Python Cookbook

Controlling XSLT Stylesheet Loading

Credit: Jürgen Hermann

Problem

You need to process XML documents and access external documents (e.g., stylesheets), but you can’t use filesystem paths (to keep documents portable) or Internet-accessible URLs (for performance and security).

Solution

4Suite’s xml.xslt package (http://www.4suite.org/) gives you all the power you need to handle XML stylesheets, including the hooks for sophisticated needs such as those met by this recipe:

# uses 4Suite Version 0.10.2 or later
from xml.xslt.Processor import Processor
from xml.xslt.StylesheetReader import StylesheetReader

class StylesheetFromDict(StylesheetReader):
    "A stylesheet reader that loads XSLT stylesheets from a python dictionary"

    def _ _init_ _(self, styles, *args):
        "Remember the dict we want to load the stylesheets from"
        StylesheetReader._ _init_ _(self, *args)
        self.styles = styles
        self._ _myargs = args

    def _ _getinitargs_ _(self):
        "Return init args for clone(  )"
        return (self.styles,) + self._ _myargs

    def fromUri(self, uri, baseUri='', ownerDoc=None, stripElements=None):
        "Load stylesheet from a dict"
        parts = uri.split(':', 1)
        if parts[0] == 'internal' and self.styles.has_key(parts[1]):
            # Load the stylesheet from the internal repository (your dictionary)
            return StylesheetReader.fromString(self, self.styles[parts[1]],
                baseUri, ownerDoc, stripElements) else: # Revert to normal behavior return StylesheetReader.fromUri(self, uri, baseUri, ownerDoc, stripElements) if _ _name_ _ == "_ _main_ _": # test ...