思路
- 输入用户名密码点击登陆
- 获取验证码的原始图片与有缺口的图片
- 找出两张图片的缺口起始处
- 拖动碎片
功能代码段
# 使用到的库
from selenium import webdriver
from selenium.webdriver.support.ui import WebDriverWait
from selenium.webdriver.common.by import By
from selenium.webdriver.support import expected_conditions as EC
from selenium.webdriver.common.action_chains import ActionChains
from PIL import Image
import time
import base64
username = '用户名'
password = '密码'
# 放在外面的原因是如果再类的内部初始化,则程序结束后浏览器会自动退出
driver = webdriver.Chrome()
初始化相关参数
# 初始化相关参数
def __init__(self):
self.url = 'https://passport.bilibili.com/login'
self.browser = driver
self.wait = WebDriverWait(self.browser, 20)
self.name = username
self.pw = password
获取按钮、输入框、碎片拖动按钮对象
def get_login_button(self):
"""
获取初始登录按钮
:return: 按钮对象
"""
button = self.wait.until(
EC.presence_of_element_located((By.XPATH, "//a[contains(@class,'btn') and contains(@class, 'btn-login')]")))
return button
def get_slider_button(self):
"""
获取拖动碎片的地方
:return: 拖动对象
"""
sliderbutton = self.wait.until(EC.presence_of_element_located((By.XPATH, "//div[@class='geetest_slider_button']")))
return sliderbutton
def get_login_input(self):
"""
获取登陆输入框(用户名/密码)
:return: 输入框对象
"""
user_login = self.wait.until(EC.presence_of_element_located((By.XPATH, "//input[@id='login-username']")))
pw_login = self.wait.until(EC.presence_of_element_located((By.XPATH, "//input[@id='login-passwd']")))
return user_login, pw_login
获取带有碎片的图片和完整图片
def save_pic(self, data, filename):
"""
解码获取到的base64再写入到文件中,保存图片
:return:
"""
data = data.split(',')[1]
data = base64.b64decode(data)
with open(filename, 'wb') as f:
f.write(data)
def get_pic(self):
"""
获取无缺口图片和有缺口图片
:return: 图片对象
"""
picName = ['full.png', 'slice.png']
# 图片对象的class
className = ['geetest_canvas_fullbg', 'geetest_canvas_bg']
# canvas标签中的图片通过js代码获取base64编码,然后再通过解码,将其写入文件才能获取到
for i in range(len(className)):
js = "var change = document.getElementsByClassName('"+className[i] + "'); return change[0].toDataURL('image/png');"
im_info = self.browser.execute_script(js)
self.save_pic(im_info, picName[i])
判断像素点是否相同
def is_pixel_equal(self, image1, image2, x, y):
"""
判断两个像素点是否是相同
:param image1: 不带缺口图片
:param image2: 带缺口图片
:param x: 像素点的x坐标
:param y: 像素点的y坐标
:return:
"""
pixel1 = image1.load()[x, y]
pixel2 = image2.load()[x, y]
threshold = 40
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
获取需要移动的距离
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:
"""
# 这个可以自行操作一下,如果发现碎片对不准,可以调整
left = 10
for i in range(left, image1.size[0]):
for j in range(image1.size[1]):
if not self.is_pixel_equal(image1, image2, i, j):
left = i
return left
return left
变速运动拖动碎片,否则容易被看出来是机器执行
def get_track(self, distance):
"""
根据偏移量获取移动轨迹
:param self:
:param distance: 偏移量
:return: 移动轨迹
"""
# 移动轨迹
track = []
# 当前位移
current = 0
# 对的不一定很准确,所以自行调整一下distance
distance = distance - 9
# 减速阈值 -> 也就是加速到什么位置的时候开始减速
mid = distance * 4 / 5
# 计算间隔
t = 0.2
# 初速度
v = 0
while current < distance:
if current < mid:
# 加速度为正2
a = 2
else:
# 加速度为负3
a = -3
v0 = v
v = v0 + a * t
move = v0 * t + 1 / 2 * a * t * t
current += move
track.append(round(move))
return track
模拟拖动碎片
def move_to_gap(self, slider, tracks, browser):
"""
拖动滑块到缺口处
:param self:
:param slider: 滑块
:param tracks: 轨迹
:return:
"""
# click_and_hold()点击鼠标左键,不松开
ActionChains(self.browser).click_and_hold(slider).perform()
for x in tracks:
# move_by_offset()鼠标从当前位置移动到某个坐标
ActionChains(self.browser).move_by_offset(xoffset=x, yoffset=0).perform()
time.sleep(0.5)
# release()在某个元素位置松开鼠标左键
ActionChains(self.browser).release().perform()
配置执行
def test(self):
# 输入用户名和密码
self.browser.get(self.url)
user_login, pw_login = self.get_login_input()
user_login.send_keys(self.name)
pw_login.send_keys(self.pw)
# 点击按钮对象
button = self.get_login_button()
button.click()
# 这里设置等待是为了使得滑动验证码能出现,之后才能通过toDataURL获取
time.sleep(3)
self.get_pic()
image1 = Image.open('full.png')
image2 = Image.open('slice.png')
left = self.get_gap(image1, image2)
track = self.get_track(left)
slider = self.get_slider_button()
self.move_to_gap(slider, track, self.browser)
完整代码
TIP
如果出现碎片移动存在一定对不准的情况,可以自行调整一下left和distance的值。
from selenium import webdriver
from selenium.webdriver.support.ui import WebDriverWait
from selenium.webdriver.common.by import By
from selenium.webdriver.support import expected_conditions as EC
from selenium.webdriver.common.action_chains import ActionChains
from PIL import Image
import time
import base64
username = '用户名'
password = '密码'
driver = webdriver.Chrome()
class Start:
def __init__(self):
self.url = 'https://passport.bilibili.com/login'
self.browser = driver
self.wait = WebDriverWait(self.browser, 20)
self.name = username
self.pw = password
def get_login_button(self):
"""
获取初始登录按钮
:return: 按钮对象
"""
button = self.wait.until(
EC.presence_of_element_located((By.XPATH, "//a[contains(@class,'btn') and contains(@class, 'btn-login')]")))
return button
def get_slider_button(self):
"""
获取拖动碎片的地方
:return: 拖动对象
"""
sliderbutton = self.wait.until(EC.presence_of_element_located((By.XPATH, "//div[@class='geetest_slider_button']")))
return sliderbutton
def get_login_input(self):
"""
获取登陆输入框(用户名/密码)
:return: 输入框对象
"""
user_login = self.wait.until(EC.presence_of_element_located((By.XPATH, "//input[@id='login-username']")))
pw_login = self.wait.until(EC.presence_of_element_located((By.XPATH, "//input[@id='login-passwd']")))
return user_login, pw_login
def save_pic(self, data, filename):
"""
解码获取到的base64再写入到文件中,保存图片
:return:
"""
data = data.split(',')[1]
data = base64.b64decode(data)
with open(filename, 'wb') as f:
f.write(data)
def get_pic(self):
"""
获取无缺口图片和有缺口图片
:return: 图片对象
"""
# 图片对象的类名
# 首先需要这个东西已经出现了,我们才能去执行相关的js代码
picName = ['full.png', 'slice.png']
className = ['geetest_canvas_fullbg', 'geetest_canvas_bg']
# canvas标签中的图片通过js代码获取base64编码
for i in range(len(className)):
js = "var change = document.getElementsByClassName('"+className[i] + "'); return change[0].toDataURL('image/png');"
im_info = self.browser.execute_script(js)
self.save_pic(im_info, picName[i])
def is_pixel_equal(self, image1, image2, x, y):
"""
判断两个像素点是否是相同
:param image1: 不带缺口图片
:param image2: 带缺口图片
:param x: 像素点的x坐标
:param y: 像素点的y坐标
:return:
"""
pixel1 = image1.load()[x, y]
pixel2 = image2.load()[x, y]
threshold = 40
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_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:
"""
# 这个可以自行操作一下,如果发现碎片对不准,可以调整
left = 10
for i in range(left, image1.size[0]):
for j in range(image1.size[1]):
if not self.is_pixel_equal(image1, image2, i, j):
left = i
return left
return left
def get_track(self, distance):
"""
根据偏移量获取移动轨迹
:param self:
:param distance: 偏移量
:return: 移动轨迹
"""
# 移动轨迹
track = []
# 当前位移
current = 0
# 因为老对不的不准确,所以自行调整一下distance
distance = distance - 9
# 减速阈值 -> 也就是加速到什么位置的时候开始减速
mid = distance * 4 / 5
# 计算间隔
t = 0.2
# 初速度
v = 0
while current < distance:
if current < mid:
# 加速度为正2
a = 2
else:
# 加速度为负3
a = -3
v0 = v
v = v0 + a * t
move = v0 * t + 1 / 2 * a * t * t
current += move
track.append(round(move))
return track
def test(self):
# 输入用户名和密码
self.browser.get(self.url)
user_login, pw_login = self.get_login_input()
user_login.send_keys(self.name)
pw_login.send_keys(self.pw)
# 点击按钮对象
button = self.get_login_button()
button.click()
# 保存图片
time.sleep(3)
self.get_pic()
image1 = Image.open('full.png')
image2 = Image.open('slice.png')
left = self.get_gap(image1, image2)
track = self.get_track(left)
slider = self.get_slider_button()
self.move_to_gap(slider, track, self.browser)
def move_to_gap(self, slider, tracks, browser):
"""
拖动滑块到缺口处
:param self:
:param slider: 滑块
:param tracks: 轨迹
:return:
"""
# click_and_hold()点击鼠标左键,不松开
ActionChains(self.browser).click_and_hold(slider).perform()
for x in tracks:
# move_by_offset()鼠标从当前位置移动到某个坐标
ActionChains(self.browser).move_by_offset(xoffset=x, yoffset=0).perform()
time.sleep(0.5)
# release()在某个元素位置松开鼠标左键
ActionChains(self.browser).release().perform()
Start().test()
原文地址:https://www.cnblogs.com/NFii/p/11707159.html
时间: 2024-11-10 11:09:27