今日最新！脚本编写的一个通用框架

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今日最新！脚本编写的一个通用框架

2023-03-21 13:15:51 来源：博客园

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总结基于python写脚本的框架

# -*- encoding:utf-8 -*-# @time: 2023/3/21 12:11# @author: Maxs_hu"""基于python实现小脚本的基础框架:    1. 利用yield生成器函数, 将读取到的数据返回    2. 线程池执行主线任务    3. try...except捕获异常. 也可做日志记录    4. 利用 函数式迭代 可以避免代码重复    5. send_sms() -> for循环+try...except. 没报错直接return. 报错就继续循环执行. 最后如果十次都没有成功. 返回异常"""from concurrent.futures import ThreadPoolExecutordef geetest():    try:        # 实现滑动验证        return True, 123    except Exception as e:        return False, str(e)def send_sms():    for i in range(5):        try:            # 发送短信            return True, 123        except Exception as e:            pass    return False, "短信发送异常了"def login():    while True:        try:            # 发送短信            return True, 123        except Exception as e:            passdef task(mobile, pwd):    try:        # 滑动验证        status, data_or_error = geetest()        if not status:            # 错误记录， mobile+pwd+data_or_error + 滑动验证            return        # 发送短信        status, data_or_error = send_sms()        if not status:            # 错误记录， mobile+pwd+data_or_error + 发送短信步骤            return        # 登录        status, data_or_error = login()        if not status:            # 错误记录， mobile+pwd+data_or_error + 登录失败            return        func_list = [geetest, send_sms, login]        for func in func_list:            status, data = func()            if not status:                return    except Exception as e:        print("异常", e)  # 记录下来def read_file():    # 手机号,密码 500    # 手机号,密码    # 手机号,密码     v1,v2 = [11,22]    with open("xxxx.txt", mode="r", encoding="utf-8") as f:        for line in f:            line = line.strip()            mobile, pwd = line.split(",")            yield mobile, pwddef run():    pool = ThreadPoolExecutor(10)    gen = read_file()  # 接受生成器函数    for mobile, pwd in gen:        # print(mobile, pwd)        pool.submit(task, mobile, pwd)    pool.shutdown()  # 主线程等待    print("5000个任务全部执行完毕")if __name__ == "__main__":    run()

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