WebScrapperJS
WebScrapperJS - Get Content/HTML of any website without being blocked by CORS even using JavaScript by WhollyAPI
Grab the CDN or Download the JavaScript File
<script src="https://cdn.jsdelivr.net/gh/SH20RAJ/WebScrapperJS/WebScrapper.js" ></script>
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.
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.
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.
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
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);
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.
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);
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 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
Other Features
WebScrapper.getparam() get URL Parameters
Assuming your Current URL is https://example.com/?id=7.
let id = getparam('id');
console.log(id);//Will Return "7" .
Use Custom string instead of current URL
let id = getparam('id','https://example.com/?id=20');
console.log(id);//Will Return "20" .
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