当img弹出时，Python在某处单击

Question

我想写一个关于图像可能随时弹出的程序，操作是关闭图像，而不是点击图像。我有一些关于代码的想法，但不成功：（图像不是弹出式的，但仍要等待1秒钟，然后单击右上角的conner…）

{<1分$ import pyautogui, time while ture: pyautogui.click(pyautogui.center(pyautogui.locateOnScreen(r'C:\Users\Lawrence\Desktop\PyTest\image.png'))) time.sleep(1) #I want to check image every sec break pyautogui.click(1880,15) # after checking the screen for every sec, the image popup, and click top right conner to close it, finish

Answer 1

试试这个。你太早就打破了，点击了错误的东西。在

import pyautogui, time
while True:
    #I want to check image every sec
    time.sleep(.1) #put time interval for checking here
    if pyautogui.locateOnScreen(r'C:\Users\Lawrence\Desktop\PyTest\image.png'):
       pyautogui.click(1880,15)  
       break