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Javascriptprogramming~20 mins

Promise chaining in Javascript - Mini Project: Build & Apply

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Promise chaining
📖 Scenario: Imagine you are ordering a pizza online. First, you select the pizza, then you pay for it, and finally, you get a confirmation message. Each step happens one after the other.
🎯 Goal: You will create three functions that return promises for each step: selectPizza, payForPizza, and confirmOrder. Then, you will chain these promises together to show the full order process.
📋 What You'll Learn
Create a function called selectPizza that returns a promise resolving to the string 'Pizza selected'.
Create a function called payForPizza that takes a string and returns a promise resolving to 'Payment done'.
Create a function called confirmOrder that takes a string and returns a promise resolving to 'Order confirmed'.
Chain the three functions using .then() to run them one after another.
Print the final confirmation message.
💡 Why This Matters
🌍 Real World
Promise chaining is used in real life when you need to do tasks one after another that take time, like loading data from the internet or processing payments.
💼 Career
Understanding promise chaining is important for web developers to handle asynchronous operations smoothly, improving user experience and app reliability.
Progress0 / 4 steps
1
Create the selectPizza function
Write a function called selectPizza that returns a promise which resolves to the string 'Pizza selected'.
Javascript
Hint

Use new Promise and call resolve with the string 'Pizza selected'.

2
Create the payForPizza function
Write a function called payForPizza that takes one parameter called message and returns a promise which resolves to the string 'Payment done'.
Javascript
Hint

The function takes message but you only need to resolve with 'Payment done'.

3
Create the confirmOrder function
Write a function called confirmOrder that takes one parameter called message and returns a promise which resolves to the string 'Order confirmed'.
Javascript
Hint

Similar to the other functions, resolve with 'Order confirmed'.

4
Chain the promises and print the final message
Use selectPizza() and chain it with payForPizza and confirmOrder using .then(). Finally, print the result of confirmOrder using console.log.
Javascript
Hint

Chain the functions with .then() and print the last result with console.log.

Practice

(1/5)
1. What does promise chaining allow you to do in JavaScript?
easy
A. Run multiple synchronous tasks at the same time
B. Stop all asynchronous tasks immediately
C. Run asynchronous tasks one after another, passing results along
D. Convert promises into synchronous functions

Solution

  1. Step 1: Understand the purpose of promise chaining

    Promise chaining is designed to run async tasks in sequence, where each step waits for the previous to finish and can use its result.
  2. Step 2: Compare options with the concept

    Only Run asynchronous tasks one after another, passing results along correctly describes running async tasks one after another and passing results along.
  3. Final Answer:

    Run asynchronous tasks one after another, passing results along -> Option C
  4. Quick Check:

    Promise chaining = sequential async tasks [OK]
Hint: Promise chaining runs async tasks in order [OK]
Common Mistakes:
  • Thinking promises run synchronously
  • Believing chaining stops tasks immediately
  • Confusing promise chaining with parallel execution
2. Which of the following is the correct syntax to chain two promises p1 and p2?
easy
A. p1.then().then(p2).then(result => console.log(result));
B. p1.then(() => p2).then(result => console.log(result));
C. p1.then(p2).then(result => console.log(result));
D. p1.then(result => p2).catch(console.log);

Solution

  1. Step 1: Understand promise chaining syntax

    To chain promises, the first promise's then should return the second promise function call, so it waits for it.
  2. Step 2: Analyze each option

    p1.then(() => p2).then(result => console.log(result)); correctly returns p2 inside then. p1.then(p2).then(result => console.log(result)); passes p2 directly, which is incorrect. p1.then().then(p2).then(result => console.log(result)); has an empty then() which is invalid. p1.then(result => p2).catch(console.log); uses catch prematurely and returns p2 incorrectly.
  3. Final Answer:

    p1.then(() => p2).then(result => console.log(result)); -> Option B
  4. Quick Check:

    Return next promise in then = correct chaining [OK]
Hint: Return next promise inside then() for chaining [OK]
Common Mistakes:
  • Passing promise directly to then instead of a function
  • Leaving then() empty
  • Misplacing catch before chain ends
3. What will be the output of this code?
Promise.resolve(5)
  .then(x => x * 2)
  .then(x => Promise.resolve(x + 3))
  .then(console.log);
medium
A. Error
B. 10
C. Promise { 13 }
D. 13

Solution

  1. Step 1: Trace each then step

    First, Promise.resolve(5) resolves with 5. Then x => x * 2 returns 10. Next, x => Promise.resolve(x + 3) returns a promise resolving to 13.
  2. Step 2: Understand final then output

    The last then(console.log) logs the resolved value 13, not a promise object.
  3. Final Answer:

    13 -> Option D
  4. Quick Check:

    Promise chaining unwraps nested promises [OK]
Hint: Promises unwrap nested promises automatically [OK]
Common Mistakes:
  • Thinking console.log prints a Promise object
  • Ignoring that then unwraps promises
  • Confusing multiplication and addition order
4. Find the error in this promise chain:
fetchData()
  .then(data => processData(data))
  .then(result => console.log(result))
  .catch(error => console.error(error));

function processData(data) {
  throw new Error('Processing failed');
}
medium
A. Error is caught by catch because thrown inside then chain
B. catch should be before then to catch errors
C. Missing return statement in processData causes error
D. Error is not caught because processData is synchronous

Solution

  1. Step 1: Understand error throwing in promise chain

    Throwing an error inside processData causes the promise to reject, which is caught by the catch at the end.
  2. Step 2: Analyze catch placement and behavior

    The catch at the end catches any error thrown in previous then callbacks, including synchronous throws.
  3. Final Answer:

    Error is caught by catch because thrown inside then chain -> Option A
  4. Quick Check:

    Thrown errors in then are caught by catch [OK]
Hint: Errors thrown in then are caught by catch at chain end [OK]
Common Mistakes:
  • Thinking synchronous errors are not caught
  • Placing catch before then incorrectly
  • Assuming missing return causes error
5. You want to run three async tasks task1, task2, and task3 in order, where each depends on the previous result. Which code correctly chains them and catches any error once?
hard
A. task1() .then(result1 => task2(result1)) .then(result2 => task3(result2)) .catch(error => console.error(error));
B. task1() .then(() => task2()) .then(() => task3()) .catch(error => console.error(error));
C. task1() .then(result1 => task2(result1)) .catch(error => console.error(error)) .then(result2 => task3(result2));
D. task1() .catch(error => console.error(error)) .then(result1 => task2(result1)) .then(result2 => task3(result2));

Solution

  1. Step 1: Understand dependency and error handling

    Each task depends on the previous result, so each then must pass the result to the next task function. A single catch at the end handles all errors.
  2. Step 2: Analyze each option

    task1() .then(result1 => task2(result1)) .then(result2 => task3(result2)) .catch(error => console.error(error)); correctly chains tasks with result passing and one catch at the end. task1() .then(() => task2()) .then(() => task3()) .catch(error => console.error(error)); passes task2 and task3 without arguments, which may fail if they expect parameters. task1() .then(result1 => task2(result1)) .catch(error => console.error(error)) .then(result2 => task3(result2)); places catch too early, breaking the chain. task1() .catch(error => console.error(error)) .then(result1 => task2(result1)) .then(result2 => task3(result2)); places catch before then, so errors in later tasks are not caught.
  3. Final Answer:

    task1().then(result1 => task2(result1)).then(result2 => task3(result2)).catch(error => console.error(error)); -> Option A
  4. Quick Check:

    Chain with single catch at end and pass results [OK]
Hint: Chain with one catch at end, pass results explicitly [OK]
Common Mistakes:
  • Placing catch too early breaking chain
  • Not passing results to dependent tasks
  • Multiple catches causing partial error handling
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10 questions~5 mineasy hard