在本地 http 路径上读取图片

Question

我想打开一个图片，它是另一个程序的输出http://192.168.1.100:8123/pic/2021-03-31/snap/238371315.jpg

我遇到问题的代码：

def get_pic(data):
    img_path='http://192.168.1.100:8123/pic/2021-03-31/snap/238371315.jpg'
    pic={'image': open(img_path, mode='rb')}
    return pic

错误：

Traceback (most recent call last):
  File "c:\Users\USER\Desktop\linebot-client\client_script.py", line 144, in <module>
    main()
  File "c:\Users\USER\Desktop\linebot-client\client_script.py", line 134, in main
    final_pic =get_pic(data1)
  File "c:\Users\USER\Desktop\linebot-client\client_script.py", line 107, in get_pic
    pic={'image': open(img_path, mode='rb')}
OSError: [Errno 22] Invalid argument: 'http://192.168.1.100:8123/pic/2021-04-01/snap/939450165.jpg'

有什么我可以改变的吗？