Conditional component rendering in React - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ component_behavior

intermediate

What is the output of this React component?

Consider this React component that conditionally renders a message based on a boolean state.

import React, { useState } from 'react';

function Welcome() {
  const [isLoggedIn, setIsLoggedIn] = useState(false);
  return (
    <div>
      {isLoggedIn ? <h1>Welcome back!</h1> : <h1>Please sign in.</h1>}
    </div>
  );
}

What will be rendered when this component first appears?

React

import React, { useState } from 'react';

function Welcome() {
  const [isLoggedIn, setIsLoggedIn] = useState(false);
  return (
    <div>
      {isLoggedIn ? <h1>Welcome back!</h1> : <h1>Please sign in.</h1>}
    </div>
  );
}

A <div><h1>Please sign in.</h1></div>

B <div><h1>Welcome back!</h1></div>

C <div></div>

D SyntaxError: Unexpected token

💡 Hint

Look at the initial value of isLoggedIn in the useState hook.

The state isLoggedIn starts as false. The ternary operator renders the second option, which is <h1>Please sign in.</h1>.

📝 Syntax

intermediate

Which option correctly uses conditional rendering with && operator?

In React, you want to render a <p>Hello User</p> only if isUser is true. Which code snippet correctly does this?

React

function Greeting({ isUser }) {
  return (
    <div>
      {/* Your code here */}
    </div>
  );
}

A {isUser || <p>Hello User</p>}

B {isUser && <p>Hello User</p>}

C {if (isUser) <p>Hello User</p>}

D {isUser ? <p>Hello User</p> : null}

💡 Hint

Remember that JSX expressions must be valid JavaScript expressions.

Option B uses the logical AND operator to conditionally render the paragraph only if isUser is true. Option B is also valid but the question asks for the && operator specifically. Option B is invalid syntax inside JSX. Option B renders the paragraph when isUser is false, which is incorrect.

❓ state_output

advanced

What is the rendered output after clicking the button twice?

Given this React component, what will be the displayed message after clicking the button two times?

import React, { useState } from 'react';

function ToggleMessage() {
  const [show, setShow] = useState(true);
  return (
    

      {show && 
Visible
}
      
    
  );
}

React

import React, { useState } from 'react';

function ToggleMessage() {
  const [show, setShow] = useState(true);
  return (
    <div>
      {show && <p>Visible</p>}
      <button onClick={() => setShow(!show)}>Toggle</button>
    </div>
  );
}

A <div><button>Toggle</button></div>

B <div><p>Visible</p><p>Visible</p><button>Toggle</button></div>

C <div><p>Toggle</p><button>Visible</button></div>

D <div><p>Visible</p><button>Toggle</button></div>

💡 Hint

Each click toggles the show state between true and false.

Initially, show is true, so <p>Visible</p> is shown. Clicking once sets show to false, hiding the paragraph. Clicking again sets it back to true, showing the paragraph again.

🔧 Debug

advanced

Why does this conditional rendering cause an error?

Examine this React component:

function Example({ show }) {
  return (
    <div>
      {show && if (true) <p>Hello</p>}
    </div>
  );
}

What error will this code cause?

React

function Example({ show }) {
  return (
    <div>
      {show && if (true) <p>Hello</p>}
    </div>
  );
}

A SyntaxError: Unexpected token 'if'

B TypeError: Cannot read property 'p' of undefined

C No error, renders <p>Hello</p> when show is true

D ReferenceError: show is not defined

💡 Hint

JSX expressions must be valid JavaScript expressions, not statements.

The if statement is not allowed inside JSX expressions. This causes a syntax error.

🧠 Conceptual

expert

Which option best describes React's conditional rendering behavior?

In React, when you use conditional rendering with expressions like {condition && <Component />}, what happens if condition is false?

A React renders the component but hides it with CSS

B React throws an error because false is not a valid React element

C React renders nothing for that expression, outputting false which is ignored in rendering

D React renders a placeholder element instead of the component

💡 Hint

Think about how React treats boolean values in JSX.

When the condition is false, React renders nothing for that expression. The false value is ignored and does not produce any DOM output or error.

Practice

1. What does conditional rendering in React allow you to do?

easy

A. Create new components on the fly without JSX

B. Change the CSS styles of components dynamically

C. Automatically fetch data from APIs

D. Show or hide parts of the UI based on data or state

Solution

1. Step 1: Understand the purpose of conditional rendering

   Conditional rendering controls what parts of the UI appear based on conditions like state or props.

2. Step 2: Differentiate from other React features

   Changing styles or fetching data are different tasks; conditional rendering specifically controls visibility.

3. Final Answer:

   Show or hide parts of the UI based on data or state -> Option D

4. Quick Check:

   Conditional rendering = Show/hide UI [OK]

Hint: Conditional rendering controls UI visibility based on conditions [OK]

Common Mistakes:

• Confusing conditional rendering with styling changes
• Thinking it automatically fetches data
• Believing it creates components without JSX

2. Which of the following is the correct syntax for conditional rendering using a ternary operator in React JSX?

easy

A. if (condition) <ComponentA /> else <ComponentB />

B. {condition ? <ComponentA /> : <ComponentB />}

C. {if condition then <ComponentA /> else <ComponentB />}

D. {condition && <ComponentA /> else <ComponentB />}

Solution

1. Step 1: Recall JSX expression syntax

   JSX uses curly braces {} to embed JavaScript expressions, including ternary operators.

2. Step 2: Identify correct ternary operator usage

   The ternary operator syntax is condition ? exprIfTrue : exprIfFalse, which matches {condition ? <ComponentA /> : <ComponentB />}.

3. Final Answer:

   {condition ? <ComponentA /> : <ComponentB />} -> Option B

4. Quick Check:

   Ternary in JSX = condition ? true : false [OK]

Hint: Use curly braces and ? : for ternary in JSX [OK]

Common Mistakes:

• Using if-else statements directly inside JSX
• Writing invalid JavaScript syntax inside JSX
• Confusing logical AND with ternary operator

3. Given the code below, what will be rendered if isLoggedIn is false?

function Greeting({ isLoggedIn }) {
  return (
    <div>
      {isLoggedIn ? <h1>Welcome back!</h1> : <h1>Please sign up.</h1>}
    </div>
  );
}

medium

A. Welcome back!

B. Syntax error

C. Please sign up.

D. Nothing will render

Solution

1. Step 1: Check the value of isLoggedIn

   The value is false, so the condition in the ternary operator is false.

2. Step 2: Evaluate the ternary operator

   Since condition is false, the expression after the colon is rendered: <h1>Please sign up.</h1>.

3. Final Answer:

   <h1>Please sign up.</h1> -> Option C

4. Quick Check:

   isLoggedIn false = show sign up message [OK]

Hint: False condition shows expression after colon in ternary [OK]

Common Mistakes:

• Assuming true branch renders when condition is false
• Thinking nothing renders on false
• Confusing syntax with runtime error

4. Identify the error in this React component that tries to conditionally render a message:

function Message({ show }) {
  return (
    <div>
      {show && <p>Hello World!</p> else <p>No message</p>}
    </div>
  );
}

medium

A. Using else inside JSX expression is invalid

B. Missing parentheses around JSX elements

C. Incorrect use of logical AND operator

D. Component must return a string, not JSX

Solution

1. Step 1: Review JSX expression syntax

   JSX expressions inside curly braces must be valid JavaScript expressions; 'else' cannot be used inside them.

2. Step 2: Identify invalid use of else

   The code uses 'else' after a logical AND expression, which is not valid JavaScript syntax.

3. Final Answer:

   Using else inside JSX expression is invalid -> Option A

4. Quick Check:

   JSX expressions cannot contain else keyword [OK]

Hint: No else keyword inside JSX expressions; use ternary instead [OK]

Common Mistakes:

• Trying to use else with logical AND
• Forgetting ternary operator syntax
• Assuming JSX can contain statements like else

5. You want to render a list of notifications only if the list is not empty. Which React pattern correctly handles this conditional rendering?

const notifications = [];

function NotificationList() {
  return (
    <div>
      {/* Choose the correct conditional rendering here */}
      {notifications.length > 0 ? (
        notifications.map((note, index) => <p key={index}>{note}</p>)
      ) : (
        <p>No notifications</p>
      )}
    </div>
  );
}

hard

A. Use ternary operator checking notifications.length > 0

B. Use only logical AND: {notifications.length && notifications.map(...)}

C. Use if statement inside JSX to check length

D. Render notifications.map(...) without any condition

Solution

1. Step 1: Understand the need for else case

   We want to show a message if there are no notifications, so we need both true and false cases.

2. Step 2: Evaluate each option

   Use ternary operator checking notifications.length > 0 uses a ternary operator to handle both cases correctly. Use only logical AND: {notifications.length && notifications.map(...)} only renders when length is truthy but shows nothing if empty. Use if statement inside JSX to check length is invalid syntax inside JSX. Render notifications.map(...) without any condition always renders map, which errors if empty.

3. Final Answer:

   Use ternary operator checking notifications.length > 0 -> Option A

4. Quick Check:

   Ternary handles both cases in JSX [OK]

Hint: Use ternary to handle both empty and non-empty lists [OK]

Common Mistakes:

• Using if statements directly inside JSX
• Using logical AND without else case
• Rendering map on empty array without check