Java SAX parser. How to get the raw XML code of the currently parsingevent

[PatlaDJ]

Java SAX parser, please need a clue how to get the raw XML code of the
currently parsing event... needed for logging, debugging purposes.

Here's and example, letting me clarify exactly what i need: (see the
comments in source)

public void startElement(St ring uri, String localName, String qName,
Attributes attributes) throws SAXException {
//..Here... or maybe somewhere elsewhere I need on my disposal the raw
XML code of every XML tags received from the XML stream.
//.. I need simply to write them down in a log file, for debugging
purposes, while parsing.
//.. Can anyone give me a suggestion how can i do that logging while
the SAX parser only returns me the tagname and attributes. While
parsing I want to log the XML code for every tag in its 'pure form',
like it is comming from the server directly on the socket's input
reader.

if ("p".equals(qNa me)) {
//Do
something...
}

}

[Martin Honnen]

Re: Java SAX parser. How to get the raw XML code of the currentlyparsin g event

PatlaDJ wrote:

Java SAX parser, please need a clue how to get the raw XML code of the
currently parsing event... needed for logging, debugging purposes.
>
Here's and example, letting me clarify exactly what i need: (see the
comments in source)
>
public void startElement(St ring uri, String localName, String qName,
Attributes attributes) throws SAXException {
//..Here... or maybe somewhere elsewhere I need on my disposal the raw
XML code of every XML tags received from the XML stream.

I don't think the SAX API provides access to the raw XML.

--

Martin Honnen

http://JavaScript.FAQTs.com/

[PatlaDJ]

Re: Java SAX parser. How to get the raw XML code of the currentlyparsin g event

On 3 Þëè, 15:30, Martin Honnen <mahotr...@yaho o.dewrote:

PatlaDJ wrote:

Java SAX parser, please need a clue how to get the raw XML code of the
currently parsing event... needed for logging, debugging purposes.

>

Here's and example, letting me clarify exactly what i need: (see the
comments in source)

>

public void startElement(St ring uri, String localName, String qName,
Attributes attributes) throws SAXException {
//..Here... or maybe somewhere elsewhere I need on my disposal the raw
XML code of every XML tags received from the XML stream.

>
I don't think the SAX API provides access to the raw XML.
>
--
>
        Martin Honnen
       http://JavaScript.FAQTs.com/

Yes... now i'm pretty sure that it doesn't provide such access. Too
bad l;(

Right now i'm working towards clonning my input stream. Currently i'm
trying to clone the bufferedReader i pass as an inputsourse to the
parser ....i can't get it done :(

Let me explain: I read from a Socket ... the chain sequence is as
follows:

Socket -InputStreamRead er -BufferedReader -InputSource -then
it goes to .parse(InputSou rce)....of the SAX.

What node on this chain can be cloned so I can read the data two
times.....once for the parser, once for my debugging log.

Am I on the right track, or maybe not ?

[Joseph J. Kesselman]

Re: Java SAX parser. How to get the raw XML code of the currentlyparsin g event

Socket -InputStreamRead er -BufferedReader -InputSource -then

it goes to .parse(InputSou rce)....of the SAX.
>
What node on this chain can be cloned so I can read the data two
times.....once for the parser, once for my debugging log.

I'd suggest dealing with it at the Java level. Find or write a "wrapper"
reader implementation which saves a copy of the data passing through it
off to a data structure or scratch file for rereading, or perhaps which
writes the data direct to your log as it passes through. Plug that in
somewhere between the InputStreamRead er and the InputSource -- whether
upstream, downstream, or in place of the BufferedReader depends on the
details of how this tee adapter is written.

[PatlaDJ]

Re: Java SAX parser. How to get the raw XML code of the currentlyparsin g event

On 3 àÌÉ, 17:34, "Joseph J. Kesselman" <keshlam-nos...@comcast. net>
wrote:

Socket -InputStreamRead er -BufferedReader -InputSource -then
it goes to .parse(InputSou rce)....of the SAX.

>

What node on this chain can be cloned so I can read the data two
times.....once for the parser, once for my debugging log.

>
I'd suggest dealing with it at the Java level. Find or write a "wrapper"
reader implementation which saves a copy of the data passing through it
off to a data structure or scratch file for rereading, or perhaps which
writes the data direct to your log as it passes through. Plug that in
somewhere between the InputStreamRead er and the InputSource -- whether
upstream, downstream, or in place of the BufferedReader depends on the
details of how this tee adapter is written.

YES!

I've solved my problem using class RecordingInputS tream that wraps the
InputStream
here is the class source code:

import java.io.ByteArr ayOutputStream;
import java.io.FilterI nputStream;
import java.io.InputSt ream;
import java.io.IOExcep tion;

/**
*
* @author Unknown
*
*/

class RecordingInputS tream extends FilterInputStre am {
protected ByteArrayOutput Stream sink;

RecordingInputS tream(InputStre am in) {
this(in, new ByteArrayOutput Stream());
}

RecordingInputS tream(InputStre am in, ByteArrayOutput Stream sink)
{
super(in);
this.sink = sink;
}

public synchronized int read() throws IOException {
int i = in.read();
sink.write(i);
return i;
}

public synchronized int read(byte[] buf, int off, int len) throws
IOException {
int l = in.read(buf, off, len);
sink.write(buf, off, l);
return l;
}

public synchronized int read(byte[] buf) throws IOException {
return read(buf, 0, buf.length);
}

public synchronized long skip(long len) throws IOException {
long l = 0;
int i = 0;
byte[] buf = new byte[1024];
while (l < len) {
i = read(buf, 0, (int)Math.min(( long)buf.length , len -
l));
if (i == -1) break;
l += i;
}
return l;
}

byte[] getBytes() {
return sink.toByteArra y();
}

void resetSink() {
sink.reset();
}
}

[jimmy Zhang]

Re: Java SAX parser. How to get the raw XML code of the currently parsing event

SAX is a bit outdated... try either Pull or VTD-XML (http://vtd-xml.sf.net)

"PatlaDJ" <patladj@gmail. comwrote in message
news:2aaf86a4-5a78-4334-a607-8cdc1a4d9082@a7 0g2000hsh.googl egroups.com...

Java SAX parser, please need a clue how to get the raw XML code of the
currently parsing event... needed for logging, debugging purposes.
>
Here's and example, letting me clarify exactly what i need: (see the
comments in source)
>
public void startElement(St ring uri, String localName, String qName,
Attributes attributes) throws SAXException {
//..Here... or maybe somewhere elsewhere I need on my disposal the raw
XML code of every XML tags received from the XML stream.
//.. I need simply to write them down in a log file, for debugging
purposes, while parsing.
//.. Can anyone give me a suggestion how can i do that logging while
the SAX parser only returns me the tagname and attributes. While
parsing I want to log the XML code for every tag in its 'pure form',
like it is comming from the server directly on the socket's input
reader.
>
if ("p".equals(qNa me)) {
//Do
something...
}
>
}