opencv selenium滑动验证码 基于opencv的selenium滑动验证码的实现

from selenium import webdriver
from selenium.webdriver.common.action_chains import ActionChains

artice = browser.find_element_by_class_name('geetest_slider_button') # 滑动按钮
action = ActionChains(browser)
action.click_and_hold(artice).perform() #按住按钮不放
action.reset_actions() 
action.pause(0.01).move_by_offset(step, 0).perform() #step 为滑动的水平距离
action.release(artice).perform() # 松开按钮

import cv2 as cv
import numpy as np
import math

# 寻找直线
def FindLines(image):
 image = cv.cvtColor(image, cv.COLOR_BGR2GRAY) # 二值化
 blurred = cv.GaussianBlur(image, (5, 5), 0) # 高斯模糊
 canny = cv.Canny(blurred, 200, 400) # canny边缘检测
 lines = cv.HoughLinesP(canny, 1, np.pi / 180, 20, minLineLength=15, maxLineGap=8) # 霍夫变换寻找直线
 return lines[:, 0, :] # 返回直线


# 这里对直线进行过滤
def FindResultLises(lines):
 resultLines = []
 for x1, y1, x2, y2 in lines:
  if (abs(y2 - y1) < 5 or abs(x2 - x1) < 5) and min(x1, x2) > 60: # 只要垂直于坐标轴的直线并且起始位置在60像素以上
   resultLines.append([x1, y1, x2, y2])
 return resultLines


# 判断点是否在直线上
def distAbs(point_exm, list_exm):
 x, y = point_exm
 x1, y1, x2, y2 = list_exm
 dist_1 = math.sqrt(abs((y2 - y1) + (x2 - x1) + 1)) # 直线的长度
 dist_2 = math.sqrt(abs((y1 - y) + (x1 - x) + 1)) + math.sqrt(abs((y2 - y) + (x2 - x) + 1)) # 点到两直线两端点距离和
 return abs(dist_2 - dist_1) 


# 交点函数 y = kx + b 求交点位置
def findPoint(line1, line2):
 poit_status = False
 x1, y1, x2, y2 = line1
 x3, y3, x4, y4 = line2
 x = y = 0

 if (x2 - x1) == 0: # 垂直x轴
  k1 = None
  b1 = 0
 else:
  k1 = 1.0 * (y2 - y1) / (x2 - x1)
  b1 = y1 * 1.0 - k1 * x1 * 1.0

 if (x4 - x3) == 0:
  k2 = None
  b2 = 0
 else:
  k2 = 1.0 * (y4 - y3) / (x4 - x3)
  b2 = y3 * 1.0 - k2 * x3 * 1.0

 if k1 is None:
  if not k2 is None:
   x = x1
   y = k2 * x1 + b2
   poit_status = True
 elif k2 is None:
  x = x3
  y = k1 * x3 + b1
  poit_status = True
 elif k1 != k2:
  x = (b2 - b1) * 1.0 / (k1 - k2)
  y = k1 * x * 1.0 + b1 * 1.0
  poit_status = True

 return poit_status, [x, y]


# 求交点
def linePoint(resultLines):
 for x1, y1, x2, y2 in resultLines:
  for x3, y3, x4, y4 in resultLines:
   point_is_exist, [x, y] = findPoint([x1, y1, x2, y2], [x3, y3, x4, y4]) # 两线是否有交点
   if point_is_exist:
    dist_len1 = distAbs([x, y], [x1, y1, x2, y2])
    dist_len2 = distAbs([x, y], [x3, y3, x4, y4])
    if dist_len1 < 5 and dist_len2 < 5: # 如果误差在5内我们认为点在直线上
     # 判断交点在行直线中是左端点还是右端点
     if abs(y2 - y1) < 5:
      # x1是行直线
      if abs(x1 - x) + abs(y1 - y) < 5: # 左端点
       return -1, [x, y]
      else:
       return 1, [x, y]
     else:
      # x2是行直线
      if abs(x3 - x) + abs(y3 - y) < 5:
       return -1, [x, y]
      else:
       return 1, [x, y]
 return 0, [0, 0]

if __name__ == '__main__':
 img = cv.imread(r'C:\Users\Administrator\Desktop\opencv\temImg.png')
 lines = FindLines(img)
 lines = FindResultLises(lines)
 L_or_R, point_x = linePoint(lines) # L_or_R 用于判断交点在行直线左边还是右边 后面拖动要用到
 xoffset = point_x[0]
 yoffset = point_x[1]
 cv.circle(img, (int(xoffset), int(yoffset)), 5, (0, 0, 255), 3)
 cv.imshow('circle', img)
 cv.waitKey(0)
 cv.destroyAllWindows()

if L_or_R == 1:
 x_offset = xoffset - 20 # 20是阴影快一半的长度 可根据实际情况调整
else:
 x_offset = offset + 20

import ramdom

# 通过加速减速模拟滑动轨迹
def moveTrack(xoffset):
 updistance = xoffset*4/5
 t = 0.2
 v = 0
 steps_list = []
 current_offset = 0
 while current_offset<xoffset:
  if current_offset<updistance:
   a = 2 + random.random() * 2
  else:
   a = -random.uniform(12,13)
  vo = v
  v = vo + a * t
  x = vo * t + 1 / 2 * a * (t * t)
  x = round(x, 2)
  current_offset += abs(x)
  steps_list.append(abs(x))
 # 上面的 sum(steps_list) 会比实际的大一点，所以再模拟一个往回拉的动作，补平多出来的距离
 disparty = sum(steps_list)-xoffset 
 last1 = round(-random.random() - disparty, 2)
 last2 = round(-disparty-last1, 2)
 steps_list.append(last1)
 steps_list.append(last2)
 
 return steps_list

for step in steps_list:
 action.reset_actions()
 action.pause(0.01).move_by_offset(step, 0).perform()
action.release(artice).perform()