如何从Java控制台应用程序中的扫描仪读取字符串?

时间:2013-07-17 04:32:20

标签: java

import java.util.Scanner;
class MyClass
{
    public static void main(String args[])
    {
        Scanner scanner = new Scanner(System.in);
        int employeeId, supervisorId;
        String name;
        System.out.println("Enter employee ID:");
        employeeId = scanner.nextInt();
        System.out.println("Enter employee name:");
        name = scanner.next();
        System.out.println("Enter supervisor ID:");
        supervisorId = scanner.nextInt();
    }
}

我在尝试输入名字和姓氏时遇到了这个异常。

Enter employee ID:
101
Enter employee name:
firstname lastname
Enter supervisor ID:
Exception in thread "main" java.util.InputMismatchException
    at java.util.Scanner.throwFor(Unknown Source)
    at java.util.Scanner.next(Unknown Source)
    at java.util.Scanner.nextInt(Unknown Source)
    at java.util.Scanner.nextInt(Unknown Source)
    at com.controller.Menu.<init>(Menu.java:61)
    at com.tests.Employeetest.main(Employeetest.java:17)

但如果我只输入名字,它就会工作。

Enter employee ID:
105
Enter employee name:
name
Enter supervisor ID:
501

我想要的是阅读完整字符串,无论是name还是firstname lastname。这有什么问题?

4 个答案:

答案 0 :(得分:23)

Scanner scanner = new Scanner(System.in);
int employeeId, supervisorId;
String name;
System.out.println("Enter employee ID:");
employeeId = scanner.nextInt();
scanner.nextLine(); //This is needed to pick up the new line
System.out.println("Enter employee name:");
name = scanner.nextLine();
System.out.println("Enter supervisor ID:");
supervisorId = scanner.nextInt();

调用nextInt()是一个问题,因为它没有拿到新行(当你点击输入时)。因此,在此之后调用scanner.nextLine()可以完成工作。

答案 1 :(得分:1)

您可以使用分隔符作为新行。直到你按回车键,你就可以把它读成字符串。

Scanner sc = new Scanner(System.in);
sc.useDelimiter(System.getProperty("line.separator"));

希望这会有所帮助。

答案 2 :(得分:1)

替换:

System.out.println("Enter EmployeeName:");
                 ename=(scanner.next());

使用:

System.out.println("Enter EmployeeName:");
                 ename=(scanner.nextLine());

这是因为next()只抓取下一个标记,空格充当标记之间的分隔符。通过这个,我的意思是扫描程序读取输入:“firstname lastname”作为两个单独的标记。因此,在您的示例中,ename将设置为firstname,扫描程序正在尝试将supervisorId设置为lastname

答案 3 :(得分:0)

您正在为nextInt输入空值,如果您给出空值,它将失败...

我在代码段中添加了空检查

试试这段代码:

import java.util.Scanner;
class MyClass
{
     public static void main(String args[]){

                Scanner scanner = new Scanner(System.in);
                int eid,sid;
                String ename;
                System.out.println("Enter Employeeid:");
                     eid=(scanner.nextInt());
                System.out.println("Enter EmployeeName:");
                     ename=(scanner.next());
                System.out.println("Enter SupervisiorId:");
                    if(scanner.nextLine()!=null&&scanner.nextLine()!=""){//null check
                     sid=scanner.nextInt();
                     }//null check
        }
}