Java开关状态，数组搜索 //带有菜单和对象数组的TestEmployeesProgram驱动程序。 导入java.util.*； 公共类测试员工计划{ 公共静态扫描仪控制台=新扫描仪（System.in）； 公共静态void main（字符串[]args）{ 最终int最大值=7； 员工[]=新员工[最大值]； int选择，k； 字符串名； 布尔未找到； 员工[0]=新经理（“杰里·布洛格斯”，“gfr”，5,38.5）； 员工[1]=新经理（“乔·布洛格斯”，“民主德国”，4,32.5）； 员工[2]=新管理人员（“玛丽·詹宁斯”，“nnv”，35.3,88.5,34.3）； 员工[3]=新职员（“Brian Jones”，“bbl”，42.4,78.5,23.5,45.3）； 员工[4]=新经理（“John Bloggs”，“gvr”，5,33.5）； 员工[5]=新管理员（“Bridget Jennings”，“nvv”，45.3,98.5,36.3）； 雇员[6]=新职员（“杰克·琼斯”，“bbb”，43.4,78.5,23.5,47.3）； //初读 choice=showMenu（）； //继续，直到4/4退出 while（选项！=MAX）{ 开关（选择）{ 案例1://经理 System.out.println（）； System.out.printf（“%s%-16s%-10s%6s”，“姓名”，“Id”，“工时”，“工资”）； System.out.println（“\n===================================================================”）； 对于（k=0；k

我正在运行程序，案例4中的姓名搜索直接进入默认的“未找到员工姓名”，不允许用户输入。我查看了代码，但找不到错误，有什么提示或帮助吗

//TestEmployeesProgram driver with menu & object array.
import java.util.*;
public class TestEmployeesProgram {

public static Scanner console = new Scanner(System.in);

public static void main(String[] args) {

    final int MAX = 7;

    Employee employee[] = new Employee[MAX];

    int choice,k;
    String name;
    boolean notFound;

    employee[0] = new Manager("Jerry Bloggs","gfr",5,38.5);
    employee[1] = new Manager("Joe Bloggs","gdr",4,32.5);
    employee[2] = new Admin("Mary Jennings","nnv",35.3,88.5,34.3);
    employee[3] = new Clerk("Brian Jones","bbl",42.4,78.5,23.5,45.3);
    employee[4] = new Manager("John Bloggs","gvr",5,33.5);
    employee[5] = new Admin("Bridget Jennings","nvv",45.3,98.5,36.3);
    employee[6] = new Clerk("Jack Jones","bbb",43.4,78.5,23.5,47.3);

    //Initial Read
    choice = showMenu();

    //Continue Until 4/Exit
    while (choice != MAX) {

        switch (choice) {
        case 1://Manager

            System.out.println();
            System.out.printf("%s %-16s %-10s %6s","Name","Id","Hours Worked","Pay");
            System.out.println("\n==================================================");

            for (k = 0; k < MAX; ++k)
            {
                if (employee[k] instanceof Manager){ //use of string method instance of.


                    System.out.println(employee[k].toString());
                }
            }
            break;

        case 2://Administration

            System.out.println();
            System.out.printf("%s %-16s %-10s %6s %-19s","Name","Id","Hours Worked","Pay","Admin Quota");
            System.out.println("\n==================================================");

            for (k = 0; k < MAX; ++k)
            {
                if (employee[k] instanceof Admin){
                System.out.println(employee[k].toString());

                }
            }
            break;

        case 3://Clerk

            System.out.println();
            System.out.printf("%s %-16s %-10s %6s %-19s","Name","Id","Hours Worked","Pay","Admin Quota","Units Sold");
            System.out.println("\n==================================================");

            for (k = 0; k < MAX; ++k)
            {
                if (employee[k] instanceof Clerk){
                System.out.println(employee[k].toString());
                }
            }
            break;

case 4://名称搜索
系统输出打印（“输入员工姓名：”）；
name=console.nextLine（）；
k=-1；
notFound=true；
而（（k

有两个错误。第一点是：

        case 4://Name search

            System.out.print("Enter employee name: ");
            name = console.nextLine();

            k = -1;
            notFound = true;



            while ((k < MAX-1) && (notFound))
            {
                ++k;
                if (name == employee[k].getName()){

                    System.out.println();
                    System.out.printf("%s %-16s %-10s %6s %-19s","Name","Id","Hours Worked","Pay","Admin Quota","Units Sold");
                    System.out.println("\n==================================================");

                    System.out.println(employee[k].toString());
                    System.out.println();
                    notFound = false;
                }


            }//end of case 4 while.
             if (notFound){
                System.out.println("Employee name not found\n");
            }
            break;

        case 7://exit
            System.out.println("Program exiting...");
            System.exit(0);

        default:
            System.out.println("Invalid menu choice 1..3 of 7 to Exit");



        }//end of switch

        //sub read 
        choice = showMenu();

    }//end of while 






}//end of main

//Menu method for employee selection.
public static int showMenu()
{

    int choice;
    System.out.println();

    System.out.println("Employee Program Menu");

    System.out.println("1.Show Manager pay details ");
    System.out.println("2.Show Admin pay details ");
    System.out.println("3.Show Clerk pay details ");
    System.out.println("4.Search by employee name ");
    System.out.println("7.Exit");


    System.out.print("Enter option: ");
    choice = console.nextInt();


    return choice;
}
}

nextInt

方法不使用行尾字符。请尝试以下方法：

System.out.print("Enter option: ");
choice = console.nextInt();

 if (name == employee[k].getName())   

第二个错误是，在这里您应该使用

equals

而不是

=

来比较字符串：

System.out.print("Enter option: ");
String line = console.nextLine();
choice = Integer.parseInt(line);

请尝试以下方法：

System.out.print("Enter option: ");
choice = console.nextInt();

 if (name == employee[k].getName())   

---

=

操作符测试两个字符串是否为同一对象（即字符串在内存中的相同位置）。

有两个错误。第一点是：

        case 4://Name search

            System.out.print("Enter employee name: ");
            name = console.nextLine();

            k = -1;
            notFound = true;



            while ((k < MAX-1) && (notFound))
            {
                ++k;
                if (name == employee[k].getName()){

                    System.out.println();
                    System.out.printf("%s %-16s %-10s %6s %-19s","Name","Id","Hours Worked","Pay","Admin Quota","Units Sold");
                    System.out.println("\n==================================================");

                    System.out.println(employee[k].toString());
                    System.out.println();
                    notFound = false;
                }


            }//end of case 4 while.
             if (notFound){
                System.out.println("Employee name not found\n");
            }
            break;

        case 7://exit
            System.out.println("Program exiting...");
            System.exit(0);

        default:
            System.out.println("Invalid menu choice 1..3 of 7 to Exit");



        }//end of switch

        //sub read 
        choice = showMenu();

    }//end of while 






}//end of main

//Menu method for employee selection.
public static int showMenu()
{

    int choice;
    System.out.println();

    System.out.println("Employee Program Menu");

    System.out.println("1.Show Manager pay details ");
    System.out.println("2.Show Admin pay details ");
    System.out.println("3.Show Clerk pay details ");
    System.out.println("4.Search by employee name ");
    System.out.println("7.Exit");


    System.out.print("Enter option: ");
    choice = console.nextInt();


    return choice;
}
}

nextInt

方法不使用行尾字符。请尝试以下方法：

System.out.print("Enter option: ");
choice = console.nextInt();

 if (name == employee[k].getName())   

第二个错误是，在这里您应该使用

equals

而不是

=

来比较字符串：

System.out.print("Enter option: ");
String line = console.nextLine();
choice = Integer.parseInt(line);

请尝试以下方法：

System.out.print("Enter option: ");
choice = console.nextInt();

 if (name == employee[k].getName())   

=

操作符测试两个字符串是否为同一对象（即字符串在内存中的相同位置）

将此更改为

if (name == employee[k].getName())

将此更改为

if (name == employee[k].getName())

---

这是一个普遍的问题。当使用nextLine（）读取整数值时，只读取整数值字符，\n并将其保留在缓冲区中，之后调用nextLine（）时，它不会提示用户输入，因为它已经有一个“\n”

要避免这种情况，请在showMenu方法中执行以下操作

if (name.equals(employee[k].getName()))

使用nextInt（）获得选择后

showMenu的最后一部分是：

 console.nextLine();

您应该使用.equals（）方法来比较字符串，如下所示

 choice=console.nextInt();
 console.nextLine();
 return choice;

---

祝你好运：）

这是一个常见的问题。当使用nextLine（）读取整数值时，只读取整数值字符，\n并将其保留在缓冲区中，之后调用nextLine（）时，它不会提示用户输入，因为它已经有一个“\n”

为了避免这种情况,，