如何在python中更改url查询的值？

Question

 url = "http://www.example.com?type=a&type1=b&type2=c"
 urllist = get_urllist(url)
 trigger = ["'or '1'='1'"," 'OR '1'='2'","'OR a=a"]

def get_urllist(url): 
    url_parsed = urlparse.urlparse(url)
    #extract the query parameters of the URL 
    query =  urlparse.parse_qs(url_parsed.query)
    #get the list of query 
    query_list = query_list(query)
    #Get Base url 
    url = urlparse._replace(query=None).geturl()
    #modify url to get url_list 
    for query in query_list : 
       # change the original query to get the expected result 


 return url_list 


def query_list(query):
     for t in trigger:
         for key, value in query.items():
            query[key] += t
         query_list.append(query) 

     return query_list

如何通过更改查询参数值返回 URL 列表？

原始网址 = " http://www.example.com?type=a&type1=b&type2=c "

预期结果：

Url_list= [" http://www.example.com?type=a 'OR '1'='1'&type1=b'OR '1'='1'&type2=c'OR '1'='1' "," http://www.example.com?type=a 'OR '1'='2'&type1=b'OR '1'='2'&type2=c'OR '1'='2'" ," http://www.example.com?type=a 'OR a=a&type1=b'OR a=a&type2=c''OR a=a" ]

Answer 1

在 Python2.x 中

您可以使用 urlparse.urlparse函数和ParseResult._replace方法：

import urlparse
url = "http://www.example.com?type=a&type1=b&type2=c"
trigger = ["'or '1'='1'"," 'OR '1'='2'","'OR a=a"]

parsed = urlparse.urlparse(url)
querys = parsed.query.split("&")
result = []
for pairs in trigger:
    new_query = "&".join([ "{}{}".format(query, pairs) for query in querys])
    parsed = parsed._replace(query=new_query)
    result.append(urlparse.urlunparse(parsed))

笔记

该urlparse模块被重命名为urllib.parse中Python 3。2to3在将您的源代码转换为 Python 3 时，该工具将自动调整导入。

在 Python3.x 中

您也可以使用urlparse.urlparse函数。

import urllib.parse as urlparse
url = "http://www.example.com?type=a&type1=b&type2=c"
trigger = ["'or '1'='1'"," 'OR '1'='2'","'OR a=a"]

parsed = urlparse.urlparse(url)
querys = parsed.query.split("&")
result = []
for pairs in trigger:
    new_query = "&".join([ "{}{}".format(query, pairs) for query in querys])
    parsed = parsed._replace(query=new_query)
    result.append(urlparse.urlunparse(parsed))

演示输出：

["http://www.example.com?type=a'or '1'='1'&type1=b'or '1'='1'&type2=c'or '1'='1'", "http://www.example.com?type=a 'OR '1'='2'&type1=b 'OR '1'='2'&type2=c 'OR '1'='2'", "http://www.example.com?type=a'OR a=a&type1=b'OR a=a&type2=c'OR a=a"]

Answer 2

您可以使用包foll。

from furl import furl

url = furl("http://www.example.com?type=a&type1=b&type2=c")
url.set({"type": "a'or '1'='1'"})
url.url

给出输出：http://www.example.com?type=a%27or+%271%27%3D%271%27

并解码：http://www.example.com?type=a'or '1'='1'

Answer 3

这是一个简单的例子：

def patch_url(url, **kwargs):
    from urllib.parse import urlparse, urlencode, parse_qsl
    return urlparse(url)._replace(query=urlencode(
        dict(parse_qsl(urlparse(url).query), **kwargs))).geturl()


assert patch_url("https://httpbin.org/get?hello=world", hello="human") \
       == "https://httpbin.org/get?hello=human"