使用 python 和 XML XPath 时获取密钥类型错误

Question

输入 XML 属性时出现类型错误

XML 从中获取数据的文件

 <data>
        <system>
            <vulnerability name = "Crack Passwords"/>
            <vulnerability type="reverse">
                <input updated="yes">2</input>
                <generator type = "encoder_diff">
                    <input into="strings_to_join">
                        <datastore access="0">passwords</datastore>
                    </input>
                </generator>
                <encoder type="md5" />
                <value>141100</value>
            </vulnerability>
    
            <vulnerability type="pwn">
                <input updated="yes">5</input>
                <generator>2011</generator>
                <encoder type="sha256" direction="N"/>
                <input into="strings_to_encode">
                    <value>So, you think you are an expert huh? I wonder if you can figure out my password.</value>
                    <value>This account password is also a flag. For example, if the password is "123456" the flag is: flag{123456}</value>
                    <value>Here is a flag for finding this message:</value>
                    <generator type="flag_generator"/>
                </input>
            </vulnerability>
    
            <vulnerability type="pcap_file">
                <input updated="yes">69</input>
                <generator>2011</generator>
                <encoder type="Base 64 encoder" />
                <value>13600</value>     
            </vulnerability>
        </system>
    </data>

Python代码的第一部分接受用户的输入并匹配XML中的属性以显示文件

第二部分Python代码到modify/change错误所在的属性

ch = input("\nEnter Tag you want to display : ") #First part
    for x in root.findall(".//vulnerability[@type = '{}']".format(ch)):
        print(x.tag,"--->",x.attrib)

changes = input("\nEnter Your changed tag : ") #Second Part
    for x in root.findall(".//vulnerability"):
        if  x.attrib['type'] == ch: #ERROR
            x.attrib['type'] = changes  #Error
    print(x.tag,x.attrib)

Answer 1

检查下面的代码：

import xml.etree.ElementTree as ET

xml = """<?xml version="1.0"?>
<data>
    <system>
        <vulnerability name = "Crack Passwords"/>
        <vulnerability type="reverse">
            <input updated="yes">2</input>
            <generator type = "encoder_diff">
                <input into="strings_to_join">
                    <datastore access="0">passwords</datastore>
                </input>
            </generator>
            <encoder type="md5" />
            <value>141100</value>
        </vulnerability>

        <vulnerability type="pwn">
            <input updated="yes">5</input>
            <generator>2011</generator>
            <encoder type="sha256" direction="N"/>
            <input into="strings_to_encode">
                <value>So, you think you are an expert huh? I wonder if you can figure out my password.</value>
                <value>This account password is also a flag. For example, if the password is "123456" the flag is: flag{123456}</value>
                <value>Here is a flag for finding this message:</value>
                <generator type="flag_generator"/>
            </input>
        </vulnerability>

        <vulnerability type="pcap_file">
            <input updated="yes">69</input>
            <generator>2011</generator>
            <encoder type="Base 64 encoder" />
            <value>13600</value>     
        </vulnerability>
    </system>
</data>
"""

tree = ET.fromstring(xml)

第一种方法.attrib():

ch = input('\nEnter Tag you want to display : ')
for x in tree.findall(f'.//vulnerability[@type="{ch}"]'):
  print(x.tag,"--->",x.attrib) # To display selected Tag
  changes = str(input('\nEnter Your changed tag : '))
  x.attrib['type'] = f'{changes}'

print(ET.tostring(tree))

第二种方法.get()和.set()：

ch = input('\nEnter Tag you want to display : ')
for x in tree.findall(f'.//vulnerability[@type="{ch}"]'):
  print(x.tag,"--->",x.attrib) # To display selected Tag
  changes = str(input('\nEnter Your changed tag : '))
  x.get(f'{ch}')
  x.set('type', f'{changes}')

print(ET.tostring(tree))

使用 python 和 XML XPath 时获取密钥类型错误

Getting Key type Error in using python and XML XPath

python

xml

xpath

xml-parsing