Scanner is skipping nextLine() after using next() or nextFoo()?

Java Io Java.util.scanner

If you have encountered a problem where the nextLine() method of the Scanner class is being skipped after using the next() or nextFoo() methods, you are not alone. This is a common issue that many Java developers face when reading input from the user. In this article, we will explore why this problem occurs and how to solve it.

Problem Description

Let's first understand the problem in more detail. The issue arises when you use the nextInt() or any other nextFoo() method of the Scanner class, and then immediately use the nextLine() method to read a string. In this scenario, the nextLine() method is skipped and the program directly moves to the next line, resulting in unexpected behavior.

Why does this problem occur?

This problem occurs due to the behavior of the nextFoo() methods in the Scanner class. When you call the nextInt() method, it reads the integer value from the input, but it leaves the newline character (\n) in the input buffer. The nextLine() method then reads this leftover newline character, assuming it as the input and moves to the next line, resulting in the skipped input.

Solution

Fortunately, there are a few different ways to solve this problem. Let's explore some of the solutions:

Solution 1: Clear the input buffer after using nextFoo()

You can manually clear the newline character (\n) from the input buffer after using the nextInt() or any other nextFoo() method. You can do this by adding an extra input.nextLine() before reading the string using nextLine(), like this:


                System.out.println("Enter numerical value");    
                int option;
                option = input.nextInt(); // Read numerical value from input
                input.nextLine(); // Clear newline character from input buffer
                System.out.println("Enter 1st string"); 
                String string1 = input.nextLine(); // Read 1st string
                System.out.println("Enter 2nd string");
                String string2 = input.nextLine(); // Read 2nd string

Solution 2: Use nextLine() to read the numerical value

Alternatively, you can use the nextLine() method to read the numerical value as a string and then convert it to an integer using the parseInt() method of the Integer class. This way, you won't have to deal with the newline character in the input buffer. Here's an example:


                System.out.println("Enter numerical value");    
                String inputString = input.nextLine(); // Read numerical value as string
                int option = Integer.parseInt(inputString); // Convert string to int
                System.out.println("Enter 1st string"); 
                String string1 = input.nextLine(); // Read 1st string
                System.out.println("Enter 2nd string");
                String string2 = input.nextLine(); // Read 2nd string

Solution 3: Use nextLine() for all inputs

Another approach is to use the nextLine() method for all inputs, including the numerical value. You can then parse the string to the desired data type using the appropriate parsing methods. Here's an example:


                System.out.println("Enter numerical value");    
                String inputString = input.nextLine(); // Read numerical value as string
                int option = Integer.parseInt(inputString); // Convert string to int
                System.out.println("Enter 1st string"); 
                String string1 = input.nextLine(); // Read 1st string
                System.out.println("Enter 2nd string");
                String string2 = input.nextLine(); // Read 2nd string

Conclusion

The issue of the nextLine() method being skipped after using the next() or nextFoo() methods of the Scanner class can be frustrating, but it can be easily solved by following the solutions mentioned in this article. By understanding the underlying cause and implementing the appropriate solution, you can ensure that your input is read correctly, without any unexpected behavior.

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