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Node.jsframework~10 mins

URL class for parsing in Node.js - Interactive Code Practice

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Practice - 5 Tasks
Answer the questions below
1fill in blank
easy

Complete the code to create a new URL object from a string.

Node.js
const myUrl = new URL([1]);
Drag options to blanks, or click blank then click option'
AURL('https://example.com')
Bnew URL('https://example.com')
Chttps://example.com
D'https://example.com'
Attempts:
3 left
💡 Hint
Common Mistakes
Forgetting quotes around the URL string.
Passing the URL constructor itself as an argument.
2fill in blank
medium

Complete the code to get the hostname from a URL object.

Node.js
const hostname = myUrl.[1];
Drag options to blanks, or click blank then click option'
Aorigin
Bhostname
Chref
Dhost
Attempts:
3 left
💡 Hint
Common Mistakes
Using host which includes port number.
Using href which is the full URL string.
3fill in blank
hard

Fix the error in accessing the pathname of a URL object.

Node.js
const path = myUrl.[1];
Drag options to blanks, or click blank then click option'
Apathname
BgetPathname
Cpath
DgetPath
Attempts:
3 left
💡 Hint
Common Mistakes
Calling pathname as a function with parentheses.
Using non-existent methods like getPath.
4fill in blank
hard

Fill both blanks to create a URL object with a base URL.

Node.js
const myUrl = new URL([1], [2]);
Drag options to blanks, or click blank then click option'
A'/page.html'
B'https://example.com'
C'page.html'
D'http://localhost'
Attempts:
3 left
💡 Hint
Common Mistakes
Using a full URL as the first argument and base URL as second.
Omitting the base URL when using a relative path.
5fill in blank
hard

Fill all three blanks to extract the protocol, hostname, and port from a URL object.

Node.js
const protocol = myUrl.[1];
const hostname = myUrl.[2];
const port = myUrl.[3];
Drag options to blanks, or click blank then click option'
Aprotocol
Bhostname
Cport
Dhost
Attempts:
3 left
💡 Hint
Common Mistakes
Using host instead of hostname and port separately.
Forgetting the colon ':' in protocol value (it is included automatically).

Practice

(1/5)
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