首先，我想说我是Java的新手，来自C ++背景。我永远无法与我的老师联系，所以我想尝试在这里发布这个问题，我一直想知道（希望我能说得对）：
如何在不使用static的情况下创建方法？我知道我可能需要为它上课，但我该怎么做呢？只是一个没有变量和函数的类？我是否创建了第二个包含除.java文件后面的类以外的主类的类？例如：
package musiclibrary;
import java.util.Scanner;
/**
* This class implements a user to create a playlist from a selection of artists and songs
* @author ahb5190
*/
public class MusicLibrary {
static String divider = "*****************************************************";
//Scanner class
static Scanner input = new Scanner(System.in);
/**
* Welcome menu
*/
public static void welcomeMenu()
{
System.out.println(divider);
System.out.println();
System.out.println("Welcome to Art's personal music library!");
System.out.println();
System.out.println("Choose an option:");
System.out.println("1) Create Playlist");
System.out.println("2) Delete Playlist");
System.out.println("3) Add Selection to Playlist");
System.out.println("4) Remove Selection from Playlist");
System.out.println("5) Quit");
System.out.println();
System.out.print("Your choice?: ");
}
/**
*
* @param min error check low bound
* @param max error check high bound
* @return
*/
public static int getData(int min, int max)
{
boolean goodInput = false;
int choice = 1; //Will be overwritten
while(!goodInput)
{
choice = input.nextInt();
if(choice < min || choice > max)
{
System.out.print(choice + " is not a valid choice. Please enter a number between " + min + " and " + max + ": ");
goodInput = false;
}
else
{
goodInput = true;
}
}
return choice;
}
/**
* @param args the command line arguments
*/
public static void main(String[] args)
{
//Variables
int getDataMin = 1;
int getDataMax = 5;
boolean quit = false;
int userInput;
do {
welcomeMenu();
userInput = getData(getDataMin, getDataMax);
if (userInput == 1)
{
quit = false;
}
else if (userInput == 2)
{
quit = false;
}
else if (userInput == 3)
{
quit = false;
}
else if (userInput == 4)
{
quit = false;
}
else if (userInput == 5)
{
quit = true;
}
} while(!quit);
}
}
是分配的java程序的最开始。如果我从static移除public static void welcomeMenu()，则当我尝试在主电话中调用non-static method welcomeMenu() cannot be referenced from a static context时，它会向我welcomeMenu();。
另一段代码（不是很整齐，是定时考试的一部分）：
package lalala;
/**
*
* @author ahb5190
*/
public class Lalala {
public class Movie
{
private String title;
private String genre;
private String director;
private String star;
public Movie (String t, String g, String d, String s)
{
title = t;
genre = g;
director = d;
star = s;
}
public String gettitle()
{
return title;
}
public String getGenre()
{
return genre;
}
public String getDirector()
{
return director;
}
public String getStar()
{
return star;
}
public void setTitle(String x)
{
title = x;
}
public void setGenre(String x)
{
genre = x;
}
public void setDirector(String x)
{
director = x;
}
public void setsStar(String x)
{
star = x;
}
public boolean equals(Movie otherMovie)
{
if(otherMovie == null)
{
return false;
}
else
{
return title.equals(otherMovie.title) && genre.equals(otherMovie.genre) && director.equals(otherMovie.director) && star.equals(otherMovie.star);
}
}
@Override
public String toString()
{
return(title + " " + genre + " " + director + " " + star);
}
}
/**
* @param args the command line arguments
*/
public static void main(String args[])
{
Movie a;
a = new Movie("Star Trek into Darkness", "Sci-fi", "J.J. Abrams", "Chris Pine"); //error: non-static variables this cannot be referenced from a static context
Movie b = new Movie("Star Trek", "Sci-Fi", "J.J. Abrams", "Chris Pine"); //error: non-static variables this cannot be referenced from a static context
Movie c = new Movie("Independence Day", "Action", "Roland Emmerich", "Will Smith"); //error: non-static variables this cannot be referenced from a static context
System.out.println("Movies");
System.out.println("Title: " + a.title);
System.out.println("Genre: " + a.genre);
System.out.println("Director: " + a.director);
System.out.println("Star: " + a.star);
System.out.println();
System.out.println("Title: " + b.title);
System.out.println("Genre: " + b.genre);
System.out.println("Director: " + b.director);
System.out.println("Star: " + b.star);
System.out.println();
System.out.println("Title: " + c.title);
System.out.println("Genre: " + c.genre);
System.out.println("Director: " + c.director);
System.out.println("Star: " + c.star);
System.out.println();
a.equals(b);
}
}
给出了与之前相同的静态变量错误，如上面的代码所述。在那种情况下，我如何让它“工作”就是从static删除public static void main(String args[])。
真的想学习正确的Java方法，任何帮助都会受到赞赏。我也希望这会遇到MCV。
答案 0 :(得分：1)
要访问类的非静态成员，您需要该类的实例。
所以new Movie("a", "b", "c", "d").getGenre()是合法的。
从main中删除static关键字是不合法的，因为它是程序的入口点，因此必须存在。
修改： 在第一个源中，main（）方法不会创建MusicLibrary的任何实例，这就是您需要在所有成员上使用静态的原因。
添加MusicLibrary lib = new MusicLibrary();，然后调用lib.welcomeMenu();，您就可以摆脱静态关键字。
答案 1 :(得分：0)