检查无限多个自生成的URL的有效性，如果有效，则安全响应（http 200）

Question

我想检查一个无限数量的自我生成的网址的有效性，如果有效的安全主体的响应在一个文件。URL看起来是这样的：https://mydomain.com/+随机字符串（例如https://mydomain.com/ake3t），我想用字母表“abcdefghijklmnopqrstuvwxyz0123456789”生成它们，然后用暴力尝试所有可能的方法。在

我用python编写了一个脚本，但由于我是一个绝对的初学者，它非常慢！因为我需要的东西非常快，所以我试着用了，因为我认为它是专门为这种工作。在

现在的问题是我不知道如何动态地生成url，我不能预先生成它们，因为它不是固定数量的url。在

有人能告诉我如何做到这一点，或者向我推荐另一种更适合这项工作的工具或库吗？在

更新： 这是我用的脚本，但我觉得它很慢。最让我担心的是，如果我使用多个线程（在threadsNr中指定），它会变慢

import threading, os
import urllib.request, urllib.parse, urllib.error 

threadsNr   = 1                                    
dumpFolder    = '/tmp/urls/'               
charSet     = 'abcdefghijklmnopqrstuvwxyz0123456789_-' 
Url_pre    = 'http://vorratsraum.com/'
Url_post    = 'alwaysTheSameTail'

# class that generate the words
class wordGenerator ():

    def __init__(self, word, charSet):
        self.currentWord = word
        self.charSet = charSet

    # generate the next word set that word as currentWord and return the word
    def nextWord (self):
        self.currentWord = self._incWord(self.currentWord)
        return self.currentWord

    # generate the next word
    def _incWord(self, word):
        word = str(word)                        # convert to string

        if word == '':                          # if word is empty 
            return self.charSet[0]              # return first char from the char set
        wordLastChar = word[len(word)-1]        # get the last char
        wordLeftSide = word[0:len(word)-1]      # get word without the last char
        lastCharPos  = self.charSet.find(wordLastChar)  # get position of last char in the char set

        if (lastCharPos+1) < len(self.charSet):         # if position of last char is not at the end of the char set
            wordLastChar = self.charSet[lastCharPos+1]  # get next char from the char set

        else:                                           # it is the last char
            wordLastChar = self.charSet[0]              # reset last char to have first character from the char set
            wordLeftSide = self._incWord(wordLeftSide)  # send left site to be increased

        return wordLeftSide + wordLastChar      # return the next word


class newThread(threading.Thread):
    def run(self):
        global exitThread
        global wordsTried
        global newWord
        global hashList

        while exitThread == False:
            part = newWord.nextWord()                # generate the next word to try
            url = Url_pre + part + Url_post

            wordsTried = wordsTried + 1
            if wordsTried == 1000: # just for testing how fast it is
                exitThread = True
            print( 'trying ' + part)          # display the word
            print( 'At URL ' + url)

            try:
                req = urllib.request.Request(url)
                req.addheaders = [('User-agent', 'Mozilla/5.0')]
                resp = urllib.request.urlopen(req)
                result = resp.read()
                found(part, result)
            except urllib.error.URLError as err:
                if err.code == 404:
                    print('Page not found!')
                elif err.code == 403:
                    print('Access denied!')
                else:
                    print('Something happened! Error code', err.code)
            except urllib.error.URLError as err:
                print('Some other error happened:', err.reason)
        resultFile.close()

def found(part, result):
    global exitThread
    global resultFile

    resultFile.write(part +"\n")

    if not os.path.isdir(dumpFolder + part):
        os.makedirs(dumpFolder + part)

    print('Found Part = '  + part)

wordsTried = 0                            
exitThread = False                              # flag to kill all threads
newWord = wordGenerator('',charSet);           # word generator

if not os.path.isdir(dumpFolder):
    os.makedirs(dumpFolder)

resultFile = open(dumpFolder + 'parts.txt','a')      # open file for append    

for i in range(threadsNr):
    newThread().start()

Answer 1

你想要暴力还是随机？下面是一个具有重复字符的顺序暴力方法。速度将在很大程度上取决于您的服务器响应。还要注意，这很可能会很快产生拒绝服务条件。在

import itertools
import url

pageChars = 5
alphabet = "abcdefghijklmnopqrstuvwxyz0123456789_-"

#iterate over the product of alphabet with <pageChar> elements
#this assumes repeating characters are allowed
# Beware this generates len(alphabet)**pageChars possible strings
for chars in itertools.product(alphabet,repeat=pageChars):
    pageString = ''.join(chars)

    urlString = 'https://mydomain.com/' + pageString

    try:
        url = urllib2.urlopen(url)

    except urllib2.HTTPError:
        print('No page at: %s' % urlString)
        continue     

    pageDate = url.read()
    #do something with page data

Answer 2

你不能检查“无限数量的网址”而不是“非常慢”，初学者或没有

scraper占用的时间几乎肯定是由您访问的服务器的响应时间决定的，而不是由脚本的效率决定的。在

你到底想干什么？在