笔者在本科阶段想学却一致没有学的Python爬虫,没有想到研究僧阶段刚进实验室的第一周就被安排学习了。这周笔者主要学习的有:UA黑名单饶过、JS混淆和验证码认证。其中,验证码认证是花费时间最长的,问题及代码如下:
用户根据图片输入相应的数字和字母,这种验证码出现相对较早,也较为普遍,对于Python爬虫来说,也较为简单。
解决办法式用Python的第三方库Tesserocr-OCR,代码如下:
from PIL import Image
import tesserocr
image = Image.open('./1.png')
result = tesserocr.image_to_text(image)
print(result)
虽然代码简单,但是准确率却非常受限制,比如当图片背景有很多线条的时候,识别准确率是比较低的。这个时候的解决办法是对图片转灰度再进行二值化处理,以此提高识别率,代码如下:
image = Image.open('./1.png')
image.show()
image = image.convert('L')
threshold = 127
table = []
for i in range(256):
if i < threshold:
table.append(0)
else:
table.append(1)
image = image.point(table,'1')
image.show()
result = tesserocr.image_to_text(image)
print(result)
但是这个办法也有限制,当背景纹理和字符的RGB都大于127,或者都小于127时(就是亮度接近时),准确率会很低。所以笔者觉得刚好深度学习比较火,用深度学习训练个模型,这样的识别率就会高很多。
滑动式验证码最为典型的是B站的登录界面。
解决思路是存三张图片,分别是完整的图、有缺口的图和缺口图。首先识别缺口在图中的位置,然后计算滑动的距离和轨迹。最后用selenium进行模拟操作。代码如下:
from selenium import webdriver
from selenium.webdriver.support.wait import WebDriverWait
from selenium.webdriver.support import expected_conditions as EC
from selenium.webdriver.common.by import By
from selenium.webdriver import ActionChains
from selenium.webdriver.common.keys import Keys
import time
import random
from PIL import Image
web='http://literallycanvas.com/'
#初始化
def init():
#定义全局变量
global url, browser, username, password, wait
url = 'https://passport.bilibili.com/login'
browser = webdriver.Chrome()
username = '************'
password = '************'
wait = WebDriverWait(browser, 20)
#登录
def login():
browser.get(url)
user = wait.until(EC.presence_of_element_located((By.ID, 'login-username')))
passwd = wait.until(EC.presence_of_element_located((By.ID, 'login-passwd')))
user.send_keys(username)
passwd.send_keys(password)
#通过输入回车键模仿用户登录
#passwd.send_keys(Keys.ENTER)
login_btn=wait.until(EC.presence_of_element_located((By.CSS_SELECTOR,'a.btn.btn-login')))
#随机延时点击
time.sleep(random.random()*3)
login_btn.click()
#设置元素的可见性用于截图
def show_element(element):
browser.execute_script("arguments[0].style = arguments[1]", element, "display: block;")
def hide_element(element):
browser.execute_script("arguments[0].style = arguments[1]", element, "display: none;")
#截图
def save_pic(obj, name):
try:
pic_url = browser.save_screenshot('.\\bilibili.png')
#开始获取元素位置信息
left = obj.location['x']
top = obj.location['y']
right = left + obj.size['width']
bottom = top + obj.size['height']
im = Image.open('.\\bilibili.png')
im = im.crop((left, top, right, bottom))
file_name = 'bili' + name + '.png'
im.save(file_name)
except BaseException as msg:
print("截图失败:%s" % msg)
def cut():
c_background = wait.until(EC.presence_of_element_located((By.CSS_SELECTOR, 'canvas.geetest_canvas_bg.geetest_absolute')))
c_slice = wait.until(EC.presence_of_element_located((By.CSS_SELECTOR,'canvas.geetest_canvas_slice.geetest_absolute')))
c_full_bg = wait.until(EC.presence_of_element_located((By.CSS_SELECTOR,'canvas.geetest_canvas_fullbg.geetest_fade.geetest_absolute')))
hide_element(c_slice)
save_pic(c_background, 'back')
show_element(c_slice)
save_pic(c_slice, 'slice')
show_element(c_full_bg)
save_pic(c_full_bg, 'full')
#判断元素是否相同
def is_pixel_equal(bg_image, fullbg_image, x, y):
#bg_image是缺口的图片
#fullbg_image是完整图片
bg_pixel = bg_image.load()[x, y]
fullbg_pixel = fullbg_image.load()[x, y]
threshold = 60
if (abs(bg_pixel[0] - fullbg_pixel[0] < threshold) and abs(bg_pixel[1] - fullbg_pixel[1] < threshold) and abs(bg_pixel[2] - fullbg_pixel[2] < threshold)):
return True
else:
return False
#计算滑块移动的距离
def get_distance(bg_image, fullbg_image):
distance = 57
for i in range(distance, fullbg_image.size[0]):
for j in range(fullbg_image.size[1]):
if not is_pixel_equal(fullbg_image, bg_image, i, j):
return i
#构造滑动轨迹
def get_trace(distance):
#distance是缺口离滑块的距离
trace = []
faster_distance = distance*(4/5)
start, v0, t = 0, 0, 0.2
while start < distance:
if start < faster_distance:
a = 1.5
else:
a = -3
move = v0 * t + 1 / 2 * a * t * t
v = v0 + a * t
v0 = v
start += move
trace.append(round(move))
return trace
#模拟拖动
def move_to_gap(trace):
slider=wait.until(EC.presence_of_element_located((By.CSS_SELECTOR,'div.geetest_slider_button')))
# 使用click_and_hold()方法悬停在滑块上,perform()方法用于执行
ActionChains(browser).click_and_hold(slider).perform()
for x in trace:
# 使用move_by_offset()方法拖动滑块,perform()方法用于执行
ActionChains(browser).move_by_offset(xoffset=x, yoffset=0).perform()
time.sleep(0.5)
ActionChains(browser).release().perform()
def slide():
distance=get_distance(Image.open('.\\bili_back.png'),Image.open('.\\bili_full.png'))
trace = get_trace(distance-5)
move_to_gap(trace)
time.sleep(3)
init()
login()
cut()
slide()
最常见的点击式验证码有12306、简书等。此处笔者以简书为例。解决思路的为:获取点击式图片的信息——调用第三方识别库——获取第三方返回的坐标——用selenium模拟用户点击。笔者用的第三方识别是超级鹰,这是一个付费的软件,但是注册后关注公众号有免费的测试额度,足够做测试使用了。代码分为两个部分,一个是超级鹰API接口,另一个是上述一系列操作。代码如下:
import requests
from hashlib import md5
class Chaojiying(object):
def __init__(self, username, password, soft_id):
self.username = username
self.password = md5(password.encode('utf-8')).hexdigest()
self.soft_id = soft_id
self.base_params = {
'user': self.username,
'pass2': self.password,
'softid': self.soft_id,
}
self.headers = {
'Connection': 'Keep-Alive',
'User-Agent': 'Mozilla/4.0 (compatible; MSIE 8.0; Windows NT 5.1; Trident/4.0)',
}
def post_pic(self, im, codetype):
"""
im: 图片字节
codetype: 题目类型 参考 http://www.chaojiying.com/price.html
"""
params = {
'codetype': codetype,
}
params.update(self.base_params)
files = {'userfile': ('ccc.jpg', im)}
r = requests.post('http://upload.chaojiying.net/Upload/Processing.php', data=params, files=files,
headers=self.headers)
return r.json()
# 验证不通过,请求该函数 , 后台 则对该次判断不做扣分处理
def report_error(self, im_id):
"""
im_id:报错题目的图片ID
"""
params = {
'id': im_id,
}
params.update(self.base_params)
r = requests.post('http://upload.chaojiying.net/Upload/ReportError.php', data=params, headers=self.headers)
return r.json()
import time
from PIL import Image
from selenium import webdriver
from selenium.webdriver import ActionChains
def crack():
# 保存网页截图
browser.save_screenshot('222.jpg')
# 获取 验证码确定按钮
button = browser.find_element_by_xpath(xpath='//div[@class="geetest_panel"]/a/div')
# 获取 验证码图片的 位置信息
img1 = browser.find_element_by_xpath(xpath='//div[@class="geetest_widget"]')
location = img1.location
size = img1.size
top, bottom, left, right = location['y'], location['y'] + size['height'], location['x'], location['x'] + size[
'width']
print('图片的宽:', img1.size['width'])
print(top, bottom, left, right)
# 根据获取的验证码位置信息和网页图片 对验证码图片进行裁剪 保存
img_1 = Image.open('222.jpg')
capcha1 = img_1.crop((left, top, right, bottom-54))
capcha1.save('tu1-1.png')
# 接入超级鹰 API 获取图片中的一些参数 (返回的是一个字典)
cjy = Chaojiying('*********', '************', '900751')
im = open('tu1-1.png', 'rb').read()
content = cjy.post_pic(im, 9004)
print(content)
# 将图片中汉字的坐标位置 提取出来
positions = content.get('pic_str').split('|')
locations = [[int(number)for number in group.split(",")] for group in positions]
print(positions)
print(locations)
# 根据获取的坐标信息 模仿鼠标点击验证码图片
for location1 in locations:
print(location1)
ActionChains(browser).move_to_element_with_offset(img1 , location1[0],location1[1]).click().perform()
time.sleep(1)
button.click()
time.sleep(1)
# 失败后重试
lower = browser.find_element_by_xpath('//div[@class="geetest_table_box"]/div[2]').text
print('判断', lower)
if lower != '验证失败 请按提示重新操作'and lower != None:
print('登录成功')
time.sleep(3)
else:
time.sleep(3)
print('登录失败')
# 登录失败后 , 调用 该函数 , 后台 则对该次判断不做扣分处理
pic_id = content.get('pic_id')
print('图片id为:',pic_id)
cjy = Chaojiying('********', '**********', '900751')
cjy.report_error(pic_id)
crack()
if __name__ == '__main__':
patn = 'chromedriver.exe'
browser = webdriver.Chrome(patn)
browser.get('https://www.jianshu.com/sign_in')
browser.save_screenshot('lodin.png')
# 填写from表单 点击登陆 获取验证码 的网页截图
login = browser.find_element_by_id('sign-in-form-submit-btn')
username = browser.find_element_by_id('session_email_or_mobile_number')
password = browser.find_element_by_id('session_password')
username.send_keys('***********')
password.send_keys('***********')
login.click()
time.sleep(5)
crack()
宫格验证码主要是指微博曾经使用的四宫格验证码,但是现在该验证码已经取消了。笔者仍然了解了一些这种验证码的破解办法——枚举。因为四宫格只用24种可能,先用代码获取所有的情况后,再手动输入对应每张图片的滑动顺寻,最后再用selenium模拟。