JavaScript: URL

Url.js

Here you will have to implement the Url class that could extract parts of it from the HTTP address represented by a string. Export this class as default.

The class must contain a constructor and the following methods:

• constructor takes an HTTP address as a string
• getScheme() returns the data transfer protocol (without a colon)
• getHostName() returns the hostname
• getQueryParams() returns the request parameters as object key-value pairs
• getQueryParam() retrieves the value of the request parameter by name. If the given parameter does not exist, the method returns the value specified by the second parameter (null by default)

While doing the exercise, you will need to work through the documentation and explore the capabilities of the URL class in order to parse the HTTP address string representation.

Examples

const url = new Url('http://google.com:80?key=value&key2=value2');
url.getScheme(); // 'http'
url.getHostName(); // 'google.com'
url.getQueryParams();
// {
//   key: 'value',
//   key2: 'value2',
// };
url.getQueryParam('key'); // 'value'
// the second parameter is a default value
url.getQueryParam('key2', 'lala'); // 'value2'
url.getQueryParam('new', 'ehu'); // 'ehu'
url.getQueryParam('new'); // null

Tips

• Do not use deprecated language methods (parse, format, etc.)
• Examine the URLSearchParams class methods to handle the query string