Iterating over objects in Javascript - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to loop over the keys of the object.

Javascript

const obj = {a: 1, b: 2, c: 3};
for (const [1] in obj) {
  console.log([1]);
}

Drag options to blanks, or click blank then click option'

Aitem

Bvalue

Ckey

Dindex

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'value' instead of 'key' in the for...in loop variable.

Trying to use 'for...of' directly on an object.

The 'for...in' loop iterates over the keys of the object, so the variable should be named 'key' to represent each key.

2fill in blank

medium

Complete the code to get the values of the object using Object.values().

Javascript

const obj = {x: 10, y: 20};
const values = Object.[1](obj);
console.log(values);

Drag options to blanks, or click blank then click option'

Avalues

Bkeys

Centries

Dget

Attempts:

3 left

💡 Hint

Common Mistakes

Using Object.keys() instead of Object.values().

Trying to call a non-existent method like Object.get().

Object.values(obj) returns an array of the object's values.

3fill in blank

hard

Fix the error in the code to correctly iterate over object entries.

Javascript

const obj = {a: 1, b: 2};
for (const [key, [1]] of Object.entries(obj)) {
  console.log(key, [1]);
}

Drag options to blanks, or click blank then click option'

Aentry

Bitem

Cindex

Dvalue

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'item' or 'index' which do not represent the value.

Trying to destructure incorrectly.

Object.entries(obj) returns [key, value] pairs, so the second variable should be 'value'.

4fill in blank

hard

Fill both blanks to create a new object with keys in uppercase and values doubled.

Javascript

const obj = {a: 1, b: 2};
const newObj = Object.fromEntries(
  Object.entries(obj).map(([[1], [2]]) => [[1].toUpperCase(), [2] * 2])
);
console.log(newObj);

Drag options to blanks, or click blank then click option'

Akey

Bvalue

Citem

Dentry

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping key and value in destructuring.

Using unclear variable names like 'item' or 'entry'.

When destructuring entries, the first element is the key and the second is the value.

5fill in blank

hard

Fill all three blanks to filter object entries with values greater than 10 and create a new object.

Javascript

const obj = {a: 5, b: 15, c: 25};
const filteredObj = Object.fromEntries(
  Object.entries(obj)
    .filter(([[1], [2]]) => [2] [3] 10)
);
console.log(filteredObj);

Drag options to blanks, or click blank then click option'

Akey

Bvalue

C>

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<' instead of '>' in the filter condition.

Swapping key and value in destructuring.

Destructure entries as [key, value], then filter where value > 10.

Practice

1. Which of the following is the correct way to loop over all keys in a JavaScript object obj?

easy

A. obj.forEach(key => console.log(key));

B. for (let i = 0; i < obj.length; i++) { console.log(obj[i]); }

C. for (let key in obj) { console.log(key); }

D. for (let key of obj) { console.log(key); }

Solution

1. Step 1: Understand looping over object keys

   The for...in loop is designed to iterate over all enumerable keys in an object.

2. Step 2: Check each option's validity

   for (let key in obj) { console.log(key); } uses for...in correctly. for (let i = 0; i < obj.length; i++) { console.log(obj[i]); } tries to use a numeric index on an object, which doesn't work. for (let key of obj) { console.log(key); } uses for...of which works on iterable objects like arrays, not plain objects. obj.forEach(key => console.log(key)); tries to use forEach on an object, which is invalid.

3. Final Answer:

   for (let key in obj) { console.log(key); } -> Option C

4. Quick Check:

   Use for...in to loop keys [OK]

Hint: Use for...in to loop keys in objects [OK]

Common Mistakes:

• Using for...of on plain objects
• Trying to use numeric indexes on objects
• Using forEach directly on objects

2. Which syntax correctly accesses the value of each key inside a for...in loop over object data?

easy

A. for (let key in data) { console.log(data[key]); }

B. for (let key in data) { console.log(key.value); }

C. for (let key in data) { console.log(data.key); }

D. for (let key in data) { console.log(data->key); }

Solution

1. Step 1: Understand how to access object values by key

   Inside a for...in loop, key is a string representing the property name. To get the value, use bracket notation: data[key].

2. Step 2: Evaluate each option

   for (let key in data) { console.log(data.key); } uses dot notation with a variable, which looks for a property literally named 'key'. for (let key in data) { console.log(data[key]); } correctly uses bracket notation. for (let key in data) { console.log(key.value); } tries to access key.value, but key is a string, not an object. for (let key in data) { console.log(data->key); } uses invalid syntax.

3. Final Answer:

   for (let key in data) { console.log(data[key]); } -> Option A

4. Quick Check:

   Access values with data[key] inside for...in [OK]

Hint: Use bracket notation data[key] to get values [OK]

Common Mistakes:

• Using dot notation with variable key
• Treating key as an object
• Using invalid syntax like arrows

3. What will be the output of the following code?

const obj = {a: 1, b: 2, c: 3};
let result = [];
for (let key in obj) {
  result.push(key + obj[key]);
}
console.log(result);

medium

A. ['1', '2', '3']

B. ['1a', '2b', '3c']

C. ['a', 'b', 'c']

D. ['a1', 'b2', 'c3']

Solution

1. Step 1: Understand the loop and concatenation

   The loop iterates over keys 'a', 'b', 'c'. For each key, it concatenates the key string with its value (number converted to string).

2. Step 2: Trace the result array

   First iteration: 'a' + 1 = 'a1', second: 'b' + 2 = 'b2', third: 'c' + 3 = 'c3'. These are pushed into the array.

3. Final Answer:

   ['a1', 'b2', 'c3'] -> Option D

4. Quick Check:

   Keys + values concatenated = ['a1', 'b2', 'c3'] [OK]

Hint: Concatenate key + value inside for...in loop [OK]

Common Mistakes:

• Confusing order of concatenation
• Expecting numeric addition instead of string concat
• Using keys or values alone

4. Identify the error in this code snippet:

const user = {name: 'Alice', age: 25};
for (let key of user) {
  console.log(key + ': ' + user[key]);
}

medium

A. Using for...of on a plain object causes a TypeError

B. Missing declaration of key variable

C. Incorrect string concatenation syntax

D. Accessing user[key] is invalid

Solution

1. Step 1: Check the loop type and object

   for...of works on iterable objects like arrays or strings. Plain objects are not iterable, so this causes a TypeError.

2. Step 2: Confirm other parts are correct

   The variable key is declared correctly. String concatenation and accessing user[key] are valid if the loop worked.

3. Final Answer:

   Using for...of on a plain object causes a TypeError -> Option A

4. Quick Check:

   for...of needs iterable, plain object is not [OK]

Hint: Use for...in or Object.keys() for objects, not for...of [OK]

Common Mistakes:

• Using for...of on plain objects
• Confusing variable declaration errors
• Thinking user[key] is invalid

5. Given the object const scores = {alice: 10, bob: 0, carol: 5, dave: null};, which code snippet correctly creates a new object with only keys having truthy values?

hard

A. const filtered = {};
for (let key in scores) {
if (scores.key) filtered.key = scores.key;
}

B. const filtered = {};
for (let key in scores) {
if (scores[key]) filtered[key] = scores[key];
}

C. const filtered = Object.keys(scores).filter(k => scores[k]).reduce((obj, k) => { obj[k] = scores[k]; }, {});

D. const filtered = Object.values(scores).filter(v => v);

Solution

1. Step 1: Understand filtering truthy values in an object

   We want to keep keys where the value is truthy (not 0, null, undefined, false, '').

2. Step 2: Analyze each option

   const filtered = {}; for (let key in scores) { if (scores[key]) filtered[key] = scores[key]; } uses a for...in loop and checks scores[key] truthiness correctly, adding to filtered. const filtered = {}; for (let key in scores) { if (scores.key) filtered.key = scores.key; } incorrectly uses dot notation with variable key. const filtered = Object.keys(scores).filter(k => scores[k]).reduce((obj, k) => { obj[k] = scores[k]; return obj; }, {}); tries to use Object.keys() with filter and reduce, but the code is missing a return statement in the arrow function (syntax error). const filtered = Object.values(scores).filter(v => v); filters values only, losing keys and returning an array, not an object.

3. Final Answer:

   const filtered = {}; for (let key in scores) { if (scores[key]) filtered[key] = scores[key]; } -> Option B

4. Quick Check:

   Use for...in with if (scores[key]) to filter [OK]

Hint: Use for...in and if (obj[key]) to filter truthy keys [OK]

Common Mistakes:

• Using dot notation with variable keys
• Confusing Object.values() with keys
• Missing return in reduce arrow function