If conditions in JSX in React - Cheat Sheet & Quick Revision

Recall & Review

beginner

How do you write a simple if condition inside JSX to show a message only if a variable isLoggedIn is true?

You can use the logical AND operator:
{isLoggedIn && <p>Welcome back!</p>}. This shows the message only when isLoggedIn is true.

beginner

What is the purpose of the ternary operator in JSX?

The ternary operator condition ? expr1 : expr2 lets you choose between two things to show in JSX depending on a condition. For example: {isLoggedIn ? <LogoutButton /> : <LoginButton />}.

beginner

Why can't you use a regular if statement directly inside JSX?

JSX expects expressions, but a regular if statement is a statement, not an expression. So you use expressions like ternary operators or logical AND inside JSX instead.

intermediate

How can you conditionally render multiple elements in JSX?

You can wrap multiple elements in a fragment <> ... </> and use a ternary operator or logical AND outside it. For example: {isLoggedIn ? <><Welcome /><LogoutButton /></> : <LoginButton />}.

intermediate

What is a clean way to handle complex if conditions in JSX?

You can move the if logic outside JSX into variables or functions, then use those variables inside JSX. This keeps JSX clean and easy to read.

Which of these is a valid way to conditionally render a message in JSX when showMessage is true?

A {showMessage && <p>Hello!</p>}

B if(showMessage) { <p>Hello!</p> }

C {if(showMessage) <p>Hello!</p>}

D <p>{if(showMessage) Hello!}</p>

Only option A uses a valid JSX expression with logical AND. Regular if statements can't be used directly inside JSX.

What does this JSX expression do? {isAdmin ? <AdminPanel /> : <UserPanel />}

A Always shows both AdminPanel and UserPanel

B Shows AdminPanel if isAdmin is true, else UserPanel

C Shows nothing

D Throws an error

The ternary operator chooses one component to show based on the condition.

Why might you move if conditions outside JSX?

A To make the code run faster

B Because JSX does not allow any conditions

C To avoid using React hooks

D To keep JSX clean and easier to read

Moving logic outside JSX helps keep the JSX simple and readable.

Which operator is commonly used for simple conditional rendering in JSX?

A ++ (increment)

B || (logical OR)

C && (logical AND)

D ?? (nullish coalescing)

Logical AND is often used to render something only if a condition is true.

What will this render if isLoggedIn is false? {isLoggedIn && <p>Welcome!</p>}

A Nothing (renders null)

B The paragraph with Welcome!

C An error

D Undefined

If the condition is false, logical AND returns false which React treats as nothing to render.

Explain how to use if conditions inside JSX to show different content based on a variable.

Think about expressions vs statements in JSX.

Describe best practices for handling complex conditional rendering in React JSX.

Consider code readability and maintainability.

Practice

1. Which of the following is the correct way to conditionally render a message in JSX when isLoggedIn is true?

easy

A. {if (isLoggedIn) <p>Welcome back!</p>}

B. {isLoggedIn && <p>Welcome back!</p>}

C. {isLoggedIn ? <p>Welcome back!</p>}

D. {isLoggedIn || <p>Welcome back!</p>}

Solution

1. Step 1: Understand JSX conditional rendering with &&

   In JSX, condition && expression renders the expression only if the condition is true.

2. Step 2: Check each option's syntax

   {isLoggedIn && <p>Welcome back!</p>} uses {isLoggedIn && <p>Welcome back!</p>}, which is correct. {if (isLoggedIn) <p>Welcome back!</p>} uses an invalid if statement inside JSX. {isLoggedIn ? <p>Welcome back!</p>} uses an incomplete ternary operator without the false branch (`: expression`), causing a syntax error. {isLoggedIn || <p>Welcome back!</p>} uses ||, which is incorrect for this purpose.

3. Final Answer:

   {isLoggedIn && <p>Welcome back!</p>} -> Option B

4. Quick Check:

   Use && for simple true condition rendering [OK]

Hint: Use && to show content only if true [OK]

Common Mistakes:

• Trying to use if statements directly inside JSX
• Using || instead of && for conditional rendering
• Confusing ternary with && when only one condition is needed

2. Which JSX snippet correctly uses a ternary operator to show <p>Hello</p> if showHello is true, otherwise <p>Goodbye</p>?

easy

A. {showHello && <p>Hello</p> else <p>Goodbye</p>}

B. {showHello ? <p>Goodbye</p> : <p>Hello</p>}

C. {if (showHello) <p>Hello</p> else <p>Goodbye</p>}

D. {showHello ? <p>Hello</p> : <p>Goodbye</p>}

Solution

1. Step 1: Understand ternary syntax in JSX

   The ternary operator syntax is condition ? expressionIfTrue : expressionIfFalse.

2. Step 2: Match the expressions to the condition

   {showHello ? <p>Hello</p> : <p>Goodbye</p>} correctly shows Hello if showHello is true, else Goodbye. {showHello && <p>Hello</p> else <p>Goodbye</p>} uses invalid syntax with else inside JSX. {if (showHello) <p>Hello</p> else <p>Goodbye</p>} tries to use if statement inside JSX, which is invalid. {showHello ? <p>Goodbye</p> : <p>Hello</p>} reverses the messages.

3. Final Answer:

   {showHello ? <p>Hello</p> : <p>Goodbye</p>} -> Option D

4. Quick Check:

   Ternary = condition ? true : false [OK]

Hint: Ternary: condition ? true : false for two choices [OK]

Common Mistakes:

• Using if/else statements directly inside JSX
• Swapping true and false expressions
• Using && when two choices are needed

3. What will be rendered by this React component?

function Greeting({ isMorning }) {
  return (
    <div>
      {isMorning ? <h1>Good morning!</h1> : <h1>Good evening!</h1>}
      {isMorning && <p>Have a nice day!</p>}
    </div>
  );
}

If isMorning is false?

medium

A. Good evening!

B. Good evening!Have a nice day!

C. Good morning!Have a nice day!

D. Good morning!

Solution

1. Step 1: Evaluate the ternary expression with isMorning = false

   The ternary isMorning ? <h1>Good morning!</h1> : <h1>Good evening!</h1> renders <h1>Good evening!</h1> because isMorning is false.

2. Step 2: Evaluate the && expression with isMorning = false

   The expression isMorning && <p>Have a nice day!</p> renders nothing because isMorning is false.

3. Final Answer:

   <div><h1>Good evening!</h1></div> -> Option A

4. Quick Check:

   False && anything = nothing [OK]

Hint: False && expression renders nothing; ternary picks false branch [OK]

Common Mistakes:

• Assuming && renders content when condition is false
• Mixing up ternary true/false branches
• Expecting both expressions to render always

4. Identify the error in this JSX snippet:

return (
  <div>
    {isActive ? <p>Active</p>}
  </div>
);

medium

A. No error, code is valid

B. Using if statement inside JSX

C. Missing else part in ternary operator

D. Incorrect use of && operator

Solution

1. Step 1: Check ternary operator syntax

   The ternary operator requires both true and false expressions: condition ? exprIfTrue : exprIfFalse. Here, the false part is missing.

2. Step 2: Understand JSX requirements

   Without the false expression, JSX will throw a syntax error because the ternary is incomplete.

3. Final Answer:

   Missing else part in ternary operator -> Option C

4. Quick Check:

   Ternary needs both ? and : parts [OK]

Hint: Ternary must have both ? and : parts [OK]

Common Mistakes:

• Omitting the else (false) part in ternary
• Confusing ternary with && operator
• Trying to use if statements inside JSX

5. You want to render a list of user names, but only show the name if the user is active. Which JSX code correctly implements this using map and conditional rendering?

hard

A. {users.map(user => user.isActive ? <li key={user.id}>{user.name}</li> : null)}

B. {users.map(user => if (user.isActive) return <li key={user.id}>{user.name}</li> )}

C. {users.map(user => user.isActive || <li>{user.name}</li>)}

D. {users.map(user => user.isActive ? <li>{user.name}</li> : <li>Inactive</li>)}

Solution

1. Step 1: Understand conditional rendering inside map

   To render only active users, use a ternary that returns the list item or null.

2. Step 2: Check key usage and syntax

   {users.map(user => user.isActive ? <li key={user.id}>{user.name}</li> : null)} correctly uses key={user.id} and returns null for inactive users, which React ignores.

3. Step 3: Analyze other options

   {users.map(user => if (user.isActive) return <li key={user.id}>{user.name}</li> )} uses invalid syntax: `if` statement directly after arrow function requires block `{}` braces. {users.map(user => user.isActive || <li>{user.name}</li>)} uses `||` which renders items for inactive users and misses keys. {users.map(user => user.isActive ? <li>{user.name}</li> : <li>Inactive</li>)} shows inactive users with text, which is not desired.

4. Final Answer:

   {users.map(user => user.isActive ? <li key={user.id}>{user.name}</li> : null)} -> Option A

5. Quick Check:

   Use ternary with key and null for conditional list rendering [OK]

Hint: Use ternary with key and null to skip items [OK]

Common Mistakes:

• Missing key prop in list items
• Using if statements inside map without return
• Rendering false or undefined instead of null