WebScrapperJS - Get Content/HTML of any website without being blocked by CORS even using JavaScript by WhollyAPI

WebScrapperJS

WebScrapperJS - Get Content/HTML of any website without being blocked by CORS even using JavaScript by WhollyAPI

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Website :- https://sh20raj.github.io/WebScrapperJS/

GitHub | Repl.it | Dev.to Article

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Grab the CDN or Download the JavaScript File

<script src="https://cdn.jsdelivr.net/gh/SH20RAJ/WebScrapperJS/WebScrapper.min.js" ></script>

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To Get HTML/Text Content of Any Website WebScrapper.gethtml() or WebScrapper.get()

var url = 'https://google.com/';
var html = WebScrapper.gethtml(url);//html of the url will be stored in this variable
console.log(html);

WebScrapper.gethtml() or WebScrapper.get() both are similar.

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Intialise own WebScrapper with URL new scrapper()

let MyWebScrapper = new scrapper('https://example.com/');
//You can now directly call gethtml() instead of passing a url into it.

console.log(MyWebScrapper.gethtml()); //Grab https://example.com/ and print on console

Still you can Use new created scrapper MyWebScrapper for grabbing new URLs. Like

let MyWebScrapper = new scrapper('https://example.com/');
//You can now directly call gethtml() instead of passing a url into it.

console.log(MyWebScrapper.gethtml()); //Grab https://example.com/ and print on console

console.log(MyWebScrapper.gethtml('https://example.com/')); //Grab https://youtube.com/ and print on console

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You can also fetch JSON Using WebScrapperJS

var json = WebScrapper.getjson('https://jsonplaceholder.typicode.com/todos/1');//Return result direct in json format
console.log(json);

Try This

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Getting Result more Faster

Use the Below codes/methods only if the origin or feching URL is not blocked by CORS Like this

if your origin is not blocking you then you must use the below fetch() code instead of gethtml() directly.
because it returns the results faster without using API.It will directly fetch origin using AJAX.

Use WebScrapper.fetch() to get the html/text

We will use this url https://webscrapperjs.sh20raj.repl.co/ because it is not blocked.

var html = WebScrapper.fetch('https://webscrapperjs.sh20raj.repl.co/');//This will be return the HTML/Text inside the webpage
console.log(html);

Try this

Use WebScrapper.fetchjson() to get the Parsed JSON

var json = WebScrapper.fetchjson('https://webscrapperjs.sh20raj.repl.co/sample.json');//This will be return the JSON inside the webpage. 
console.log(json);

Try this

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Try this on Codepen

Sample Code | Codepen :- https://codepen.io/SH20RAJ/pen/VwrwjXJ?editors=1001

<div id="scrappedcontent"></div>

<script src="https://cdn.jsdelivr.net/gh/SH20RAJ/WebScrapperJS/WebScrapper.min.js" ></script> 
<script>
  let MyWebScrapper = new scrapper('https://google.com/');
//You can now directly call gethtml() instead of passing a url into it.

console.log(MyWebScrapper.gethtml()); //Grab https://example.com/ and print on console
var html = MyWebScrapper.gethtml('https://example.com/');

console.log(html); //Grab https://youtube.com/ and print on console

document.getElementById('scrappedcontent').innerHTML = html;
</script>

See Results Here