思路
- 计算出图案中滑块的位置就可以使用selenium来模拟登陆
分析
- 检查分析网页代码找到了图片所在div
- 获得该div的class用来截取该图片与有缺口的图做对求出图片需要滑动的距离就可以了 -代码
import time
from io import BytesIO
from PIL import Image
from selenium import webdriver
from selenium.webdriver.common.action_chains import ActionChains
from selenium.webdriver.common.by import By
from selenium.webdriver.support import expected_conditions as EC
from selenium.webdriver.support.ui import WebDriverWait
EMAIL = '14785137962'
PASSWORD = 'ybdqq1607474154'
BORDER = 10
INIT_LEFT = 60
class CrackGeetest():
def __init__(self):
self.url = 'https://passport.bilibili.com/login?gourl=https%3A%2F%2Faccount.bilibili.com%2Faccount%2Fbig'
self.browser = webdriver.Chrome()
self.wait = WebDriverWait(self.browser, 20)
self.email = EMAIL
self.password = PASSWORD
def __del__(self):
self.browser.close()
# def get_geetest_button(self):
# """
# 获取滑块按钮
# :return: button
# """
# button = self.wait.until(EC.element_to_be_clickable((By.CLASS_NAME, 'gt_slider_knob')))
# return button
def get_position(self):
"""
获取图片验证码在网页中的坐标位置
:return: 验证码位置元组
"""
time.sleep(3)
img = self.wait.until(EC.presence_of_element_located((By.CLASS_NAME, 'gt_box')))
location = img.location
size = img.size
top, bottom, left, right = location['y'], location['y'] + size['height'], location['x'], location['x'] + size[
'width']
return (top, bottom, left, right)
def get_screenshot(self):
"""
获取网页截图
:return: 截图对象
"""
screenshot = self.browser.get_screenshot_as_png()
screenshot = Image.open(BytesIO(screenshot))
return screenshot
def get_slider(self):
"""
获取滑块
:return: 滑块对象
"""
slider = self.wait.until(EC.element_to_be_clickable((By.CLASS_NAME, 'gt_slider_knob')))
return slider
def get_geetest_image(self, name='captcha.png'):
"""
获取验证码图片
:return: 图片对象
"""
# 掉用方法获得验证码坐标
top, bottom, left, right = self.get_position()
print('验证码位置', top, bottom, left, right)
# 截取整个网页
screenshot = self.get_screenshot()
# 根据坐标截取出验证码图片
captcha = screenshot.crop((left, top, right, bottom))
print(left, top, right, bottom)
# 保存验证码图片
captcha.save(name)
return captcha
def open(self):
"""
打开网页输入用户名密码
:return: None
"""
self.browser.get(self.url)
email = self.wait.until(EC.presence_of_element_located((By.ID, 'login-username')))
password = self.wait.until(EC.presence_of_element_located((By.ID, 'login-passwd')))
email.send_keys(self.email)
password.send_keys(self.password)
def get_gap(self, image1, image2):
"""
获取缺口偏移量
:param image1: 不带缺口图片
:param image2: 带缺口图片
:return:
"""
# 从图片中横坐标位65处开始便利图片像素点
left = 65
time.sleep(2)
for i in range(left, image1.size[0]):
print(i)
for j in range(35,image1.size[1]-35):
if not self.is_pixel_equal(image1, image2, i, j):
print(i,j)
left = i
return left
return left
def is_pixel_equal(self, image1, image2, x, y):
"""
判断两个像素是否相同
:param image1: 图片1
:param image2: 图片2
:param x: 位置x
:param y: 位置y
:return: 像素是否相同
"""
# 取两个图片的像素点
pixel1 = image1.load()[x, y]
pixel2 = image2.load()[x, y]
threshold = 60
# 判断像素点之间的RGB值是不是在可接受范围
if abs(pixel1[0] - pixel2[0]) < threshold and abs(pixel1[1] - pixel2[1]) < threshold and abs(
pixel1[2] - pixel2[2]) < threshold:
return True
else:
return False
def get_track(self, distance):
"""
根据偏移量获取移动轨迹(使用加速度与减速度来模拟,当然并不是每个网站都可以使用加速度来解决的,
如有妖气使用的顶象验证还会判断是否是存在加速度与加速度,毕竟人手动的速度是有波动的)
:param distance: 偏移量
:return: 移动轨迹
"""
# 移动轨迹
track = []
# 当前位移
current = 0
# 减速阈值
mid = distance * 4 / 5
# 计算间隔
t = 0.3
# 初速度
v = 0
while current < distance:
if current < mid:
# 加速度为正4,实验多次得到的较为准确的速度
a = 4
else:
# 加速度为负5
a = -5
# 初速度v0
v0 = v
# 当前速度v = v0 + at
v = v0 + a * t
# 移动距离x = v0t + 1/2 * a * t^2
move = v0 * t + 1 / 2 * a * t * t
# 当前位移
current += move
# 加入轨迹
track.append(round(move))
return track
def move_to_gap(self, slider, track):
"""
拖动滑块到缺口处
:param slider: 滑块
:param track: 轨迹
:return:
"""
# 点击不放开
ActionChains(self.browser).click_and_hold(slider).perform()
# 滑动
for x in track:
ActionChains(self.browser).move_by_offset(xoffset=x, yoffset=0).perform()
time.sleep(0.5)
# 对齐松开
ActionChains(self.browser).release().perform()
def login(self):
"""
登录
:return: None
"""
submit = self.wait.until(EC.element_to_be_clickable((By.CLASS_NAME, 'login-btn')))
submit.click()
time.sleep(1)
print('登录成功')
def crack(self):
# 输入用户名密码
self.open()
# 鼠标移动到验证按钮
slider = self.get_slider()
# button = self.get_geetest_button()
ActionChains(self.browser).move_to_element(slider).perform()
# 获取验证码图片
image1 = self.get_geetest_image('captcha1.png')
# 点按呼出缺口
slider.click()
# 获取带缺口的验证码图片
time.sleep(1)
image2 = self.get_geetest_image('captcha2.png')
# 获取缺口位置
gap = self.get_gap(image1, image2)
print('缺口位置', gap)
# 减去缺口位移
gap -= BORDER
# 获取移动轨迹
track = self.get_track(gap)
print('滑动轨迹', track)
# 拖动滑块
self.move_to_gap(slider, track)
time.sleep(30)
if __name__ == '__main__':
crack = CrackGeetest()
crack.crack()
- 在截取图片后面我从65开始计算图片的像素,是为了避开移动的滑块即天蓝色部分不进行计算
- 计算得到红色部分的横坐标,就是滑块最终移动到的位置(返回图片红色的部分的最终横坐标)
- 但滑块本身距离图片边缘就有一定距离大概10像素
- 最后移动距离是红色-绿色