[Python 原创] python自动打开微信，登陆，关闭窗口注释是用CodeGeeX.rar

import pygetwindow as gw import subprocess import pyautogui import time def activate_window(window_title): try: window = gw.getWindowsWithTitle(window_title)[0] window.activate() return window except IndexError: return None except Exception as e: print("激活窗口异常:", str(e)) return None def click_window_point(x, y, clicks=1, interval=0.1): # 定义一个函数，参数为x,y,clicks,interval，用于模拟鼠标点击操作 pyautogui.moveTo(x, y, duration=0.01) for _ in range(clicks): pyautogui.click() time.sleep(interval) def main(): # 定义应用程序路径 app_path = r'D:\Program\WeChat\WeChat.exe' # 启动应用程序 subprocess.Popen(app_path) print('打开VX') # 循环检查应用程序是否已经打开 while True: # 获取所有应用程序窗口标题 all_windows = gw.getAllTitles() # 检查应用程序是否已经打开 if "微信" in all_windows: # 激活微信窗口 wechat_window = activate_window("微信") # 检查应用程序是否已经打开 if wechat_window is not None: # 获取微信窗口宽度 window_width = wechat_window.width # 获取微信窗口高度 window_height = wechat_window.height # 检查微信窗口高度是否小于600，宽度是否小于500 if window_height < 600 and window_width < 500: # 获取微信窗口左上角x坐标 target_x = wechat_window.left + 200 # 获取微信窗口左上角y坐标 target_y = wechat_window.top + 420 # 点击微信窗口左上角坐标 click_window_point(target_x, target_y) # 等待2.5秒 time.sleep(2.5) print("等待微信登录") # 检查微信窗口高度是否大于600，宽度是否大于500 elif window_height > 600 and window_width > 500: # 初始化vx_window vx_window = None # 获取所有应用程序窗口标题 windows = pyautogui.getWindowsWithTitle("微信") # 遍历应用程序窗口标题 for window in windows: # 检查应用程序窗口标题是否为微信 if "微信" in window.title: # 记录微信窗口 vx_window = window # 跳出循环 break # 检查微信窗口是否已经找到 if vx_window is not None: # 激活微信窗口 vx_window.activate() # 获取微信窗口右上角x坐标 right_top_x = vx_window.left + vx_window.width - 10 # 获取微信窗口右上角y坐标 right_top_y = vx_window.top + 10 # 移动鼠标到微信窗口右上角坐标 pyautogui.moveTo(right_top_x - 10, right_top_y + 20) # 等待0.25秒 time.sleep(0.25) # 点击鼠标 pyautogui.click() else: # 检查微信窗口是否已经找到 print("微信窗口未找到。请确认微信窗口已经打开。") print("关闭微信窗口") # 跳出循环 break if __name__ == "__main__": main()