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“java scanner string nextline after nextint” Code Answer

java scanner string nextline after nextint

java by DevPedrada on Aug 24 2020 Donate

//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();

Source: stackoverflow.com

java scanner string nextline after nextint

java by Grieving Gannet on May 07 2020 Donate

int option = input.nextInt();
input.nextLine();  // Consume newline left-over
String str1 = input.nextLine();

Source: stackoverflow.com

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“java scanner string nextline after nextint” Code Answer

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