如何基于子元素中的文本删除XML文件中的父节点？使用Python 3.6

Question

现在我正试图在Python3.6中使用lxml。我想删除包含对冲的“程序”，如果没有一个程序包含“keep”，则完全删除“Request”。 xml的结构如下所示：

<Requests>
   <Request>
        <ProgramSelection>
            <Program> <![CDATA[hedge]]> </Program>
            <Program> <![CDATA[keep]] </Program>
        </ProgramSelection>
    </Request>
</Requests>

import lxml.etree


file_name = r'C:filename.xml'
parser = lxml.etree.XMLParser(strip_cdata=False)
tree = lxml.etree.parse(file_name, parser)
root = tree.getroot()

for elem in tree.xpath("./Request[ProgramSelection/Program='hedge']"):
    root.remove(elem)

Answer 1

你很接近。以下两个XPath选择与删除条件匹配的元素

import lxml.etree

file_name = r'test.xml'
parser = lxml.etree.XMLParser(strip_cdata=False)
tree = lxml.etree.parse(file_name, parser)
root = tree.getroot()

# remove <Request> lacking a <Program>keep</Program>
for request in tree.xpath(
        "Request[not(ProgramSelection/Program[contains(text(),keep)])]"):
    request.getparent().remove(request)

# remove <Program>hedge</Program>
for program in tree.xpath(
        "Request/ProgramSelection/Program[contains(text(), hedge)]"):
    program.getparent().remove(program)

print(lxml.etree.tostring(tree, pretty_print=True).decode())

您可以将它们组合成可读性稍差的“或”

import lxml.etree

file_name = r'test.xml'
parser = lxml.etree.XMLParser(strip_cdata=False)
tree = lxml.etree.parse(file_name, parser)
root = tree.getroot()

# remove <Request> lacking a <Program>keep</Program>
# remove <Program>hedge</Program>
for elem in tree.xpath("Request[
        not(ProgramSelection/Program[contains(text(),keep)])]"
        "|"        
        "Request/ProgramSelection/Program[contains(text(), hedge)]"):
    elem.getparent().remove(elem)

print(lxml.etree.tostring(tree, pretty_print=True).decode())