String manipulation types (Uppercase, Lowercase) in Typescript - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to convert the string to uppercase.

Typescript

const text = "hello";
const upperText = text.[1]();
console.log(upperText);

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

Auppercase

BtoUpperCase

CtoLowerCase

Dlowercase

Attempts:

3 left

💡 Hint

Common Mistakes

Using toLowerCase() instead of toUpperCase()

Using incorrect method names like uppercase()

The toUpperCase() method converts all characters in a string to uppercase.

2fill in blank

medium

Complete the code to convert the string to lowercase.

Typescript

const text = "WORLD";
const lowerText = text.[1]();
console.log(lowerText);

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

AtoUpperCase

Buppercase

Clowercase

DtoLowerCase

Attempts:

3 left

💡 Hint

Common Mistakes

Using toUpperCase() instead of toLowerCase()

Using incorrect method names like lowercase()

The toLowerCase() method converts all characters in a string to lowercase.

3fill in blank

hard

Fix the error in the code to convert the string to uppercase.

Typescript

const greeting = "hello world";
const shout = greeting.[1]();
console.log(shout);

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

AtoUpperCase

Btouppercase

CtoupperCase

DToUpperCase

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong casing like touppercase or ToUpperCase

Misspelling the method name

The correct method name is toUpperCase() with exact casing.

4fill in blank

hard

Fill both blanks to create a new string with the first letter uppercase and the rest lowercase.

Typescript

const word = "jAVAscript";
const fixed = word.charAt(0).[1]() + word.[2](1).toLowerCase();
console.log(fixed);

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

Aslice

BtoUpperCase

CtoLowerCase

Dsubstring

Attempts:

3 left

💡 Hint

Common Mistakes

Using toLowerCase() for the first blank

Choosing substring instead of slice for the second blank (both work)

Confusing the blanks

charAt(0).toUpperCase() capitalizes the first letter, and slice(1).toLowerCase() lowercases the rest.

5fill in blank

hard

Fill all three blanks to create a function that capitalizes the first letter of a string and makes the rest lowercase.

Typescript

function capitalize(str: string): string {
  return str.[1](0).[2]() + str.[3](1).toLowerCase();
}

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

AcharAt

BtoUpperCase

Cslice

Dsubstring

Attempts:

3 left

💡 Hint

Common Mistakes

Using substring instead of slice (both work but slice is preferred here)

Forgetting to call toUpperCase() as a function

The function gets the first character with charAt(0), converts it to uppercase with toUpperCase(), then adds the rest of the string sliced from index 1 with slice(1) converted to lowercase.

Practice

1. Which TypeScript string method converts all characters in a string to uppercase?

easy

A. convertCase()

B. toLowerCase()

C. toTitleCase()

D. toUpperCase()

Solution

1. Step 1: Understand string case methods

   TypeScript strings have built-in methods to change letter cases, including toUpperCase() and toLowerCase().

2. Step 2: Identify method for uppercase

   The method toUpperCase() changes all letters to uppercase, while toLowerCase() changes to lowercase.

3. Final Answer:

   toUpperCase() -> Option D

4. Quick Check:

   Uppercase method = toUpperCase() [OK]

Hint: Uppercase method always has 'Upper' in its name [OK]

Common Mistakes:

• Confusing toUpperCase() with toLowerCase()
• Thinking toTitleCase() exists in TypeScript
• Assuming convertCase() is a valid method

2. Which of the following is the correct syntax to convert a string variable name to lowercase in TypeScript?

easy

A. name.toLowerCase()

B. toLowerCase(name)

C. name.lowercase()

D. lowerCase(name)

Solution

1. Step 1: Recall method call syntax on strings

   In TypeScript, string methods are called on the string variable using dot notation.

2. Step 2: Identify correct method name and usage

   The method to convert to lowercase is toLowerCase(), called as name.toLowerCase(). Other options are invalid syntax or incorrect method names.

3. Final Answer:

   name.toLowerCase() -> Option A

4. Quick Check:

   Lowercase method called on variable = name.toLowerCase() [OK]

Hint: String methods use dot notation like variable.method() [OK]

Common Mistakes:

• Using function style like toLowerCase(name)
• Using incorrect method names like lowercase()
• Missing parentheses after method name

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

const text = "Hello World";
console.log(text.toUpperCase());

medium

A. "hello world"

B. SyntaxError

C. "HELLO WORLD"

D. "Hello World"

Solution

1. Step 1: Understand the input string and method

   The string is "Hello World" and the method toUpperCase() converts all letters to uppercase.

2. Step 2: Apply toUpperCase() to the string

   Applying toUpperCase() changes "Hello World" to "HELLO WORLD".

3. Final Answer:

   "HELLO WORLD" -> Option C

4. Quick Check:

   toUpperCase() output = "HELLO WORLD" [OK]

Hint: toUpperCase() makes all letters big exactly [OK]

Common Mistakes:

• Expecting lowercase output instead
• Thinking original string is unchanged in output
• Confusing method name causing syntax error

4. Identify the error in this TypeScript code snippet:

let greeting = "Hi There";
console.log(greeting.toLowercase());

medium

A. Missing semicolon after string declaration

B. toLowercase() is not a valid method; correct is toLowerCase()

C. Variable greeting should be const, not let

D. Console.log should be console.Log

Solution

1. Step 1: Check method name spelling

   The method to convert string to lowercase is toLowerCase() with capital 'C'. The code uses toLowercase() with lowercase 'c', which is invalid.

2. Step 2: Verify other syntax elements

   Semicolon is optional in TypeScript, let is valid for variable, and console.log is case-sensitive and correct as written.

3. Final Answer:

   toLowercase() is not a valid method; correct is toLowerCase() -> Option B

4. Quick Check:

   Method name case matters: toLowerCase() correct [OK]

Hint: Method names are case-sensitive; watch capital letters [OK]

Common Mistakes:

• Ignoring case sensitivity in method names
• Thinking semicolon is mandatory error
• Confusing console.log capitalization

5. You have an array of strings: const fruits = ["Apple", "BANANA", "Cherry"];
How do you create a new array where all fruit names are lowercase using TypeScript?

hard

A. fruits.map(fruit => fruit.toLowerCase())

B. fruits.toLowerCase()

C. fruits.forEach(fruit => fruit.toLowerCase())

D. fruits.map(fruit => toLowerCase(fruit))

Solution

1. Step 1: Understand array transformation

   To change each string in an array, use map() to create a new array with transformed values.

2. Step 2: Apply toLowerCase() inside map callback

   Inside map, call fruit.toLowerCase() to convert each string to lowercase. This returns a new array with all lowercase fruits.

3. Final Answer:

   fruits.map(fruit => fruit.toLowerCase()) -> Option A

4. Quick Check:

   Use map() with toLowerCase() for array lowercase [OK]

Hint: Use map() with toLowerCase() to transform arrays [OK]

Common Mistakes:

• Calling toLowerCase() directly on array
• Using forEach which returns void, not new array
• Trying to call toLowerCase() as a standalone function