Reading string input in Java - Mini Project: Build & Apply

Reading string input

📖 Scenario: You are creating a simple Java program that asks a user for their name and then greets them.

🎯 Goal: Build a Java program that reads a user's name from the keyboard and prints a greeting message.

📋 What You'll Learn

Use Scanner to read input from the user

Store the input in a variable called name

Print a greeting message using the name variable

💡 Why This Matters

🌍 Real World

Reading user input is essential for interactive programs like games, forms, and command-line tools.

💼 Career

Many programming jobs require handling user input safely and effectively, especially in Java backend or desktop applications.

1

Set up Scanner to read input

Write a line to create a Scanner object called scanner to read from System.in.

Java

import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        // Create Scanner object called scanner
        
    }
}

Hint

Use Scanner scanner = new Scanner(System.in); to read input from the keyboard.

2

Read the user's name

Write a line to read a full line of text from the user using scanner.nextLine() and store it in a variable called name.

Java

import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        // Read user input line into variable name
        
    }
}

Hint

Use String name = scanner.nextLine(); to read the user's full input line.

3

Create a greeting message

Write a line to create a variable called greeting that stores the text "Hello, " followed by the name variable and an exclamation mark.

Java

import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        String name = scanner.nextLine();
        // Create greeting message using name
        
    }
}

Hint

Use string concatenation with "Hello, " + name + "!" to create the greeting.

4

Print the greeting message

Write a line to print the greeting variable using System.out.println().

Java

import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        String name = scanner.nextLine();
        String greeting = "Hello, " + name + "!";
        // Print the greeting message
        
    }
}

Hint

Use System.out.println(greeting); to show the greeting on the screen.

Practice

1. What is the correct way to read a full line of text input from the user in Java?

easy

A. Use Scanner.nextLine() method

B. Use Scanner.nextInt() method

C. Use System.out.println() method

D. Use Scanner.read() method

Solution

1. Step 1: Understand Scanner methods

   The Scanner class has different methods to read input. nextLine() reads the entire line as a string.

2. Step 2: Identify the correct method for full line input

   nextInt() reads only integers, System.out.println() prints output, and Scanner.read() does not exist. So, nextLine() is correct.

3. Final Answer:

   Use Scanner.nextLine() method -> Option A

4. Quick Check:

   Read full line = nextLine() [OK]

Hint: Use nextLine() to read full text line input [OK]

Common Mistakes:

• Using nextInt() to read strings
• Confusing output methods with input
• Forgetting to import Scanner

2. Which of the following is the correct syntax to create a Scanner object to read input from the keyboard?

easy

A. Scanner input = new Scanner(System.out);

B. Scanner input = new Scanner(System.file);

C. Scanner input = new Scanner(System.err);

D. Scanner input = new Scanner(System.in);

Solution

1. Step 1: Understand Scanner constructor

   The Scanner constructor takes an InputStream to read from. System.in is the standard input stream (keyboard).

2. Step 2: Identify correct stream for keyboard input

   System.out and System.err are output streams, and System.file does not exist. So, new Scanner(System.in) is correct.

3. Final Answer:

   Scanner input = new Scanner(System.in); -> Option D

4. Quick Check:

   Keyboard input = System.in [OK]

Hint: Use System.in to read keyboard input with Scanner [OK]

Common Mistakes:

• Using System.out or System.err instead of System.in
• Trying to read from non-existent System.file
• Missing parentheses in Scanner constructor

3. What will be the output of the following Java code if the user inputs: Hello World

import java.util.Scanner;
public class Main {
  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    String text = sc.nextLine();
    System.out.println(text);
    sc.close();
  }
}

medium

A. Hello

B. Hello World

C. World

D. Compilation error

Solution

1. Step 1: Understand nextLine() behavior

   The nextLine() method reads the entire line including spaces until the user presses Enter.

2. Step 2: Analyze the input and output

   User inputs "Hello World" and presses Enter. The program reads the full line and prints it exactly.

3. Final Answer:

   Hello World -> Option B

4. Quick Check:

   nextLine() reads full line = Hello World [OK]

Hint: nextLine() reads entire line including spaces [OK]

Common Mistakes:

• Assuming nextLine() reads only first word
• Confusing next() with nextLine()
• Forgetting to close Scanner

4. Identify the error in the following code snippet that reads a string input from the user:

import java.util.Scanner;
public class Test {
  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    String name = sc.next();
    System.out.println(name);
    sc.close();
  }
}

What is the problem if the user inputs "John Doe"?

medium

A. The program does not compile

B. The program throws a runtime exception

C. The program reads only "John" and ignores "Doe"

D. The program reads the full line "John Doe" correctly

Solution

1. Step 1: Understand difference between next() and nextLine()

   next() reads input only until the first space, so it reads "John" only.

2. Step 2: Analyze input and output behavior

   User inputs "John Doe" but next() stops at space, so "Doe" is ignored in this read.

3. Final Answer:

   The program reads only "John" and ignores "Doe" -> Option C

4. Quick Check:

   next() reads until space = partial input [OK]

Hint: Use nextLine() to read full input with spaces [OK]

Common Mistakes:

• Using next() when full line is needed
• Expecting next() to read multiple words
• Not testing with multi-word input

5. You want to read multiple lines of user input until the user types "exit". Which code snippet correctly implements this using Scanner and nextLine()?

hard

A. Scanner sc = new Scanner(System.in);
String line;
do {
line = sc.nextLine();
System.out.println(line);
} while (!line.equals("exit"));
sc.close();

B. Scanner sc = new Scanner(System.in);
while (sc.nextLine() != "exit") {
System.out.println(sc.nextLine());
}
sc.close();

C. Scanner sc = new Scanner(System.in);
String line = sc.next();
while (!line.equals("exit")) {
System.out.println(line);
line = sc.next();
}
sc.close();

D. Scanner sc = new Scanner(System.in);
while (true) {
String line = sc.nextLine();
if (line == "exit") break;
System.out.println(line);
}
sc.close();

Solution

1. Step 1: Understand string comparison in Java

   Use line.equals("exit") to compare strings, not ==.

2. Step 2: Analyze loop structure and input reading

   Scanner sc = new Scanner(System.in); String line; do { line = sc.nextLine(); System.out.println(line); } while (!line.equals("exit")); sc.close(); uses a do-while loop reading input once per iteration and compares correctly. Scanner sc = new Scanner(System.in); while (sc.nextLine() != "exit") { System.out.println(sc.nextLine()); } sc.close(); reads twice per loop and uses != with strings (wrong). Scanner sc = new Scanner(System.in); String line = sc.next(); while (!line.equals("exit")) { System.out.println(line); line = sc.next(); } sc.close(); uses next() which reads only one word. Scanner sc = new Scanner(System.in); while (true) { String line = sc.nextLine(); if (line == "exit") break; System.out.println(line); } sc.close(); uses == for string comparison (wrong).

3. Final Answer:

   The do-while loop with equals("exit") -> Option A

4. Quick Check:

   Use equals() and read once per loop [OK]

Hint: Use equals() for string compare and read input once per loop [OK]

Common Mistakes:

• Using == instead of equals() for strings
• Reading input multiple times per loop
• Using next() instead of nextLine() for full lines