我正在使用NVD XML并尝试解析和拆分XML以最终进入数据库。我遇到的问题是解析的XML attrib是＆＃34;或者＆＃39;围绕价值观。我无法拆分这些字符串。我已经包含了代码和当前失败的条目。预期的输出是
product,america's_first_federal_credit_union,america's_first_fcu_mobile_banking
代码
#!/usr/bin/env python
import os
import sys
import time
from subprocess import call
import xml.etree.ElementTree
import re
range_from = 2017
range_to = 2017
def process_entry(entry):
cve = entry.attrib.get("name")
print cve
cpes = get_cpes_affected(entry)
def get_cpes_affected(entry):
child = []
for e in entry.iter():
if "}prod" in e.tag:
print e.attrib
print unichr(34)
if unichr(34) in e.attrib:
print "hey yo"
child.append("product," + str(e.attrib).split('"')[1] + "," + str(e.attrib).split('"')[3])
else:
child.append("product," + str(e.attrib).split("'")[3] + "," + str(e.attrib).split("'")[7])
#print e.tag, e.attrib
if "'prev'" in e.attrib:
child.append("version," + str(e.attrib).split("'")[7] + "," + str(e.attrib).split("'")[3])
if "}vers" in e.tag and "'prev'" not in e.attrib:
child.append("version," + str(e.attrib).split("'")[3] + ",")
#print e.tag, e.attrib
for derp in child:
print derp
for i in range(range_from, range_to+1):
os.system("wget -O tmp.zip https://nvd.nist.gov/download/nvdcve-%i.xml.zip" % i)
os.system("unzip -o tmp.zip")
e = xml.etree.ElementTree.parse('nvdcve-%i.xml' % i).getroot()
for entry in e:
process_entry(entry)
正在解析的XML条目示例
<entry type="CVE" name="CVE-2017-5916" seq="2017-5916" published="2017-05-05" modified="2017-05-16" severity="Medium" CVSS_version="2.0" CVSS_score="4.3" CVSS_base_score="4.3" CVSS_impact_subscore="2.9" CVSS_exploit_subscore="8.6" CVSS_vector="(AV:N/AC:M/Au:N/C:P/I:N/A:N)">
<desc>
<descript source="cve">The America's First Federal Credit Union (FCU) Mobile Banking app 3.1.0 for iOS does not verify X.509 certificates from SSL servers, which allows man-in-the-middle attackers to spoof servers and obtain sensitive information via a crafted certificate.</descript>
</desc>
<loss_types>
<conf/>
</loss_types>
<range>
<network/>
</range>
<refs>
<ref source="MISC" url="https://medium.com/@chronic_9612/follow-up-76-popular-apps-confirmed-vulnerable-to-silent-interception-of-tls-protected-data-64185035029f" adv="1">https://medium.com/@chronic_9612/follow-up-76-popular-apps-confirmed-vulnerable-to-silent-interception-of-tls-protected-data-64185035029f</ref>
</refs>
<vuln_soft>
<prod name="america's_first_fcu_mobile_banking" vendor="america's_first_federal_credit_union">
<vers num="3.1.0" prev="1" edition=":~~~iphone_os~~"/>
</prod>
</vuln_soft>
输入
失败{'vendor': "america's_first_federal_credit_union", 'name': "america's_first_fcu_mobile_banking"}
只是要包含一个字符串的例子，它能够分开而没有问题
{'vendor': 'emirates_nbd_bank_p.j.s.c', 'name': 'emirates_nbd_ksa'}
抱歉忘了包含错误
Traceback (most recent call last):
File "prev-version-load.py", line 49, in <module>
process_entry(entry)
File "prev-version-load.py", line 18, in process_entry
cpes = get_cpes_affected(entry)
File "prev-version-load.py", line 33, in get_cpes_affected
child.append("product," + str(e.attrib).split("'")[3] + "," + str(e.attrib).split("'")[7])
IndexError: list index out of range
答案 0 :(得分：0)
考虑替换......
if "}prod" in e.tag:
print unichr(34)
if unichr(34) in e.attrib:
print "hey yo"
child.append("product," + str(e.attrib).split('"')[1] + "," + str(e.attrib).split('"')[3])
else:
child.append("product," + str(e.attrib).split("'")[3] + "," + str(e.attrib).split("'")[7])
#print e.tag, e.attrib
if "'prev'" in e.attrib:
child.append("version," + str(e.attrib).split("'")[7] + "," + str(e.attrib).split("'")[3])
if "}vers" in e.tag and "'prev'" not in e.attrib:
child.append("version," + str(e.attrib).split("'")[3] + ",")
使用...
reg=r"\"|'(?=[^\"]*')|'(?=\W*\")"
if "prod" in e.tag:
#print(re.split(reg,str(e.attrib)))
child.append("product," + re.split(reg,str(e.attrib))[3] + "," + re.split(reg,str(e.attrib))[7])
#print e.tag, e.attrib
if "prev" in e.attrib:
child.append("version," + re.split(reg,str(e.attrib))[7] + "," + re.split(reg,str(e.attrib))[3])
if "vers" in e.tag and "prev" not in e.attrib:
child.append("version," + re.split(reg,str(e.attrib))[3] + ",")
如果有效，请告诉我，我会解释。
<强>更新
更好的解决方案如下： -
if "prod" in e.tag:
#print(e.attrib)
child.append("product," + e.attrib['name'] + "," + e.attrib['vendor'])
if "prev" in e.attrib:
child.append("version," + e.attrib['prev'] + "," + e.attrib['num'])
if "vers" in e.tag and "prev" not in e.attrib:
child.append("version," + e.attrib['num'] + ",")
对于您的所有三种情况，我的原始解决方案和更新的解决方案，您给定xml的工作示例为here。
答案 1 :(得分：0)
这与解析xml无关，而是与格式化输出的方式无关。
与shell脚本不同，大多数事情只是字符串，你可以只做字符串摆弄以获得你想要的输出，python是面向对象的语言，Python中的对象有类型。特别是e.attrib是一种字典类型，你不能在字典上进行字符串操作。
我建议使用ElementTree的findall()方法而不是做我认为你想做的事情。例如，我认为这就是你真正想要做的事情：
#!/usr/bin/env python
from xml.etree import ElementTree as ET
range_from = 2017
range_to = 2017
def process_entry(entry):
cve = entry.attrib.get("name")
print cve
cpes = get_cpes_affected(entry)
def get_cpes_affected(entry):
prods = entry.findall('nvd:vuln_soft/nvd:prod', namespaces=namespaces)
for prod in prods:
print prod.attrib
print '"'
for prod in prods:
print "product,{},{}".format(prod.attrib['vendor'], prod.attrib['name'])
for vers in prod.findall('nvd:vers', namespaces=namespaces):
if vers.get('edition'):
print "version,{},".format(vers.attrib['edition'])
elif vers.get('prev') == '1':
print "version,{},".format(vers.attrib['prev'])
else:
print "version,{},".format(vers.attrib['num'])
namespaces = {'nvd': 'http://nvd.nist.gov/feeds/cve/1.2'}
# OPTIONAL: registering namespace is useful for outputting XML with ET.tostring()/ET.dump()
#for prefix, ns in namespaces.items():
# ET.register_namespace(prefix, ns)
for i in range(range_from, range_to+1):
e = ET.parse('nvdcve-%i.xml' % i).getroot()
for entry in e:
process_entry(entry)