Python：我不明白XML迭代是如何工作的

Question

感谢brilliant help on my XML parsing problem我得到了一个让我迷失在实际处理XML元素（使用lxml）的问题。

我的数据是NMAP扫描输出，由许多的记录，像下面的：

<?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> 

我期待在产生行时

• 12345端口是开放的或
• 端口135是开放的，12345是开放

我用这个下面的代码，我与我的事情如何去理解说：

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") 

的地方我不知道在注释中高亮显示。使用此代码，我始终匹配Microsoft Windows RPC，就好像我处于端口135的记录内（它首先在端口12345之前的XML文件中）。

我相信这个问题在我了解find函数的某个地方。它可能匹配所有东西，与我所处的地点无关。换句话说，没有递归（据我所知）。

在这种情况下，我该如何编码“当您在端口12345的记录中时获取服务名称”的概念？

谢谢。

---

编辑& SOLUTION：

我发现在我的代码的问题。我转贴整个脚本，如果有人在这个问题一天绊倒（输出来自NMAP所以它可能是有趣的人重用 - 这一点，解释的代码如下:)的大块：

#!/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() 

我会还探讨了Dikei和Ignacio Vazquez-Abrams提供的xpath想法。

谢谢大家！

Answer 1

这应该是很容易使用XPath

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