URL class for parsing in Node.js - Mini Project: Build & Apply

URL class for parsing

📖 Scenario: You are building a simple tool to analyze website addresses (URLs). You want to break down a URL into parts like the protocol, hostname, and pathname.

🎯 Goal: Create a Node.js script that uses the built-in URL class to parse a given URL string and extract its components.

📋 What You'll Learn

Create a variable with a URL string

Create a URL object from the string

Extract the protocol, hostname, and pathname from the URL object

Print or store these parts for further use

💡 Why This Matters

🌍 Real World

Parsing URLs is common when working with web addresses, APIs, or routing in web applications.

💼 Career

Understanding how to use the URL class helps in backend and frontend development for handling links, redirects, and data extraction from URLs.

1

Create a URL string variable

Create a variable called myUrl and set it to the string "https://example.com/path/page.html".

Node.js

// Create a variable called myUrl with the URL string
// Your code here

Hint

Use const to create a variable named myUrl and assign the URL string to it.

2

Create a URL object from the string

Create a variable called parsedUrl and set it to a new URL object created from the myUrl string.

Node.js

const myUrl = "https://example.com/path/page.html";
// Create a URL object called parsedUrl from myUrl
// Your code here

Hint

Use new URL(myUrl) to create the URL object and assign it to parsedUrl.

3

Extract protocol, hostname, and pathname

Create three variables: protocol, hostname, and pathname. Assign each the corresponding property from parsedUrl: protocol, hostname, and pathname.

Node.js

const myUrl = "https://example.com/path/page.html";
const parsedUrl = new URL(myUrl);
// Extract protocol, hostname, and pathname from parsedUrl
// Your code here

Hint

Use dot notation to get protocol, hostname, and pathname from parsedUrl.

4

Log the extracted parts

Add three console.log statements to print protocol, hostname, and pathname each on its own line.

Node.js

const myUrl = "https://example.com/path/page.html";
const parsedUrl = new URL(myUrl);
const protocol = parsedUrl.protocol;
const hostname = parsedUrl.hostname;
const pathname = parsedUrl.pathname;
// Log protocol, hostname, and pathname
// Your code here

Hint

Use console.log() to print each variable on its own line.

Practice

1. What does the URL class in Node.js primarily help you do?

easy

A. Break down and work with parts of a web address easily

B. Create new web servers

C. Encrypt data sent over the internet

D. Manage file system paths

Solution

1. Step 1: Understand the purpose of the URL class

   The URL class is designed to parse and handle web addresses, making it easy to access parts like hostname, pathname, and query.

2. Step 2: Compare with other options

   Creating servers, encrypting data, and managing file paths are unrelated to URL parsing.

3. Final Answer:

   Break down and work with parts of a web address easily -> Option A

4. Quick Check:

   URL class = parse web address [OK]

Hint: URL class = split web address parts easily [OK]

Common Mistakes:

• Confusing URL class with server creation
• Thinking URL class encrypts data
• Mixing URL class with file system modules

2. Which of the following is the correct way to create a new URL object for the address https://example.com/path?name=abc in Node.js?

easy

A. const url = url.parse('https://example.com/path?name=abc');

B. const url = URL('https://example.com/path?name=abc');

C. const url = new URL('https://example.com/path?name=abc');

D. const url = new URL.parse('https://example.com/path?name=abc');

Solution

1. Step 1: Recall the correct syntax for creating a URL object

   The URL class requires the new keyword and a string argument: new URL(string).

2. Step 2: Check each option

   const url = new URL('https://example.com/path?name=abc'); uses correct syntax. const url = URL('https://example.com/path?name=abc'); misses new. const url = url.parse('https://example.com/path?name=abc'); uses old url.parse method, not the URL class. const url = new URL.parse('https://example.com/path?name=abc'); incorrectly combines new and URL.parse.

3. Final Answer:

   const url = new URL('https://example.com/path?name=abc'); -> Option C

4. Quick Check:

   Use new URL() to create URL objects [OK]

Hint: Always use 'new URL()' to create URL objects [OK]

Common Mistakes:

• Omitting the 'new' keyword
• Using old url.parse() instead of URL class
• Trying to call URL as a function without 'new'

3. What will be the output of this Node.js code?

const url = new URL('https://example.com:8080/path/page?query=123#section');
console.log(url.hostname);
console.log(url.port);
console.log(url.pathname);
console.log(url.hash);

medium

A. example.com

/path/page
section

B. https://example.com
8080
path/page
section

C. example.com:8080

/path/page
#section

D. example.com
8080
/path/page
#section

Solution

1. Step 1: Understand URL properties

   hostname gives domain without port, port gives port number, pathname gives path starting with '/', hash includes '#' plus fragment.

2. Step 2: Match values from the URL

   Hostname is 'example.com', port is '8080', pathname is '/path/page', hash is '#section'.

3. Final Answer:

   example.com 8080 /path/page #section -> Option D

4. Quick Check:

   URL parts match output A [OK]

Hint: hostname excludes port; hash includes '#' [OK]

Common Mistakes:

• Including port in hostname
• Missing leading slash in pathname
• Omitting '#' in hash

4. Consider this code snippet:

const url = new URL('https://example.com/path');
url.hostname = 'newsite.com';
url.port = 3000;
url.pathname = 'newpath';
console.log(url.href);

What is the output?

medium

A. https://newsite.com:3000//newpath

B. https://newsite.com:3000/newpath

C. https://newsite.com:3000/path

D. https://newsite.com:3000/newpath/

Solution

1. Step 1: Understand pathname assignment

   Setting url.pathname = 'newpath' normalizes the path by adding a leading slash, resulting in '/newpath'.

2. Step 2: Construct the full URL

   Hostname changes to 'newsite.com', port to '3000', pathname becomes '/newpath'. So full URL is 'https://newsite.com:3000/newpath'.

3. Final Answer:

   https://newsite.com:3000/newpath -> Option B

4. Quick Check:

   Pathname setter normalizes with leading '/' [OK]

Hint: pathname setter adds leading '/' automatically [OK]

Common Mistakes:

• Expecting '//newpath' without leading slash
• Assuming pathname auto-adds slash
• Confusing pathname with href

5. You want to change the query parameter id to 42 in this URL: https://shop.com/products?category=books&id=10. Which code correctly updates the URL using the URL class?

hard

A. const url = new URL('https://shop.com/products?category=books&id=10');
url.searchParams.set('id', '42');
console.log(url.href);

B. const url = new URL('https://shop.com/products?category=books&id=10');
url.query.id = 42;
console.log(url.href);

C. const url = new URL('https://shop.com/products?category=books&id=10');
url.search.id = '42';
console.log(url.href);

D. const url = new URL('https://shop.com/products?category=books&id=10');
url.setQuery('id', '42');
console.log(url.href);

Solution

1. Step 1: Identify how to update query parameters

   The URL class provides searchParams with methods like set() to update query parameters safely.

2. Step 2: Check each option's method

   const url = new URL('https://shop.com/products?category=books&id=10'); url.searchParams.set('id', '42'); console.log(url.href); uses searchParams.set(), which is correct. Options A, C, and D use invalid properties or methods not available on URL objects.

3. Final Answer:

   const url = new URL('https://shop.com/products?category=books&id=10'); url.searchParams.set('id', '42'); console.log(url.href); -> Option A

4. Quick Check:

   Use searchParams.set() to update query [OK]

Hint: Use url.searchParams.set() to change query values [OK]

Common Mistakes:

• Trying to set query directly as object
• Using non-existent methods like setQuery
• Assigning query parameters without searchParams