XPath: if child exists - get value, if not - write empty line - xpath

Question

I have three node types: without categories, with a single category and multiple categories:
<technology>
<categories>
<category></category>
</categories>
<name></name>
</technologie>
<technology>
<name></name>
</technologie>
<technology>
<categories>
<category></category>
<category></category>
<category></category>
</categories>
<name></name>
</technologie>
I want to write data from technology node like:
if there is a single category - write it,
if there is no category - write an empty line
if there are multiple categories - write the first one.
I tried things like //technology/*[descendant::category], or //technology/categories/category[1] - but i fail on getting the empty line, if category not exists, all existing categories are written one by one.
How do i get data written on the descripted way?

Answer 1

I don't know if this is the direction you want to take, but FWIW, you can handle it with lxml. If your code is:
snip = """
<technology>
<categories>
<category>Cat-A</category>
</categories>
<name>Nam-1</name>
</technology>
<technology>
<name>Nam-2</name>
</technology>
<technology>
<categories>
<category>Cat-B</category>
<category>Cat-C</category>
<category>Cat-D</category>
</categories>
<name></name>
</technology>
"""
You can use:
from lxml import etree
tree = lxml.etree.fromstring(snip, parser=lxml.etree.HTMLParser())
results = tree.xpath("*//technology")
for result in results:
for j in result.getchildren():
if j.tag == 'categories':
for m in j.itertext():
if m.strip() != '':
print(m)
break
break
else:
print('none')
break
Output:
Cat-A
none
Cat-B