如何组织不同类型的数组？

Question

嗨,

我是Java新手并试图找出如何将这些数据推入数组(6行,3列)？

x1 John 6
x2 Smith 9
x3 Alex 7
y1 Peter 8
y2 Frank 9
y3 Andy 4

之后,我将从最后一栏获取数字进行数学计算.

这是我的代码......

public class Testing {
    public static void main(String[] args) {

        Employee eh = new Employee_hour();

        Employee_hour [] eh_list = new Employee_hour[6];
        eh_list[0] = new Employee_hour("x1", "John", 6);
        eh_list[1] = new Employee_hour("x2", "Smith", 9);
        eh_list[2] = new Employee_hour("x3", "Alex", 7);
        eh_list[3] = new Employee_hour("y1", "Peter", 8);
        eh_list[4] = new Employee_hour("y2", "Frank", 9);
        eh_list[5] = new Employee_hour("y3", "Andy", 4);
        print(eh_list);
    }

    private static void print(Employee_hour[] mass){
        for (int i = 0; i < mass.length; i++) {
            System.out.print(mass[i] + " ");
        }
        System.out.println();
    }
}

但我得到这个作为输出......

testing.Employee_hour@1a752144 testing.Employee_hour@7fdb04ed testing.Employee_hour@420a52f testing.Employee_hour@7b3cb2c6 testing.Employee_hour@4dfd245f testing.Employee_hour@265f00f9

我怎样才能从最后一栏获得数字？

Answer 1

为什么不为您的记录创建特定的Java bean？

class Person {
  String id;
  String name;
  int anotherNumber;
  // constructor, getters, setters
}

然后像这样使用它:

Person [] people = new Person[10];
people[0] = new Person("x1", "John", 6);
...

或者更好地使用java.util.List而不是阵列.

现场访问

要访问单独的字段,您需要将字段设为公共(非常糟糕的想法)并简单地将它们称为object_instance.field_name,或者提供所谓的getter:

class Person {
  String id;
  String name;
  int anotherNumber;
  // constructor, getters, setters

  public int getAnotherNumber() {
     return anotherNumber;
  }
}

打印时调用它:

for (int i = 0; i < mass.length; i++) {
    System.out.print(mass[i].getAnotherNumber() + " ");
}

你为什么尝试不起作用:

System.out.println(mass[0])在您的情况下将打印整个对象表示,默认情况下它会打印它在您的情况下执行的操作.要做得更好,你需要覆盖Object的String toString()方法:

class Person {
  String id;
  String name;
  int anotherNumber;
  // constructor, getters, setters

  public String toString() {
     return "{id="+id+", name="+name+", anotherNumber="+anotherNumber+"}";
  }
}