前言
不多說，繼續猿人學的
分析
打開網站：
然後可以看到，就是一堆圖片:
然後抓包到的接口如下，還是一如既往的就拿到了：
再看，請求接口裏沒有請求參數，請求頭裏也沒有奇奇怪怪的參數：
你以爲這麼簡單？看看返回結果：
不用說，info的值就是前端展示的字段了，然後字符串有點長，直接點那個【copy】吧：
放到本地看看，格式化以後如下：
複製其中一個img標籤看看裏面有啥：
<img
src=\"data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAABUAAAAcCAYAAACOGPReAAAAAXNSR0IArs4c6QAAAARnQU1BAACxjwv8YQUAAAAJcEhZcwAADsQAAA7EAZUrDhsAAAD0SURBVEhLY5RVUPvPQGXABKUJAA2GvzP3Mry6f5PhzfI4qBhuQNjQqD6Gzxc3Mjxzk2H4CRUiBHAYKs3wP7Gd4evhSwzPWr0ZPvBBhYkEqIaaBzL8nrmS4f3FfQzP6oIY3smwM/yFSpECkAyNY/g2q4PhhZsBwxegy/5BRZlfP2HgIdbfUIAnTD8xCGyuZ5AyW8jATqmhzD9fMwjumsUgbWPKwJu3AipKGkAydC8DZ24sg5SGDQNPei8D01OoMBkAydCnDIyHTkHZlAE8YUo+GDWU+mAEGfq46RCUhQAUGypbZwdlIcBoRFEfjBpKfUADQxkYAKYHOb9g+7HMAAAAAElFTkSuQmCC\"
class=\"img_number 9a870b2668709c7b49d3dceb3079e403\" style=\"left:34.5px\">
對前端有點了解的應該都知道這個img標籤是啥了，就不多說了，src裏的base64就是圖片了，然後style就是控制顯示樣式的，它這裏的只是控制了位置，然後class就有點和網頁展示的不一樣：
網頁源碼裏顯示的class類名只有個img_number，而返回的數據裏多了一串9a870b2668709c7b49d3dceb3079e403，這個是啥東西呢？看看接口的調用棧：
關鍵點
點進去，就看到如下的，
直譯的意思就是，把有些標籤設置了不展示，再看源碼，確實有不展示的
注意這一段：
var j_key = '.' + hex_md5(btoa(data.key + data.value).replace(/=/g, ''));
$(j_key).css('display', 'none');
看下這個j_key是啥：
提示沒有data變量，這裏的data就是接口返回的結果，而，data.key和data.value其實就是剛纔我們看到的：
那行，定義了再看下：
補充下
btoa和atob是window對象的兩個函數，其中btoa是binary to ascii，用於將binary的數據用ascii碼錶示，即Base64的編碼過程，而atob則是ascii to binary，用於將ascii碼解析成binary數據
那我們只要把返回的結果裏j_key去掉，剩下的纔是實際的數據就行了，拿着這個去返回結果裏搜下看看，有18個
而總共有68個img標籤
那拿着這個j_key【b41f7c1c3d536b1b9f8afeeff0f7292f】搜有多少個，一個都沒有
那就說明，j_key並不是固定的，而是根據返回每次返回的那個key和value生成的那個編碼活動的隱藏了，那麼不出意外一定只有兩個值，一個顯示的，一個不顯示的，一個個查看之後果然就下面兩個：
第一個有39個：
第二個有29個：
加起來確實是總數的68，行，這裏完畢了。
然後的問題的是，這裏的圖片的數字怎麼拿出來呢？
你可以用ocr提取，但是這個題的規則是你不能使用ocr和ai識別，那還是得從代碼方向下手了，首先觀察可以得知肯定是10進制的：
也就是總共就0-9了，那麼那些base64的值是不是固定的呢？如果不是固定的又怎麼辦呢？
先找個0吧，點下調試工具的這個，
然後鼠標放到【0】的位置：
把src裏的數值複製出來：
data:image/png;base64,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
然後找下這個有多少個，一搜，能夠跟展示的數字對上，確實是4個
當然這裏是運氣好，假如有display:none可能就對不上了，不過這個都是小問題。
那麼這裏我們可以大膽的猜測每個數字一定是固定的，然後我們只需要找到這個base64的編碼表，對應出1到9即可
插一句，如果不是固定的怎麼辦，那得多請求幾次，然後找到一個規律，也就是這個映射表可能大一點而已，但一定也是有一套規律的
這裏經過我的發現，搞出了如下的映射表：
number_dict = {
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}
ok，數字搞定，那麼還有個問題，這個順序怎麼展示的呢？再看源碼，也就是剛纔說的那個style的left:
先補充下，這個left，值越大會靠右，越小會靠左
首先，肯定是4個或者3個數一組，注意到的是，一個數字的佔位是11px的橫向距離，也就是寬度
加上，它把css設置了relative，即相對定位
然後看left的值和順序對照着看，這個就是css的偏移量了，就差這一步了，兄弟們
首先初始是0px，即以0px作爲參照物，一個img寬度是11px，
第一個23px的8，先方下，左邊預留了2個位置的img 【8】
第二個-11.5px的6，因爲設置了relative，從8的位置開始往左放1個位置，此時 【6，8】
第3個11.5px，從最右邊的8開始往右放一個位置，【6，8，1】
最後一個-23px，從最右邊1開始往左放兩個位置，【6，0，8，1】
排列起來就是，6081
再看第二組：
從0px開始，
11.5px，先放下，左邊預留一個位置，【9】
23px，從9開始，其實此時，只要它這個值不是負數，就一定在【9】的右邊，因爲relative定位，存放右邊，【9，4】
-23px，從最右邊的4開始往左左邊兩個位置【2，9，4】
-11px，從最右邊的4開始往左邊放一個位置，【2，9，1，4】
反正就是這個規律，可能我表述得不夠清楚，雖然以前寫過前端代碼，但是並不是專門搞前端的，後面的就不跟着再分析了，直接通過這個規律，寫一個方法
再代碼確認下，請求第二頁和第三頁是否能正常返回：
確實沒毛病哈
python代碼實現
代碼
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=換成你的sessionid'
}
number_dict = {
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}
def control_css_left(number_style, num_list):
"""
通過偏移量移位，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]: # 如果該索引位置已有，向後移動一位
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)
# 轉換偏移量
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()
執行：
放到平臺提交
完美
結語
這個題實話說不難，主要你可能得懂一點前端的知識才行，有個點要注意的是，在還原實際順序的時候，如果索引位置上已有值，需要向後移一位