我有深层嵌套的xml标签Movie,我想直接使用xpath
访问。
<?xml version='1.0' encoding='utf8'?>
<collection>
<genre category="Action">
<decade years="1980s">
<movie favorite="True" title="Indiana Jones: The raiders of the lost Ark">
<format multiple="No">DVD</format>
<year>1981</year>
<rating>PG</rating>
<description>
'Archaeologist and adventurer Indiana Jones
is hired by the U.S. government to find the Ark of the
Covenant before the Nazis.'
</description>
</movie>
</decade>
</genre>
</collection>
</xml>
如何访问
movie
标记及其属性?我尝试使用
root = etee.tostring(above_xml)
print root.xpath("movie")
[]
但我什么也没得到。
最佳答案
我不确定这是您要寻找的内容,但是请尝试以下操作:
tree = lxml.etree.fromstring(xml_above, parser=lxml.etree.HTMLParser())
film = tree.xpath("//movie/*/text()")
for i in film:
print(i)
输出:
DVD
1981
PG
'Archaeologist and adventurer Indiana Jones
is hired by the U.S. government to find the Ark of the
Covenant before the Nazis.'
关于python - 如何使用xml中的xpath访问其他标签内部的标签?,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/55901637/