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[XSL-LIST Mailing List Archive Home] [By Thread] [By Date] [Recent Entries] [Reply To This Message] RE: element to root
 I remember your similar post from a few days ago and had kept it aside to look at in time; today's repost prompted me to do so. Well, the bad news is that while I think there's an elegant, recursive, etc XSL solution, I've seen it only hazily in the past half-hour. The good news is that this is readily amenable to a procedural solution, if you'll accept one, as implemented in the following jython script. The idea is that we write out each line, maintain a "stack" of open-tags we find, and write closing and re-opening tags at each <br/>. The script below doesn't use an XML parser, but could readily be fitted to use a SAX parser. In which case you wouldn't need the additional simplyfying assumption that element start-tags are on lines by themselves. Given your input below: it produces: <p>
<span style="style a">
span a text 1
<span style="style b">
span b pre br text
</span>
</span><br name="b"/> <span style="style a">
<span style="style b">
span b post br text
</span>
span a text 2
</span>
</p>which is what you asked for: And given your input from your original email (reformatted for start-tags are on lines by themselves): it produces this: <p> <strong> strong:text(top) </strong> <br/> <strong>
prefix
<span style="a style">
span a
<span style="rgb();">
span b
</span>
</span>
</strong><br/> <strong>
<span style="a style">
<span style="rgb();">
text
</span>
text
</span>
strong:text(btm)
</strong><br/> <strong> suffix </strong> <br/> <strong> </strong> Root level text with <br/> tag. </p> Here's the script (which is more illustrative than production-worthy): import re stack, b, linecount = [], 0, 0 f = open(r'e:\temp\test.xml') try:
while 1:
line = f.readline( ).rstrip()
if not line: breaklinecount += 1 if linecount == 1:
pass
elif re.match("\s*<br.+>$", line):
b = 1
elif re.match("\s*<?\w+[^>]*>$", line): # start-tag
stack.append( line )
elif re.match("\s*</\w+[^>]*>$", line): # end-tag
stack = stack[:len(stack)-1] if b:
# write closing tags
stack.reverse()
for t in stack:
m = re.match("(?P<TAB>\s*)<(?P<NAME>\w+)", t)
print "%s</%s>" % ( m.group('TAB'), m.group('NAME') )print line if b:
# write opening tags again
stack.reverse()
for t in stack:
print t
b = 0finally: f.close( ) Regards, --A _________________________________________________________________ Dont just search. Find. Check out the new MSN Search! http://search.msn.click-url.com/go/onm00200636ave/direct/01/ 
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