几种滑动验证码处理
思路:
1 找到距离
2 生成滑动轨迹
3 模拟事件
两张图片 一个缺口图片 一个带缺口的完整图
向右滑到底
1 首先获取滑动的距离 末端 - 小图标 中心点 适当多点无所谓
2 生成轨迹
#生成直线轨迹
def get_tracks_2( distance, seconds, ease_func):
tracks = [0]
offsets = [0]
for t in np.arange(0.0, seconds, 0.1):
ease = ease_func
offset = round(ease(t / seconds) * distance)
tracks.append(offset - offsets[-1])
offsets.append(offset)
return tracks
3 最后一步 模拟事件 滑到位置
def move_to_gap(self, track):
self._driver.web_driver_wait(2, self._driver.XPATH, 'xpath 路径')
slider = self._driver.find_element_by_xpath('xpath 路径')
action = ActionChains(self._driver)
action.click_and_hold(slider)
while track:
x = track.pop(0)
y = x % 2 * random.choice([4, 5, 2])
action.move_by_offset(xoffset=x, yoffset=y)
time.sleep(0.5) # 这里不加延时会导致滑块失败
action.release().perform()
另一种情况 篮球入框
图片是一个小篮球 和 一个篮球框
1 计算篮球和框的坐标
self._driver.web_driver_wait(1, self._driver.XPATH, '//div[@class="rds-drag-basket"]')
basket_style = self._driver.find_element_by_xpath('xpath 路径').get_attribute("style")
ball_style = self._driver.find_element_by_xpath('xpath 路径').get_attribute('style')
basket_left = re.findall(r"left: (.+?)px", basket_style)[0]
basket_top = re.findall(r"top: (.+?)px", basket_style)[0]
ball_left = re.findall(r"left: (.+?)px", ball_style)[0]
ball_top = re.findall(r"top: (.+?)px", ball_style)[0]
x = int(basket_left.split('.')[0]) - int(ball_left.split('.')[0]) + 23
y = int(basket_top.split('.')[0]) - int(ball_top.split('.')[0]) + 23
self.move_to_basket(x, y)
2 生成轨迹 并移动
def move_to_basket(self, x, y):
slider = self._driver.find_element_by_xpath('xpath 路径')
action = ActionChains(self._driver)
action.click_and_hold(slider)
track1 = self.get_tracks_2(x + random.choice([-1, -2, -3, -4, -5, 1, 2, 3, 4, 5]), random.randint(2, 4), self.ease_out_quart)
track2 = self.get_tracks_2(y + random.choice([-1, -2, -3, -4, -5, 1, 2, 3, 4, 5]), random.randint(2, 4), self.ease_out_quart)
while track1 or track2:
try:
x = track1.pop(0)
except:
x = 0
try:
y = track2.pop(0)
except:
y = 0
action.move_by_offset(xoffset=x, yoffset=y)
time.sleep(0.5) # 这里不加延时会导致滑块失败
action.release().perform()
最后一种 三张图片 完整的 完整带缺口的 缺口图片
1 比较两张图片 然后灰度值 算出缺口位置
def compute_gap(self, img1, img2):
"""计算缺口偏移 这种方式成功率很高"""
# 将图片修改为RGB模式
img1 = img1.convert("RGB")
img2 = img2.convert("RGB")
# 计算差值
diff = ImageChops.difference(img1, img2)
# 灰度图
diff = diff.convert("L")
# print(self.otsu_threshold(diff))
table = [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]
# 二值化 阀值table提前计算好的
diff = diff.point(table, '1')
left = 43
for w in range(diff.size[0] - 1, left, -1):
lis = []
for h in range(diff.size[1] - 2, 0, -1):
if diff.load()[w, h] == 1:
lis.append(w)
if len(lis) > 15:
self.error = ErrorLog(img1, img2, diff, w)
return w - left
接下来就是常规操作 生成轨迹 移动记好了