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/

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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.js" ></script>

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• WebScrapper.get() will return you the content of the provided url in a String.

• WebScrapper.gethtml() will return you the content of the provided url as Parsed DOM. ( Will get the html and Parse it as a DOM object . Will return you a #Document)

• WebScrapper.getjson() will return you the content of the provided url as Parsed JSON.

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To Get HTML/Text/Content of Any Website in a String.

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

This will be return the HTML/Text inside the webpage in a String.

Try this

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To Get HTML Content of Any Website in DOM Parsed Form WebScrapper.gethtml()

var url = 'https://google.com/';
var html = WebScrapper.gethtml(url);//html of the url will be Parsed and stored in this variable
console.log(html);
console.log(html.title);//As you Use document.title you can Use Like this to get the title.

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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 in a string

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 a string.
console.log(html);

This will be return the HTML/Text inside the webpage in a String.

Try this

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Use WebScrapper.fetchhtml() to get the Parsed HTML/DOM document as WebScrapper.gethtml().

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

Try this

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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);
console.log(json.id);

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

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Other Features

WebScrapper.getparam() get URL Parameters

Assuming your Current URL is https://example.com/?id=7.

let id = WebScrapper.getparam('id');
console.log(id);//Will Return "7" .

Use Custom string instead of current URL

let id = WebScrapper.getparam('id','https://example.com/?id=20');
console.log(id);//Will Return "20" .

WebScrapper.getRandomInt() get random integer in range

This function take 2 parameter WebScrapper.getRandomInt(min,max) the generated number will be in between min and max.

let id = WebScrapper.getRandomInt(10,100);
console.log(id);//Will Return a number between 10 and 100 .

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