java string next line
// its \r\n
String a = "Hello, "
String b = "there!"
System.out.println(a+b);//returns Hello, there
System.out.println(a+"\r\n"+b);//returns Hello,
//there! //it will be on the next line!java string next line
// its \r\n
String a = "Hello, "
String b = "there!"
System.out.println(a+b);//returns Hello, there
System.out.println(a+"\r\n"+b);//returns Hello,
//there! //it will be on the next line!java scanner string nextline after nextint
//The problem is with the input.nextInt() method - it only reads the int value. So when you continue reading with input.nextLine() you receive the "\n" Enter key. So to skip this you have to add the input.nextLine(). Hope this should be clear now.
//Try it like that:
System.out.print("Insert a number: ");
int number = input.nextInt();
input.nextLine(); // This line you have to add (It consumes the \n character)
System.out.print("Text1: ");
String text1 = input.nextLine();
System.out.print("Text2: ");
String text2 = input.nextLine();Copyright © 2021 Codeinu
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