前言
不多说，继续猿人学的
分析
打开网站：
然后看到
关键点
python代码实现
结语
前言
不多说，继续猿人学的
分析
打开网站：
然后可以看到，就是一堆图片:
然后抓包到的接口如下，还是一如既往的就拿到了：
再看，请求接口里没有请求参数，请求头里也没有奇奇怪怪的参数：
你以为这么简单？看看返回结果：
不用说，info的值就是前端展示的字段了，然后字符串有点长，直接点那个【copy】吧：
放到本地看看，格式化以后如下：
复制其中一个img标签看看里面有啥：
<img
src=\"data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAABUAAAAcCAYAAACOGPReAAAAAXNSR0IArs4c6QAAAARnQU1BAACxjwv8YQUAAAAJcEhZcwAADsQAAA7EAZUrDhsAAAD0SURBVEhLY5RVUPvPQGXABKUJAA2GvzP3Mry6f5PhzfI4qBhuQNjQqD6Gzxc3Mjxzk2H4CRUiBHAYKs3wP7Gd4evhSwzPWr0ZPvBBhYkEqIaaBzL8nrmS4f3FfQzP6oIY3smwM/yFSpECkAyNY/g2q4PhhZsBwxegy/5BRZlfP2HgIdbfUIAnTD8xCGyuZ5AyW8jATqmhzD9fMwjumsUgbWPKwJu3AipKGkAydC8DZ24sg5SGDQNPei8D01OoMBkAydCnDIyHTkHZlAE8YUo+GDWU+mAEGfq46RCUhQAUGypbZwdlIcBoRFEfjBpKfUADQxkYAKYHOb9g+7HMAAAAAElFTkSuQmCC\"
class=\"img_number 9a870b2668709c7b49d3dceb3079e403\" style=\"left:34.5px\">
对前端有点了解的应该都知道这个img标签是啥了，就不多说了，src里的base64就是图片了，然后style就是控制显示样式的，它这里的只是控制了位置，然后class就有点和网页展示的不一样：
网页源码里显示的class类名只有个img_number，而返回的数据里多了一串9a870b2668709c7b49d3dceb3079e403，这个是啥东西呢？看看接口的调用栈：
关键点
点进去，就看到如下的，
直译的意思就是，把有些标签设置了不展示，再看源码，确实有不展示的
注意这一段：
var j_key = '.' + hex_md5(btoa(data.key + data.value).replace(/=/g, ''));
$(j_key).css('display', 'none');
看下这个j_key是啥：
提示没有data变量，这里的data就是接口返回的结果，而，data.key和data.value其实就是刚才我们看到的：
那行，定义了再看下：
补充下
btoa和atob是window对象的两个函数，其中btoa是binary to ascii，用于将binary的数据用ascii码表示，即Base64的编码过程，而atob则是ascii to binary，用于将ascii码解析成binary数据
那我们只要把返回的结果里j_key去掉，剩下的才是实际的数据就行了，拿着这个去返回结果里搜下看看，有18个
而总共有68个img标签
那拿着这个j_key【b41f7c1c3d536b1b9f8afeeff0f7292f】搜有多少个，一个都没有
那就说明，j_key并不是固定的，而是根据返回每次返回的那个key和value生成的那个编码活动的隐藏了，那么不出意外一定只有两个值，一个显示的，一个不显示的，一个个查看之后果然就下面两个：
第一个有39个：
第二个有29个：
加起来确实是总数的68，行，这里完毕了。
然后的问题的是，这里的图片的数字怎么拿出来呢？
你可以用ocr提取，但是这个题的规则是你不能使用ocr和ai识别，那还是得从代码方向下手了，首先观察可以得知肯定是10进制的：
也就是总共就0-9了，那么那些base64的值是不是固定的呢？如果不是固定的又怎么办呢？
先找个0吧，点下调试工具的这个，
然后鼠标放到【0】的位置：
把src里的数值复制出来：
data:image/png;base64,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
然后找下这个有多少个，一搜，能够跟展示的数字对上，确实是4个
当然这里是运气好，假如有display:none可能就对不上了，不过这个都是小问题。
那么这里我们可以大胆的猜测每个数字一定是固定的，然后我们只需要找到这个base64的编码表，对应出1到9即可
插一句，如果不是固定的怎么办，那得多请求几次，然后找到一个规律，也就是这个映射表可能大一点而已，但一定也是有一套规律的
这里经过我的发现，搞出了如下的映射表：
number_dict = {
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}
ok，数字搞定，那么还有个问题，这个顺序怎么展示的呢？再看源码，也就是刚才说的那个style的left:
先补充下，这个left，值越大会靠右，越小会靠左
首先，肯定是4个或者3个数一组，注意到的是，一个数字的占位是11px的横向距离，也就是宽度
加上，它把css设置了relative，即相对定位
然后看left的值和顺序对照着看，这个就是css的偏移量了，就差这一步了，兄弟们
首先初始是0px，即以0px作为参照物，一个img宽度是11px，
第一个23px的8，先方下，左边预留了2个位置的img 【8】
第二个-11.5px的6，因为设置了relative，从8的位置开始往左放1个位置，此时 【6，8】
第3个11.5px，从最右边的8开始往右放一个位置，【6，8，1】
最后一个-23px，从最右边1开始往左放两个位置，【6，0，8，1】
排列起来就是，6081
再看第二组：
从0px开始，
11.5px，先放下，左边预留一个位置，【9】
23px，从9开始，其实此时，只要它这个值不是负数，就一定在【9】的右边，因为relative定位，存放右边，【9，4】
-23px，从最右边的4开始往左左边两个位置【2，9，4】
-11px，从最右边的4开始往左边放一个位置，【2，9，1，4】
反正就是这个规律，可能我表述得不够清楚，虽然以前写过前端代码，但是并不是专门搞前端的，后面的就不跟着再分析了，直接通过这个规律，写一个方法
再代码确认下，请求第二页和第三页是否能正常返回：
确实没毛病哈
python代码实现
代码
import requests
from bs4 import BeautifulSoup
from lxml import etree
from hashlib import md5
import base64
url = 'https://match.yuanrenxue.com/api/match/4?page=2'
headers = {
'accept': 'application/json, text/javascript, */*; q=0.01',
'accept-encoding': 'gzip, deflate, br',
'accept-language': 'zh-CN,zh;q=0.9',
'cache-control': 'no-cache',
'pragma': 'no-cache',
'user-agent': 'yuanrenxue.project',
'x-requested-with': 'XMLHttpRequest',
'cookie': 'sessionid=换成你的sessionid'
}
number_dict = {
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}
def control_css_left(number_style, num_list):
"""
通过偏移量移位，重新排序
:param number_style:
:param num_list:
:return:
"""
number_style = [round(__ / 11, 1) for __ in number_style]
order_list = [None] * len(number_style)
for index, number in enumerate(number_style):
flag = int(index + number)
temp_number = num_list[index]
if order_list[flag]: # 如果该索引位置已有，向后移动一位
order_list[flag + 1] = str(temp_number)
else:
order_list[flag] = str(temp_number)
# print(order_list)
return order_list
def get_j_key(key, value):
flag = base64.b64encode((key + value).encode('ascii')).decode('utf-8').replace('=', '')
m = md5()
m.update(flag.encode('utf-8'))
return m.hexdigest()
def parser_info(info, j_key):
html = etree.HTML(info)
data = html.xpath('//td')
cont = []
for index, item in enumerate(data):
img_flag = item.xpath('.//img')
temp_cont = []
# print('index', index)
for im in img_flag:
src = im.xpath('./@src')
src = ''.join(src) if src else ''
src_number = number_dict.get(src, None)
if src_number is None:
continue
# print(12312323, src_number)
if j_key in im.attrib.get('class'):
continue
left = im.xpath('./@style')
left = ''.join(left) if left else ''
left = left.replace('left:', '').replace('px', '')
temp_cont.append({float(left): src_number})
cont.append(temp_cont)
# print(1111,cont)
return cont
def fetch(page):
url = f'https://match.yuanrenxue.com/api/match/4?page={page}'
req = requests.get(url, headers=headers)
res = req.json()
req.close()
key = res.get('key')
value = res.get('value')
info = res.get('info')
# print(12312313, key, value, info)
j_key = get_j_key(key, value)
# print(123123, j_key)
cont = parser_info(info, j_key)
# 转换偏移量
result = []
for item in cont:
left = [list(_.keys())[0] for _ in item]
number = [list(_.values())[0] for _ in item]
temp = control_css_left(left, number)
temp = int(''.join(temp))
result.append(temp)
print(1231312, result)
return result
def get_result():
sum_res = 0
for i in range(1, 6):
result = fetch(i)
sum_res += sum(result)
print('答案是：', sum_res)
get_result()
执行：
放到平台提交
完美
结语
这个题实话说不难，主要你可能得懂一点前端的知识才行，有个点要注意的是，在还原实际顺序的时候，如果索引位置上已有值，需要向后移一位
本文章为转载内容，我们尊重原作者对文章享有的著作权。如有内容错误或侵权问题，欢迎原作者联系我们进行内容更正或删除文章。
