[XML-SIG] Re: Re: Could somebody help me?

Fredrik Lundh fredrik at pythonware.com
Fri Jan 28 11:59:21 CET 2005

"Prasad PS" wrote:

> In the code below, what I am doing is - I am opening an xml file and
> appending a node to the root document. Then I add this root document to
> the xml file
> fp = open (string.strip(self.cnfDtls.GetLogFilePath()), 'w')
> xml.dom.ext.PrettyPrint(doc, self.xmlFile)
> self.xmlFile.write("\n")
> fp.close().
> So for each appending of node, the file is opened and the new root
> document with the appeneded node is re-written to the file. This is
> taking lot of time.

> Is there any way to append a node to the xml file without re-writing the
> file?

do you have to reload the file every time?  (do you even have to save it
every time?)

a little profiling might also help; what part of the update process is taking
most of the time (loading, updating, or saving).

I would do

          load file
          while processing
                add node
                if enough time has elapsed (say, a second or two):
                    save file

if that's not efficient enough, you might have use a custom XML writer
(you cannot just append to an XML file, since the end tag must be last
in the file.  you could open the file, read backwards until you find the
start of the end tag, overwrite another node, and put the end tag back
again.  or you could use a custom writer that keeps track of where the
end tag is, and overwrites it every time a new node is added)


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