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多缺口滑块验证demo

环境

win10

Python3.9

分享一下项目中碰到的一个多缺口滑块验证，先触发一下滑块抓包分析一下是用的哪一家滑块经过分析发现不是用的顶象或数美，验证图片的接口地址是他自己的的一个接口，应该是自己写的验证，往有经验的大佬指点下。抓到滑块图片地址，是经过加密的，通过canvas绘画至页面，始终没有找到真实的url，用最笨的方法直接在页面截图

def get_v3(self):

新开一个窗口，通过执行js来新开一个窗口（有奇效，可以不被检测到）

js = 'window.open(“https://www.vivo.com.cn/service/mobilePhoneAuthenticityCheck/query”);'
self.driver.execute_script(js)

切换窗口

self.driver.switch_to.window(self.driver.window_handles[1])
IMEI = “862056063123458”

定位输入框

phone_IMEI = self.wait.until(EC.presence_of_element_located((By.XPATH, ‘//*[@id=“phone_IMEI”]’)))
phone_IMEI.send_keys(IMEI)

勾选复选框

self.wait.until(EC.presence_of_element_located((By.XPATH, ‘/html/body/main/div[2]/div/p/span’))).click()

点击立即查询

self.wait.until(EC.presence_of_element_located((By.XPATH, ‘//*[@id=“query_IMEI”]’))).click()

定位滑块图片

必须先遍历一遍页面所有元素，否则找不到新弹出的滑块元素

self.web_driver_wait_ruishu(10, “class”, ‘dx_captcha dx_captcha_loading-style-popup dx_captcha_basic dx_captcha-type-basic dx_captcha_basic-style-popup’)
print(“定位到滑块弹窗”)
dx_captcha = self.wait.until(EC.presence_of_all_elements_located((By.XPATH, ‘/html/body/div/div/div[2]/div[2]/div[1]/div[2]/div[2]’)))
print(len(dx_captcha))
if len(dx_captcha) > 1:
dx_captcha = dx_captcha[-1]
else:
dx_captcha = dx_captcha[0]

截图

dx_captcha.screenshot(self.bgImg_path)

剩下的就是识别缺口距离了，并生成移动轨迹

@staticmethod
def clear_white(img):
""“清除图片的空白区域，这里主要清除滑块的空白”"“
img = cv2.imread(img)
rows, cols, channel = img.shape
min_x = 255
min_y = 255
max_x = 0
max_y = 0
for x in range(1, rows):
for y in range(1, cols):
t = set(img[x, y])
if len(t) >= 2:
if x <= min_x:
min_x = x
elif x >= max_x:
max_x = x

if y <= min_y:
min_y = y
elif y >= max_y:
max_y = y
img1 = img[min_x:max_x, min_y: max_y]
return img1

@staticmethod
def template_match(tpl, target):
th, tw = tpl.shape[:2]
result = cv2.matchTemplate(target, tpl, cv2.TM_CCOEFF_NORMED)
min_val, max_val, min_loc, max_loc = cv2.minMaxLoc(result)
tl = max_loc
br = (tl[0] + tw, tl[1] + th)
cv2.rectangle(target, tl, br, (0, 0, 255), 2)
return tl[0], tl[1]

def calculate_distance(self, pic1_path, pic2_path):
”"“
计算滑块到缺口的距离
”""
img1 = self.clear_white(pic1_path)
img1 = cv2.cvtColor(img1, cv2.COLOR_RGB2GRAY)
slide = cv2.Canny(img1, 100, 200)

img2 = cv2.imread(pic2_path, 0)
back = cv2.Canny(img2, 100, 200)

slide_pic = cv2.cvtColor(slide, cv2.COLOR_GRAY2RGB)
back_pic = cv2.cvtColor(back, cv2.COLOR_GRAY2RGB)

滑块在图片上的位置

x, y = self.template_match(slide_pic, back_pic)

滑块到缺口的距离

dis_x = int((x + 5) * (340 / 552))
dis_y = int(y * (340 / 552))
return dis_x, dis_y

def get_tracks(self, distance, _y):
""“
获取轨迹参数
”""
tracks = list()
y, v, t, current = 0, 0, 1, 0

mid = distance * 3 / 4

exceed = random.randint(40, 90)
z = random.randint(30, 150)

while current < (distance + exceed):
if current < mid / 2:
a = 2
elif current < mid:
a = 3
else:
a = -3
a /= 2
v0 = v
s = v0 * t + 0.5 * a * (t * t)
current += int(s)
v = v0 + a * t

y += random.randint(-3, 3)
z = z + random.randint(5, 10)
tracks.append([min(current, (distance + exceed)), y, z])

while exceed > 0:
exceed -= random.randint(0, 5)
y += random.randint(-3, 3)
z = z + random.randint(5, 9)
tracks.append([min(current, (distance + exceed)), y, z])
tr = []
for i, x in enumerate(tracks):
tr.append({
‘x’: x[0],
‘y’: _y,
‘relative_time’: x[2]
})
return tr

效果

本文仅供学习交流使用，如侵立删！

【分享】Python 多缺口滑块验证demo