Am Montag, 26. September 2005 15:57 schrieb Martin Koekenberg:

Hello,

I've a xml file on the file system (the source is on an other webserver and the download is scheduled). How ca I parse this file with a xslt in Zope. ZopeXMLmethods isn't working annymore in Zope 2.8.1.

Does annyone knows a Product of method to parse xml in Zope without the zopeXMLmethods product ?

Regards,

Martin Koekenberg

I do it like this (ExternalMethod in this case): -------------------------------------------------------------------- import libxml2 import libxslt stylestring = file("/path/to/style/file.xsl").read() def xslt(data): # note: if styledoc and style are defined outside the function, # zope dumps core :-( styledoc = libxml2.parseDoc(stylestring) style = libxslt.parseStylesheetDoc(styledoc) doc = libxml2.parseDoc(data) result = style.applyStylesheet(doc, None) html = style.saveResultToString(result) style.freeStylesheet() doc.freeDoc() result.freeDoc() return html -------------------------------------------------------------------- may be not very smart, but it's working :-) Cheers, Sascha