Prasing a site how to log in through VK?

S

Sergey Yavin2020-02-03 09:33:50

Python

Sergey Yavin, 2020-02-03 09:33:50

How to parse a site where you need to log in only through VK. I use requests and bs4. When parsing the page, it gives the following result.

<html lang="en">
<head><meta charset="utf-8"/>
<meta content="width=device-width,initial-scale=1,shrink-to-fit=no" name="viewport"/>
<meta content="e1bab0387497028b" name="yandex-verification">
<meta content="#000000" name="theme-color"/>
<link href="img/ios/touch-icon-iphone-60x60.png" rel="apple-touch-icon"/>
<link href="img/ios/touch-icon-ipad-76x76.png" rel="apple-touch-icon" sizes="60x60"/>
<link href="img/ios/touch-icon-iphone-retina-120x120.png" rel="apple-touch-icon" sizes="114x114"/>
<link href="img/ios/touch-icon-ipad-retina-152x152.png" rel="apple-touch-icon" sizes="144x144"/>
<link href="/manifest.json" rel="manifest"/><link href="/favicon.ico" rel="shortcut icon"/><title>BKStat</title>
<link href="/static/css/2.8d290a45.chunk.css" rel="stylesheet"/>
<link href="/static/css/main.df53a29f.chunk.css" rel="stylesheet"/>
</meta>
</head>
<body>
<noscript>You need to enable JavaScript to run this app.</noscript>
<div id="root"></div>
<script>!function(l){function e(e){for(var r,t,n=e[0],o=e[1],u=e[2],f=0,i=[];f<n.length;f++)t=n[f],p[t]&&i.push(p[t][0]),p[t]=0;for(r in o)Object.prototype.hasOwnProperty.call(o,r)&&(l[r]=o[r]);for(s&&s(e);i.length;)i.shift()();return c.push.apply(c,u||[]),a()}function a(){for(var e,r=0;r<c.length;r++){for(var t=c[r],n=!0,o=1;o<t.length;o++){var u=t[o];0!==p[u]&&(n=!1)}n&&(c.splice(r--,1),e=f(f.s=t[0]))}return e}var t={},p={1:0},c=[];function f(e){if(t[e])return t[e].exports;var r=t[e]={i:e,l:!1,exports:{}};return l[e].call(r.exports,r,r.exports,f),r.l=!0,r.exports}f.m=l,f.c=t,f.d=function(e,r,t){f.o(e,r)||Object.defineProperty(e,r,{enumerable:!0,get:t})},f.r=function(e){"undefined"!=typeof Symbol&&Symbol.toStringTag&&Object.defineProperty(e,Symbol.toStringTag,{value:"Module"}),Object.defineProperty(e,"__esModule",{value:!0})},f.t=function(r,e){if(1&e&&(r=f(r)),8&e)return r;if(4&e&&"object"==typeof r&&r&&r.__esModule)return r;var t=Object.create(null);if(f.r(t),Object.defineProperty(t,"default",{enumerable:!0,value:r}),2&e&&"string"!=typeof r)for(var n in r)f.d(t,n,function(e){return r[e]}.bind(null,n));return t},f.n=function(e){var r=e&&e.__esModule?function(){return e.default}:function(){return e};return f.d(r,"a",r),r},f.o=function(e,r){return Object.prototype.hasOwnProperty.call(e,r)},f.p="/";var r=window.webpackJsonp=window.webpackJsonp||[],n=r.push.bind(r);r.push=e,r=r.slice();for(var o=0;o<r.length;o++)e(r[o]);var s=n;a()}([])</script><script src="/static/js/2.b66ff35e.chunk.js">
</script><script src="/static/js/main.eced3301.chunk.js">
</script>
</body>
</html>

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1 answer(s)

S

Skid330, 2020-02-03
@sjaserds

Alternatively, you can use Selenium
https://habr.com/en/post/248559/