Hi, Wednesday 19 May 2010

Library is perfect , but I have a small problem. Can someone give me an advice of how to get href = ...... - path to xsl file from the following xml-file :

<?xml version="1.0" encoding="utf-8" ?> <?xml-stylesheet href="../xsl/index.xsl" type="text/xsl" ?> <root> </root>

Try this one: ------------------ tree = etree.XML("""<?xml version="1.0" encoding="utf-8" ?> <?xml-stylesheet href="../xsl/index.xsl" type="text/xsl" ?> <root> </root> """) root = tree.getroottree() stylesheet = root.xpath("/processing-instruction('xml-stylesheet')")[0] print stylesheet.text ------------------ Probably there are other methods. :) Note, a processing instruction just knows a target (here: xml-stylesheet) and its content (everything in between "<?xml-stylesheet" and "?>"). A processing instruction has no concept of attributes. As such, you can get only text which you have to parse by yourself. This can be done by a regular expression. For example: ------------------ import re m=re.search('href\s*="([a-zA-Z0-9./\-\\+%]+)"', stylesheet.text) print m.groups() ('../xsl/index.xsl',) ------------------ Depending on what you expect in the href "pseudo-attribute", you have to improve the pattern in the above regular expression. Hope that helps, Tom