JAVA PROGRAM • LEVEL 01 — BASICS & INPUT/OUTPUT

Read a Complete Text Line in Java

Learn read a complete text line in java with a complete, compiler-ready Java 17 example.

BeginnernextLine()String input

PROBLEM UNDERSTANDING

Input and expected output

Sample input

Bhavya

Sample output

Welcome, Bhavya!

COMPLETE JAVA 17 PROGRAM

Complete Java 17 implementation

class Main {
    public static void main(String[] args) throws Exception {
        java.util.Scanner input = new java.util.Scanner(System.in);
        String name = input.nextLine();
        System.out.println("Welcome, " + name + "!");
    }
}

CURRENT STEP

Select Start to walk through the important lines.

SELECTED LINE

No line selected

EXPECTED OUTPUT FOR THE SAMPLE

Welcome, Bhavya!

0%Step 0 of 0

PROGRAM EXPLANATION

Algorithm and explanation

Read the sample input and convert it into the Java values required for read a complete text line in java.
Apply nextLine() and String input in the order shown by the program.
Print the final result and compare it with the documented sample output.

This Java 17 example demonstrates nextLine() and String input through a complete, executable program. The values are intentionally small so students can trace each statement, verify the output, and then modify the example safely in the online compiler.

EFFICIENCY

Time and space complexity

Time complexity

O(1)

Auxiliary space

O(1)

DEBUGGING CHECKLIST

Common mistakes

Check this

Create one Scanner for System.in, read values in the documented order, and use the correct next... method for each type.

Check this

Use nextLine() with compatible Java types; Java checks types and method signatures at compile time.

Check this

Java is case-sensitive, and every statement that requires a semicolon must end with one.

Try it yourself

Practice: Run the program with the sample input, predict its output, and then test one boundary case of your own.