URL class for parsing in Node.js - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output of this URL parsing code?

Consider the following Node.js code using the URL class. What will be logged to the console?

Node.js

import { URL } from 'url';
const myUrl = new URL('https://example.com:8080/path/page?name=chatgpt&lang=en#section1');
console.log(myUrl.hostname);
console.log(myUrl.port);
console.log(myUrl.pathname);
console.log(myUrl.searchParams.get('name'));
console.log(myUrl.hash);

A

"example.com"
"8080"
"path/page"
"chatgpt"
"#section1"

B

"https://example.com"
"8080"
"/path/page"
"chatgpt"
"section1"

C

"example.com"
""
"/path/page"
"chatgpt"
"#section1"

D

"example.com"
"8080"
"/path/page"
"chatgpt"
"#section1"

💡 Hint

Remember that the URL class separates hostname, port, pathname, search parameters, and hash exactly as in the URL string.

The hostname is the domain without protocol or port. The port is "8080" as specified. The pathname includes the leading slash. The searchParams.get('name') returns the value of the 'name' query parameter. The hash includes the leading '#'.

❓ component_behavior

intermediate

Which option correctly adds a query parameter to a URL?

Given a URL object, which code snippet correctly adds a new query parameter 'page=2' without overwriting existing parameters?

Node.js

import { URL } from 'url';
const url = new URL('https://example.com/search?query=nodejs');

A

url.searchParams.append('page', '2');
console.log(url.toString());

B

url.searchParams.set('page', '2');
console.log(url.toString());

C

url.searchParams = 'page=2';
console.log(url.toString());

D

url.search = '?page=2';
console.log(url.toString());

💡 Hint

Appending adds a parameter without removing existing ones. Setting replaces or adds a parameter.

append adds a new parameter preserving existing ones. set replaces the parameter if it exists or adds it if not. Options C and D overwrite the entire query string, removing existing parameters.

📝 Syntax

advanced

Which option causes a TypeError when creating a URL object?

Which of the following URL constructor usages will cause a TypeError?

A new URL('://missing.scheme.com');

B new URL('ftp://user:[e-mail]:21/path');

C new URL('http://example.com');

D new URL('https://example.com/path?query=1#frag');

💡 Hint

A valid URL must have a scheme like http or https.

Option A is missing the scheme before '://', which is required. This causes a TypeError. Other options have valid URL formats.

🔧 Debug

advanced

Why does this code throw a TypeError?

Analyze the code below. Why does it throw a TypeError?

Node.js

import { URL } from 'url';
const url = new URL('https://example.com');
url.searchParams.get('page').toUpperCase();

A Because searchParams is not a valid property of URL.

B Because URL constructor requires a base URL, missing here.

C Because 'page' parameter does not exist, get() returns null, and calling toUpperCase() on null causes TypeError.

D Because toUpperCase() is not a function on string.

💡 Hint

Check what happens when you call get() on a missing parameter.

The get method returns null if the parameter is missing. Calling toUpperCase() on null causes a TypeError.

🧠 Conceptual

expert

What is the value of url.origin after this code runs?

Consider this code snippet. What will be the value of url.origin?

Node.js

import { URL } from 'url';
const url = new URL('https://user:[e-mail]:8443/path?query=1#frag');

A "https://user:[e-mail]:8443"

B "https://example.com:8443"

C "https://example.com"

D "https://[e-mail]:8443"

💡 Hint

The origin includes scheme, hostname, and port but excludes username and password.

The origin property returns the scheme, hostname, and port. It does not include username or password.