Python+selenium破解拼图验证码的脚本
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实现思路
很多网站都有拼图验证码
1.首先要了解拼图验证码的生成原理
2.制定破解计划，考虑其可能性和成功率。
3.编写脚本
很多网站的拼图验证码都是直接借助第三方插件，也就是一类一种解法。
笔者遇到的这种拼图验证码实际上是多个小碎片经过重新组合成的一张整体，首先要在网站上抓取这种小碎片图片并下载到本地
我们先捋一捋大体思路：
获取所有碎片图片----找出他们的排列顺序逻辑-----找出他们中含有颜色深的真正位置的那个小碎块的序号-----根据每块碎片的宽度和上下和这个深色小块的序号算出距离----用selenium向右移动滑块到这个距离
核心代码
先导入必要的包：
# -*- coding:utf-8 -*-
import colorsys
import urllib,os,uuid,re,time
from PIL import Image
from selenium.webdriver.common.action_chains import ActionChains
from selenium import webdriver抓到并下载到本地
def create(locapath,fileName):
filePath=locapath+'/'+fileName
if not os.path.exists(filePath):
file=open(filePath,'a+')
file.close()
return filePath
def downloadImg():
list = openBrowser()
for i in range(2):
fileName = str(i) + '_test.jpg'
urllib.urlretrieve(list[i], create('/Users/zijiawang/Downloads/', fileName)) #下载到本地
time.sleep(3)
def openBrowser():
global wzj
wzj = webdriver.Firefox()
wzj.get('https://。。。。')
image1_url = wzj.find_elements_by_class_name('gt_cut_bg_slice')[0].get_attribute('style')
image1_url=image1_url[23:-38]
image2_url = wzj.find_elements_by_class_name('gt_cut_fullbg_slice')[0].get_attribute('style')
image2_url=image2_url[23:-38]
return [image1_url,image2_url]
分割并获取rgb颜色
ef getcolor(image):
list = [[x,y] for x in range(26) for y in range(2)]
listt = []
colors = []
for i in range(len(list)):
l = list[i][0]*12+12
w = list[i][1]*58+58
listt.append([12*0.25+i/2*12, 58*0.25+((i+10)%2)*58])
listt.append([12*0.25+i/2*12, 58*0.75+((i+10)%2)*58])
listt.append([12*0.75+i/2*12, 58*0.25+((i+10)%2)*58])
listt.append([12*0.75+i/2*12, 58*0.75+((i+10)%2)*58])
for i in range(len(listt)):
colors.append(image[listt[i][0],listt[i][1]])
return colors找出不同的小碎片序号
def getDeferent():
deferent = []
for i in range(208):
if abs(color1[i][2] - color2[i][2]) >20:
print u'B差值: ',color1[i][2]-color2[i][2]
deferent.append(i)
# print ([i,abs(color1[i][0] - color2[i][0]),abs(color1[i][1] - color2[i][1]),abs(color1[i][2] - color2[i][2])])
return deferent
def getUPDOWN():
deferent = getDeferent()
xlist = []
ylist = []
for i in deferent:
if (i / 4) % 2 == 0:
xlist.append(i)
else:
ylist.append(i)
uplist = []
downlist = []
for i in xlist:
uplist.append(i / 4)
for i in ylist:
downlist.append(i / 4)
for i in range(len(uplist)):
uplist[i] /= 2
for i in range(len(downlist)):
downlist[i] /= 2
func = lambda x, y: x if y in x else x + [y]
up = reduce(func, [[], ] + uplist)
func = lambda x, y: x if y in x else x + [y]
down = reduce(func, [[], ] + downlist)
return up,down编写好排列方法，会用的
def bubble(l):
for index in range(len(l) - 1, 0 , -1):
for two_index in range(index):
if l[two_index] > l[two_index + 1]:
l[two_index], l[two_index + 1] = l[two_index + 1], l[two_index]
return l最后的进行解析，拼装源码
def end():
orders = getUPDOWN()
orderx = orders[0]
ordery = orders[1]
print 'x:',orderx
print 'y:',ordery
orderX = [145, 157, 277, 265, 169, 181, 253, 241, 97, 109, 301, 289, 73, 85, 37, 25, 1, 13, 133, 121, 49, 61, 229,
217, 193, 205]
orderY = [157, 145, 265, 277, 181, 169, 241, 253, 109, 97, 289, 301, 85, 73, 25, 37, 13, 1, 121, 133, 61, 49, 217,
229, 205, 193]
print u'X:',orderX
print u'Y:',orderY
for x in range(26):
orderX[x] -= 1
orderX[x] /= 12
for y in range(26):
orderY[y] -= 1
orderY[y] /= 12
endX = []
endY = []
for i in range(len(orderX)):
for j in range(len(orderx)):
if orderx[j] == orderX[i]:
endX.append(i)
break
for i in range(len(orderY)):
for j in range(len(ordery)):
if ordery[j] == orderY[i]:
endY.append(i)
break
print 'endx:',endX
print 'endy:',endY
os.remove('/Users/zijiawang/Downloads/0_test.jpg')
os.remove('/Users/zijiawang/Downloads/1_test.jpg')
all = []
all = endX+endY
# print 'all:',all
#系统最多可实验5次
func = lambda x, y: x if y in x else x + [y]
all_old = reduce(func, [[], ] + all)
# print 'all_old:',all_old
all_end =bubble(all_old)
print 'all_end:',all_end
duandata = 1000
for i in range(1,len(all_end)):
if all_end[i]-1 in all_end:
pass
else:
duandata = all_end[i]
print u'断点为:',duandata
guess =[]
if duandata != 1000:
if endX != [] and endY!=[]:
print u'断点1'
guess = [duandata,all_end[0]]
elif endX ==[]:
print u'断点2'
guess = [duandata, endY[0]]
elif endY == []:
print u'断点3'
guess = [duandata, endX[0]]
else:
print u'无断点'
guess = [ all_end[0]]
end_guess =[]
print u'猜测的点:',guess
for i in guess:
end_guess.append(i * 12 - 38)
end_guess.append(i * 12 - 28)
end_guess.append(i * 12 - 20)
end_guess.append(i * 12 - 17)
end_guess.append(i * 12 - 15)
print u'猜测的位移量: ',end_guess
return end_guess主函数
if __name__ == '__main__':
downloadImg()
color1 = getcolor(Image.open('/Users/zijiawang/Downloads/0_test.jpg').load())
color2 = getcolor(Image.open('/Users/zijiawang/Downloads/1_test.jpg').load())
guess = end()
for i in guess:
print i,wzj.title
try:
ele = wzj.find_element_by_xpath('//div[@class="gt_slider_knob gt_show"]')
except:
ele = wzj.find_element_by_xpath('//div[@class="gt_slider_knob gt_show moving"]')
ActionChains(wzj).click_and_hold(ele).perform()
ActionChains(wzj).move_to_element_with_offset(ele, 0,i).perform()因每个网站的都大同小异，这里不写具体测试的网站了，以免对其造成压力。此算法属于高度定制的，应用其他网站需要改些参数，但是代码注释较小。需要注释，不明白的地方请留言哈。
以上就是Python+selenium破解拼图验证码的脚本的详细内容，更多关于Python selenium拼图验证码的资料请关注编程网其它相关文章！
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