Thursday, 27 July 2017
java - .nextLine() method does not take subsequent String inputs
Answer
Answer
I have a program that asks the user for an input of numbers and characters. For this I am using a Scanner
class. But when I take two subsequent string inputs, the console just moves and asks the user to input the second string. The user is unable to input the first string. I have attached my code below to be more specific.
import java.util.Scanner;
class XYZ {
public static void main(String args[]) {
int x, y, z, b, c;
String p, q;
Scanner sc = new Scanner(System.in);
System.out.println("Enter a number");
b = sc.nextInt();
System.out.println("Enter a new number");
c = sc.nextInt();
System.out.println("Enter a String");
p = sc.nextLine();
System.out.println("Enter a new String");
q = sc.nextLine();
System.out.println("Enter a number");
x = sc.nextInt();
System.out.println("Enter 2nd number");
y = sc.nextInt();
System.out.println("Enter 3rd number");
z = sc.nextInt();
ABC a = new ABC();
a.PQR(b, c);
a.PQR(p, q);
a.PQR(x, y, z);
}
}
class ABC {
void PQR(int a, int b) {
int res = a + b;
System.out.println("The sum is" + " " + res);
}
void PQR(String a, String b) {
System.out.println("Thye concataneted string is" + a + " " + b);
}
void PQR(int a, int b, int c) {
int res = a * b * c;
System.out.println("The product is" + " " + res);
}
}
Answer
You need to make a dummy call to nextLine()
after your call to nextInt()
. The nextInt()
isn't taking the "\n" into account.
Alternatively, and preferably, just use nextLine()
to accept all input and whichever ones need to be ints
, just use:
int i = Integer.parseInt(**string**);
Does this help?
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