Byte position of single character

Hi,
I am dealing with very big XML files and since I don't need all the information in them, I would like to have random access to the XML elements, to pick the ones I am interested in (without loading/searching the whole document every time).
So, I wanted to create a index of the XML file with xpath expressions (from the root to the element of interest) as keys and byte positions as values. I would need to parse the XML only once and record the byte position at the beginning of the start tag and at the end of the end tag. Thus if I read from the start position to the corresponding end position in the file, I would end up with proper xml that is only the part of the file that I am interested in.
E.g.: Once I have the index I could use RandomAccessFile to retrieve a byte array from start to stop position and convert it to a String using the correct encoding.
BUT: I haven't found a way to get to the needed byte position.
I tried using XMLStreamReader, since it has a method getLocation, but I have problems with that too.
( see: http://forum.java.sun.com/thread.jspa?threadID=5198055 )
And if I use a reader like InputStreamReader, I would be able to read character by character and parse the file myself, but I don't get the byte position (even using CountingInputStream, since the reader buffers the input stream).
Any ideas how to tackle this problem, hints or alternatives are highly appreciated!
Thanks!

Do you mean "how many characters before the last newline", or "how many bytes before the last non-\n (possibly also non-\r) byte", or "what is the byte position of the first byte of the last character", or something else? Those can be different things if you have a multibyte encoding.
For last byte (or ISO-8859 character): get length of file (see File javadoc), seek and read using RandomAccessFile until you find something you don't consider newline (you need to think about carriage return vs linefeed.)

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Message was edited by: chris227 regexp added

Byte position of single character

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