a file to itself?

Currently I do:
File source = new File(servlet_context.getRealPath(_xml));
transformer.transform(new StreamSource(source), new StreamResult(source));

Is there any way to do buffered reads and writes to this file? Is there a benefit to doing this?

Is there a way to use the new 1.4 non blocking I/O (java.nio)?
<random_thought> The code below should(?) work on the entire file, but could you possibly identify a section (nodeset) of an XML file and perform a transformation on it? Is this just silly?</random_thought>


I tried to do the following but I get an error in saxon 6.52 saying: (The requested operation cannot be performed on a file with a user-mapped section open)
--------
RandomAccessFile in = new RandomAccessFile(_context.getRealPath(_xml), "rw");
FileChannel channel = in.getChannel();
Charset set = Charset.forName("US-ASCII");
CharsetDecoder decoder = set.newDecoder();
MappedByteBuffer bytes = channel.map(FileChannel.MapMode.READ_WRITE, 0, (int)channel.size());
CharBuffer chars = decoder.decode(bytes);


StreamSource source = new StreamSource(new CharArrayReader(chars.array()));
StreamSource result_source = new StreamSource(new CharArrayReader(chars.array()),
servlet_context.getRealPath(_xml));
StreamResult result = new StreamResult(result_source);


transformer.transform(source, result);
-----------

thanks for any help,
-Rob


XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list



Current Thread