java - 为什么有时会跳过 input.next() ？

Question

在我的代码(如下)中，只是跳过了 input.next(); 。有人可以指出原因吗？

for (int i=0; i<empNum; i++)//for each employee they want to work with
    {
        System.out.print("\r\n\r\nPROFILE FOR EMPLOYEE #" + (i+1) + ":\r\n"
                        +"type Hourly(1), Salaried(2), Salaried plus Commission(3)\r\n"
                        +"Enter 1, 2, or 3 ==> ");//display type gathering
        int typeChooser = input.nextInt();//gather type

        System.out.print("Name ==> ");//ask for name
        String name = input.next();//get name

        System.out.print("Social Security Number ==> ");//ask for ssn
        String ssn = input.next();//THIS PART IS SKIPPED

        System.out.print("Birthday Month (1-12) ==> ");//ask for bdayMonth
        int bdayMonth = input.nextInt();//get bdayMonth

        System.out.print("Birthday bonus week (1-4) ==> ");//ask for bdayWeek
        int bdayWeek = input.nextInt();//get bdayWeek            
}

编辑:我刚刚注意到它被跳过的唯一一次是当名称中有空格时(即 Bob Smith 而不仅仅是 Bob)

Answer 1

社会安全号码是否包含空格？如果是这样，您可以尝试 nextLine(); 方法。此方法返回当前行的其余部分，不包括末尾的任何行分隔符。

System.out.print("Social Security Number ==> ");//ask for ssn
String ssn = input.nextLine();