按順序遍歷多個字符列表

Question

我想查找長度在1-4個字符之間的字母字符串。

我開始通過52個字母列表迭代：

letters = string.ascii_letters 

我則需要通過同一個目錄遍歷該字符串的接下來的三個字符，直到我發現我要找的字符串。

如果每個_代表的52個字母列表，我需要基本上做到這一點，而在每次迭代檢查匹配：

_ 
_ _ 
_ _ _ 
_ _ _ _ 

如何將我最好的結構系列迴路做到這一點？

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如果問題的前提看起來很混亂，這是針對暴力破解設置的問題。我簡單地提取了我正在努力解決的部分問題。

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編輯：這是我到目前爲止的地方。

#we know the salt is the 2-digit '50' 
#we know the key is limited to 4 alphabetical letters 
#cycle through all possibilities of the key till we match the hash 

letters = string.ascii_letters 
lcounter = 0 
i = 0 
j = 0 
k = 0 
l = 0 
tryhash = "a" 
word = [letters[i]] 

while(tryhash != hash): 
    for c in letters: 
     word = [letters[i]] #this does not work as the additional letters need to be appended to word after the first lcounter loop 
     tryword = ''.join(word) 
     tryhash = crypt.crypt(tryword, "50") 

     if (tryhash == hash): 
      print(word) 
      break 

     i += 1 

     if (lcounter > 0) and (i == 52): 
      i = 0 
      if (lcounter == 1) and (j == 0): 
       word.insert(lcounter, letters[j]) 
      j += 1 

      if (lcounter > 1) and (k == 52): 
       j = 0 
       if (lcounter == 2) and (k == 0): 
        word.insert(lcounter, letters[k]) 
       k += 1 

       if (lcounter > 2) and (k == 52): 
        k = 0 
        if (lcounter == 3) and (l == 0): 
         word.insert(lcounter, letters[l]) 
        l += 1 

    lcounter += 1 

Answer 1

這樣的事情也許是：

my_string = "some" 
for letter1 in string.ascii_letters: 
    if letter1 == my_string: 
     print("success") 
    for letter2 in string.ascii_letters: 
     if letter1 + letter2 == my_string: 
      print("success") 
     for letter3 in string.ascii_letters: 
      if letter1 + letter2 + letter3 == my_string: 
       print("success") 
      for letter4 in string.ascii_letters: 
       if letter1 + letter2 + letter3 + letter4 == my_string 
        print("success") 

Answer 2

你可以做這樣的事情：

 

    import string 
    import itertools 

    data = string.ascii_lowecase 

    for i in itertools.permutations(data, 4): 

     if i == 'your_string': 
      #do something 
     else: 
      pass