当说类没有接受String []的静态void主方法时，在何处添加main方法

Question

我正在尝试使我的名为ZawTennisPlayer的程序使用构造函数和显示方法输出以下内容。 我正在使用一个名为TestTennisPlayer2的示例代码来对其进行测试并获得所需的输出。 我当前的代码是：

xticks([1 2 3]);
xticklabels({'string 1','string 2','string 3'});

但是，当该程序编译时，却出现错误“该类没有接受String []的静态void main方法”。当我运行它时。我知道我应该在ZawTennisPlayer程序中的某个位置添加一些“公共静态void main（String [] args）”，但是我不确定在哪里。任何想法我如何可以修复程序以获得所需的输出？预先感谢！

我用来测试程序的示例代码“ TestTennisPlayer2”是：

public class ZawTennisPlayer

{
//instance variables
private String playerName;
private String country;
private int rank;
private int age;
private int wins;
private int losses;

//default constructor
public ZawTennisPlayer()
{
playerName=null;
country=null;
rank=0;
age=0;
wins=0;
losses=0;

}

//parameterized constructor
public ZawTennisPlayer(String playerName,String country)
{
this.playerName=playerName;
this.country=country;
rank=0;
age=0;
wins=0;
losses=0;

}

//parameterized constructor
public ZawTennisPlayer(String playerName,String country,int rank, int age)
{
this.playerName=playerName;
this.country=country;
this.rank=rank;
this.age=age;
wins=0;
losses=0;
}

//parameterized constructor
public ZawTennisPlayer(String playerName,String country,int rank, int age,int wins,int losses)
{
this.playerName=playerName;
this.country=country;
this.rank=rank;
this.age=age;
this.wins=wins;
this.losses=losses;

}

//all accesor and mutator method for all six fields.

public String getPlayerName()

{

return playerName;

}

public void setPlayerName(String playerName)

{

this.playerName = playerName;

}

public String getCountry()

{

return country;

}

public void setCountry(String country)

{

this.country = country;

}

public int getRank()

{

return rank;

}

public void setRank(int rank)

{

this.rank = rank;

}

public int getAge()

{

return age;

}

public void setAge(int age)

{

this.age = age;

}

public int getWins()

{

return wins;

}

public void setWins(int wins)

{

this.wins = wins;

}

public int getLosses()

{

return losses;

}

public void setLosses(int losses)

{

this.losses = losses;

}

//method to display player details
public void displayPlayer()
{

System.out.println("Player's name: " + getPlayerName());

System.out.println("Player's country: " + getCountry());

System.out.println("Player's rank: " + getRank());

System.out.println("Player's age: " + getAge());

System.out.println("Player's wins: " + getWins());

System.out.println("Player's losses: " + getLosses());

System.out.println();
}

}

Answer 1

如评论中所述，似乎您正在尝试运行ZawTennisPlayer的地方TestTennisPlayer2
如果通过命令提示符运行，请使用

>javac TestTennisPlayer2.java

>java TestTennisPlayer2 

或通过Eclipse IDE

open TestTennisPlayer2.java, right click -> run as > java application