Taking input string java

Taking input string java

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Java User Input (Scanner)

The Scanner class is used to get user input, and it is found in the java.util package.

To use the Scanner class, create an object of the class and use any of the available methods found in the Scanner class documentation. In our example, we will use the nextLine() method, which is used to read Strings:

Example

import java.util.Scanner; // Import the Scanner class class Main < public static void main(String[] args) < Scanner myObj = new Scanner(System.in); // Create a Scanner object System.out.println("Enter username"); String userName = myObj.nextLine(); // Read user input System.out.println("Username is: " + userName); // Output user input >> 

If you don’t know what a package is, read our Java Packages Tutorial.

Input Types

In the example above, we used the nextLine() method, which is used to read Strings. To read other types, look at the table below:

Method	Description
nextBoolean()	Reads a boolean value from the user
nextByte()	Reads a byte value from the user
nextDouble()	Reads a double value from the user
nextFloat()	Reads a float value from the user
nextInt()	Reads a int value from the user
nextLine()	Reads a String value from the user
nextLong()	Reads a long value from the user
nextShort()	Reads a short value from the user

In the example below, we use different methods to read data of various types:

Example

import java.util.Scanner; class Main < public static void main(String[] args) < Scanner myObj = new Scanner(System.in); System.out.println("Enter name, age and salary:"); // String input String name = myObj.nextLine(); // Numerical input int age = myObj.nextInt(); double salary = myObj.nextDouble(); // Output input by user System.out.println("Name: " + name); System.out.println("Age: " + age); System.out.println("Salary: " + salary); >> 

Note: If you enter wrong input (e.g. text in a numerical input), you will get an exception/error message (like «InputMismatchException»).

You can read more about exceptions and how to handle errors in the Exceptions chapter.

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Taking input string java

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How to take String input from console in Java?

And I have problem in the following portion of code as taking input. This is not taking any input from user skip this part. Is there any problem in this code.

 System.out.println("Do You Want more Calculation. Yes/No"); String str1=input.nextLine(); if("no".equals(str1))

2 Answers 2

The problem is because of the fact that you pick nextInt earlier. The user inputs a number and a new line character. You pick the number and the new line character is kept buffered. When you perform nextLine() it reads all characters between what is being pointed and the next EOL . It reads empty String as it’s in the buffer before \n and then another nextLine() requires the program to wait for the input.

When you put a breakpoint after String str = input.nextLine() you’ll see that it is actually empty String : «» .

 System.out.println("Do You Want more Calculation. Yes/No"); String str=input.nextLine(); System.out.println("Do You Want more Calculation. Yes/No"); String str1=input.nextLine(); 

 input.nextLine(); System.out.println("Do You Want more Calculation. Yes/No"); String str1=input.nextLine(); 

Why wouldn’t you use, input.next() ? I think they are looking for the next complete token, not the next line.

Your problem is you were using nextInt() in your cases, so your string stored is the new line. All you have to do is replace:

 System.out.println("Do You Want more Calculation. Yes/No"); String str1=input.nextLine(); if("no".equals(str1))

 System.out.println("Do You Want more Calculation. Yes/No"); input.nextLine(); // this is what I added String str1=input.nextLine(); if("no".equals(str1))

You can remove the following as those 2 lines serve no useful purpose:

 System.out.println("Do You Want more Calculation. Yes/No"); String str=input.nextLine(); 

Explanation: the input.nextLine(); that I added basically captures the new line character when you hit enter or return. Next, you will have to call input.nextLine() again and assign it to str1 to capture the string you intended to use. Also, a small note on your program: if a user were to type in No, your program still considers the value as Yes, because it will change go to false only if no is lower case. A solution to that would be to convert the string you receive to lower case by calling str1.toLowerCase() . Also, you might want to check for an explicit yes answer, because a user might type dog and your program still considers it as a yes.

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How to take String Input in Java

There are various ways to take String input in Java. In this section, we are going to discuss how to take String input in Java. There are following ways to take String input in Java:

1. By Using Java Scanner class
2. By Using Java BufferedReader class
3. By Using the Command Line argument

By Using Java Scanner class

The Scanner class is defined in java.util package that is used to take input from the user. The Scanner class provides the following two methods to take input from the user.

Let’s discuss each of the mentioned methods to take input from the user.

1. Scanner.nextLine() Method

The nextLine() method reads the input till the line gets terminated and shifts the cursor to the next line. The syntax of the nextLine() method is:

The method does not accept any parameter. It returns the string that was skipped. If the method finds no line to read, the method throws the NoSuchElementException.

FileName: StringInputExample.java

// importing the Scanner class import java.util.Scanner; public class StringInputExample < // main method public static void main(String argvs[]) < // To hold the string input by the user String str; // creating an object of the Scanner class Scanner scnr = new Scanner(System.in); System.out.print("Enter a string: "); // invoking the method nextLine() // to take input from the user str = scnr.nextLine(); // for new line System.out.println(); // displaying the entered string System.out.print("The string entered by the user is: " + str ); >>

Explanation: After displaying the message Enter a string, the cursor waits on the console for the user to enter string. The method nextLine() reads till the user hits enter and return the read stuff, which is captured by the variable str. The same read string is shown in the output. Note that hitting the enter button shows that the line has been terminated.

Scanner.next() Method

The next() method reads the input till the line gets terminated or white space is encountered. The main difference between the nextLine() and next() method is that the latter one terminates when white space is encountered, where the former one terminates only when enter is pressed. The syntax of the next() method is:

The read line is returned.

The method does not accept any parameter. It returns the string that was skipped. If the method finds no line to read, the method throws the NoSuchElementException.

FileName: StringInputExample1.java

// importing the Scanner class import java.util.Scanner; public class StringInputExample1 < // main method public static void main(String argvs[]) < // To hold the string input by the user String str; // Instantiating the Scanner class by creating its object Scanner scnnr = new Scanner(System.in); System.out.print("Enter a string: "); // invoking the method next() // to take input from the user str = scnnr.next(); // for new line System.out.println(); // displaying the entered string System.out.print("The string entered by the user is: " + str ); >>

Explanation: The user enters My Name is Khan. However, after the word “My” white space is encountered. Hence, the rest three words are not read, and the same can be confirmed by observing the output.

By using Java BufferedReader Class

Java BufferedReader Class is used to read the stream of characters. The class accepts an object of the InputStreamReader class. So, it is necessary to create a constructor of the InputStreamReader class and pass its object to the BufferedReader class as a parameter. The BufferedReader class has the readLine() method to take input from a user. The readLine() method reads one line at a time.

It returns the line entered by the user. It does not include any line-termination characters, or null if the end of the stream has been reached.

Let’s observe the following program to understand how the BufferedReader class is used for taking input from the user.

FileName: StringInputExample2.java

// Import statements import java.io.BufferedReader; import java.io.IOException; import java.io.InputStreamReader; public class StringInputExample2 < // main method public static void main(String argvs[]) throws IOException < String str; // Creating an object of the InputStreamReader class InputStreamReader inputStrObj = new InputStreamReader(System.in); // Creating an object of the BufferedReader class BufferedReader bufrObj = new BufferedReader(inputStrObj); System.out.print("Enter a string: "); // invoking the method readLine() // to take input from the user str = bufrObj.readLine(); // for new line System.out.println(); // displaying the entered string System.out.print("The string entered by the user is: " + str ); >>

Explanation: In the constructor of the InputStreamReader class, System.in is passed. It is done because the input is being taken from the keyboard. With the help of InputStreamReader class, the BufferedReader class stores the input given by the user through keyboard.

By using the Command Line Argument

Command Line argument is present in the main method (String argvs[]). String argvs[] is a string array that accepts a line as an input. Let’s understand the usage of command-line argument with the help of the following Java program.

FileName: StringInputExample3.java

public class StringInputExample3 < // main method public static void main(String argvs[]) < // To store the elements of the String array argvs String str = ""; // Calculating the size of the String array int size = argvs.length; // loop for iterating over the elements of the String array argvs for(int i = 0; i < size; i++) < // Converting the String array into a string str = str + argvs[i] + " "; >// displaying the entered string System.out.print("The string entered by the user is: " + str ); > >

Explanation: The command line argument is put when the program is executed using the java command. The number of elements in the String array argvs[] is decided by the white spaces present in the input string. In the input string (“My name is Khan”), there are three white spaces present. Hence, there are four string elements present in the array argvs (“My”, “name”, “is”, “Khan”). The array argvs is then parsed using the for-loop, and elements of the array argvs are then concatenated to build the string again, as we have shown in the above output.

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