感谢brilliant help on my XML parsing problem我在实际处理XML元素的过程中遇到了麻烦（使用lxml）。
我的数据是nmap扫描的输出，由许多记录组成，如下所示：
<?xml version="1.0"?>
<?xml-stylesheet href="file:///usr/share/nmap/nmap.xsl" type="text/xsl"?>
<nmaprun scanner="nmap" args="nmap -sV -p135,12345 -oX 10.232.0.0.16.xml 10.232.0.0/16" start="1340201347" startstr="Wed Jun 20 16:09:07 2012" version="5.21" xmloutputversion="1.03">
<host>
<status state="down" reason="no-response"/>
<address addr="10.232.0.1" addrtype="ipv4"/>
</host>
<host starttime="1340201455" endtime="1340201930">
<status state="up" reason="echo-reply"/>
<address addr="10.232.49.2" addrtype="ipv4"/>
<hostnames>
<hostname name="host1.example.com" type="PTR"/>
</hostnames>
<ports>
<port protocol="tcp" portid="135">
<state state="open" reason="syn-ack" reason_ttl="123"/>
<service name="msrpc" product="Microsoft Windows RPC" ostype="Windows" method="probed" conf="10"/>
</port>
<port protocol="tcp" portid="12345">
<state state="open" reason="syn-ack" reason_ttl="123"/>
<service name="http" product="Trend Micro OfficeScan Antivirus http config" method="probed" conf="10"/>
</port>
</ports>
<times srtt="890" rttvar="2835" to="100000"/>
</host>
</nmaprun>
我正在考虑在
时生成一条线我使用以下代码，我对我的理解进行了评论：
from lxml import etree
import time
scanTime = str(int(time.time()))
d = etree.parse("10.233.85.0.22.xml")
# find all hosts records
for el_host in d.findall("host"):
# only process hosts UP
if el_host.find("status").attrib["state"] =="up":
# here comes a piece of code which sets the variable hostname
# used later - that part works fine (removed for clarity)
# get the status of port 135 and 12345
Open12345 = Open135 = False
for el_port in el_host.findall("ports/port"):
# we are now looping thought the <port> records for a given <host>
if el_port.attrib["portid"] == "135":
Open135 = el_host.find("ports/port/state").attrib["state"] == "open"
if el_port.attrib["portid"] == "12345":
Open12345 = el_host.find("ports/port/state").attrib["state"] == "open"
# I want to get for port 12345 the description, so I search
# for <service> within a given port - only 12345 in my case
# I just search the first one as there is only one
# this is the place I am not sure I get right
el_service = el_host.find("ports/port/service")
if el_service.get("product") is not None:
Type12345 = el_host.find("ports/port/service").attrib["product"]
if Open12345:
print "%s %s \"%s\"\n" % (scanTime,hostname,Type12345)
if not Open12345 and Open135:
print "%s %s \"%s\"\n" % (scanTime,hostname,"NO_OfficeScan")
评论中突出显示了我不确定的地方。使用此代码，我总是匹配 Microsoft Windows RPC ，就像我在端口135的记录中一样（它在XML文件中首先出现在端口12345之前）。
我确信这个问题在于我理解find函数的方式。它可能与所有东西相匹配，与我所处的位置无关。换句话说，没有递归（据我所知）。
在这种情况下，如何编写“当您在12345端口的记录中时获取服务名称”的概念？
谢谢。
<小时/> 编辑＆amp;解决方案：
我在代码中发现了问题。如果总有人偶然发现这个问题，我会重新发布整个脚本（输出来自nmap，因此有人可以重用它 - 这可以解释下面的大块代码:)：
#!/usr/bin/python
from lxml import etree
import time
import argparse
parser = argparse.ArgumentParser()
parser.add_argument("file", help="XML file to parse")
args = parser.parse_args()
scanTime = str(int(time.time()))
d = etree.parse(args.file)
f = open("OfficeScanComplianceDSCampus."+scanTime,"w")
print "Parsing "+ args.file
# find all hosts records
for el_host in d.findall("host"):
# only process hosts UP
if el_host.find("status").attrib["state"] =="up":
# get the first hostname if it exists, otherwise IP
el_hostname = el_host.find("hostnames/hostname")
if el_hostname is not None:
hostname = el_hostname.attrib["name"]
else:
hostname = el_host.find("address").attrib["addr"]
# get the status of port 135 and 12345
Open12345 = Open135 = False
for el_port in el_host.findall("ports/port"):
# we are now looping thought the <port> records for a given <host>
if el_port.attrib["portid"] == "135":
Open135 = el_port.find("state").attrib["state"] == "open"
if el_port.attrib["portid"] == "12345":
Open12345 = el_port.find("state").attrib["state"] == "open"
# if port open get info about service
if Open12345:
el_service = el_port.find("service")
if el_service is None:
Type12345 = "UNKNOWN"
elif el_service.get("method") == "probed":
Type12345 = el_service.get("product")
else:
Type12345 = "UNKNOWN"
if Open12345:
f.write("%s %s \"%s\"\n" % (scanTime,hostname,Type12345))
if not Open12345 and Open135:
f.write("%s %s \"%s\"\n" % (scanTime,hostname,"NO_OfficeScan"))
if Open12345 and not Open135:
f.write("%s %s \"%s\"\n" % (scanTime,hostname,"Non-Windows with 12345"))
f.close()
我还将探讨Dikei和Ignacio Vazquez-Abrams给出的xpath想法。
谢谢大家！
答案 0 :(得分：2)
使用xpath
这应该很容易from lxml import etree
d = etree.parse("10.233.85.0.22.xml")
d.xpath('//port[@portid="12345"]/service/@name') // return name of service in portid=12345
d.xpath('//port[@portid="12345"]/service/@product') // return product in port with portid=12345