繁体   English   中英

为什么我初始化字符串后得到空值?

[英]Why do I get null after I initialized a string?

我不明白为什么 Sp1.location 返回 NULL。 如果我运行它接缝的程序,我可以成功地初始化位置。 我以类似的方式将属性编码为整数,但这没有给我带来任何问题。

public class Database {

    static Scanner userInput = new Scanner(System.in);

    public static void main(String[] args) {

        System.out.println("Add a new spawnpoint.\n");

        System.out.println("State the name of this spawnpoint: ");


        Spawnpoints Sp1 = new Spawnpoints(getSpawnName());

        System.out.println("Done");
        System.out.println("Location: " + Sp1.getLocation()); //return as null

    }

    public static String spawnName;

    public static String getSpawnName() {

        spawnName = userInput.next();
        return spawnName;
    }

    public void setSpawnName(String spawnName) {
        this.spawnName = spawnName;
    }
}


// Import libraries
import java.util.*;

这是我的另一堂课

public class Spawnpoints extends Database {


        // Define scanner, so you can accept user input
        static Scanner userInput = new Scanner(System.in);

             // Define attributes of Spawnpoints


            private String location;
            private String iniLocation;


    // Creator, method for creating a instance of Spawnpoints. Will be the actual spawnpoints
    // I include a iniLocation so no user input is asked when calling on getLocation. 

    public Spawnpoints(String spawnName) {
        getIniLocation();

    }

    // Setters & Getters getLocation
    private String getIniLocation() {
        System.out.println("State the location of this spawnpoint:\n");
        pokemon = userInput.next ();
        return iniLocation;
    }

    public void setIniLocation(String iniLocation) {
        this.iniLocation = iniLocation;
    }


    public String getLocation() {
        location = iniLocation;
        return location;
    }


    public void setLocation(String location) {
        this.location = location;
    }



    public static void main (String[] args) {


    }

}

因为您没有设置location ,所以您将输入分配给pokemon而不是iniLocation并且当您调用函数来获取位置时,您会返回iniLocation的值,该值尚未分配任何值,因此为空。 阅读代码中的注释

private String getIniLocation() {
    System.out.println("State the location of this spawnpoint:\n");
    pokemon = userInput.next (); // remove this
    iniLocation = userInput.next (); // with this
    return iniLocation;
}

如果您在构造函数中初始化scanner对象,这是一个很好的做法。

class AnyClass{
 Scanner scan;
  public AnyClass(){
    scan= new Scanner(System.in);
  }
}

暂无
暂无

声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM