我正在使用这些Scanner
方法nextInt()
并nextLine()
读取输入。
它看起来像这样:
System.out.println("Enter numerical value");
int option;
option = input.nextInt(); // Read numerical value from input
System.out.println("Enter 1st string");
String string1 = input.nextLine(); // Read 1st string (this is skipped)
System.out.println("Enter 2nd string");
String string2 = input.nextLine(); // Read 2nd string (this appears right after reading numerical value)
问题是输入数值后,第一个input.nextLine()
被跳过,第二个input.nextLine()
被执行,这样我的输出是这样的:
Enter numerical value
3 // This is my input
Enter 1st string // The program is supposed to stop here and wait for my input, but is skipped
Enter 2nd string // ...and this line is executed and waits for my input
我测试了我的应用程序,看起来问题在于使用input.nextInt()
. 如果我删除它,那么两者string1 = input.nextLine()
都string2 = input.nextLine()
将按照我的意愿执行。