Usually for personal use, it can save me about 30 minutes
- Note that python and pycharm need to be installed
- If there is an error (no module named ***, just use pip install ***)
- You can modify the filling statement yourself
- After entering the number, within 5 seconds, switch to the evaluation page and click the first evaluation option, as shown below
from pynput.keyboard import Controller as key_cl
import time
from pynput import keyboard
while 1:
message = input('请看使用说明!!!\n输入 0 自动课程评估 \n输入 1 自动教师评估 '
'\n输入后5秒内切换至浏览器对应页面\n等待自动评估完成后输入验证码\n请输入:')
try:
num = int(message)
assert num != 0 or num != 1
key = key_cl()
time.sleep(8)
if num == 0:
for i in range(21):
key.press(keyboard.KeyCode.from_vk(9))
key.release(keyboard.KeyCode.from_vk(9))
time.sleep(0.1)
key.press(keyboard.KeyCode.from_vk(32))
key.release(keyboard.KeyCode.from_vk(32))
time.sleep(0.1)
for i in range(5):
key.press(keyboard.KeyCode.from_vk(9))
key.release(keyboard.KeyCode.from_vk(9))
time.sleep(0.1)
key.type('评估课程')
key.press(keyboard.KeyCode.from_vk(9))
key.release(keyboard.KeyCode.from_vk(9))
time.sleep(0.1)
key.press(keyboard.KeyCode.from_vk(32))
key.release(keyboard.KeyCode.from_vk(32))
time.sleep(0.1)
key.press(keyboard.KeyCode.from_vk(9))
key.release(keyboard.KeyCode.from_vk(9))
time.sleep(0.1)
key.press(keyboard.KeyCode.from_vk(32))
key.release(keyboard.KeyCode.from_vk(32))
time.sleep(0.1)
print('\n此次课程评估完成,请按需要继续输入\n')
if num == 1:
for i in range(20):
key.press(keyboard.KeyCode.from_vk(9))
key.release(keyboard.KeyCode.from_vk(9))
time.sleep(0.1)
key.press(keyboard.KeyCode.from_vk(32))
key.release(keyboard.KeyCode.from_vk(32))
time.sleep(0.1)
for i in range(2):
key.press(keyboard.KeyCode.from_vk(9))
key.release(keyboard.KeyCode.from_vk(9))
time.sleep(0.1)
key.type('评估教师')
print('\n此次教师评估完成,请按需要继续输入\n')
except:
print('your input has to be 0 or 1 \n 请重新输入,如想退出直接关闭')