Question

我对Python非常陌生并创建了一个小密码破解程序，它使用强力攻击，我试图让我的进度条在程序运行时输出，这里是＃s到目前为止我所拥有的：

import zipfile
import sys
import time


def progress_bar(sleep_time):
    for i in range(101):
        time.sleep(sleep_time)
        sys.stdout.write("\r[{0}] {1}%".format('#'*(i/10), i))
        sys.stdout.flush()


def obtain_password(path_to_zip_file):
    password = None

    zip_file = zipfile.ZipFile(path_to_zip_file)

    with open('wordlist.txt', 'r') as dict:
        for line in dict.readlines():
            possible = line.strip("\n")
            try:
                zip_file.extractall(pwd=possible)
                password = "Password found {}".format(possible)
            except:
                    pass

    return password

所以我的问题是如何在obtain_password方法运行时输出进度条？我是否需要稍微改变进度条方法？

Answer 1

你想要做的事情不会奏效，你必须记住你只有一个线索。

你可以做的是获取你的词汇表中的行数，并进行数学计算。顺便说一句，它肯定比计时器精确得多。

我没有对代码进行测试，但是根据这些内容，你会得到你想要的东西：

import zipfile
import sys
import time

def obtain_password(path_to_zip_file):
    password = None
    zip_file = zipfile.ZipFile(path_to_zip_file)
    with open('wordlist.txt', 'r') as f:
        lines = f.readlines()
        total = len(lines) # get number of lines
        current = 0
        for line in lines:
            current += 1
            if current % 1000 == 0: # every 1000 lines, shows the progress
                print('%.2f %%' % float(current / total * 100))
            possible = line.strip("\n")
            try:
                zip_file.extractall(pwd=possible)
                #password = "Password found {}".format(possible)
                print(possible)
                sys.exit()
            except:
                pass

另外，我建议您了解extractall引发的异常并正确捕捉它们。抓住这样的一切：except:不是一个好习惯。

Answer 2

有办法做你想做的事。

让您的密码破解者暂时更新进度条

 import time

 # Stores the time between updates in seconds.
 time_between_updates = 10
 last_update = 0

 def your_expensive_operation():
     for i in range(10000000):
         time.sleep(1)            # Emulate an expensive operation
         if time.time() - last_update > time_between_updates:
             print("\r" + (int(i/10000000.0 * 79) * "#"), end='')

 your_expensive_operation()

使用线程

import time
import threading

# Stores your current position in percent.
current_position = 0
done = False

def paint_thread():
    while not done:
        print("\r" + (int(current_position * 79) * "#"), end='')
        # Make it update once a second.
        time.sleep(1)

thread = threading.Thread(target=paint_thread)
thread.start()

for i in range(10000000):
     time.sleep(1)            # Emulate an expensive operation
     current_position = i/10000000.0

done = True

使用进度条输出

2 个答案: