Question

我试图得到它，以便当用户输入例如吉他模型时，它将从我的ArrayList返回该特定模型的所有细节。但是，当我运行程序时没有任何结果，我没有错误。我不知道该怎么做，任何帮助将不胜感激。

P.S。在我的交换机案例中，我一直尝试多种不同的方式来获得没有运气的结果，所以如果它到处都是请忽略。

package guitarsrentalmanagementsystem;

import StockManagement.Guitar;
import java.util.ArrayList;
import java.util.Scanner;

public class GuitarsRentalManagementSystem {

    public static void main(String[] args) {
        // TODO code application logic here        
        ArrayList<Guitar> data = new ArrayList<>();
        data.add(new Guitar(100, 150.99, "Gibson", "Les Paul", 
                "Acoustic", "INDIAN_ROSEWOOD", "MAHOGANY", 1995, 8.99,  "Red", 
        true));
        data.add(new Guitar(101, 180.00, "Fender", "Stratocaster", 
                "Electric", "BRAZILIAN_ROSEWOOD", "MAHOGANY", 2001, 10.99, "Brown", 
        true));
        data.add(new Guitar(102, 110.50, "Martin", "Noisemaker", 
                "Acoustic", "INDIAN_ROSEWOOD", "BRAZILIAN_ROSEWOOD", 2005, 5.99, "Black", 
        true));

        for(Guitar g : data){
            g.printDetail();
        }

        //Show data from the adobe arraylist        
        System.out.println("Valid search parameters are as follows :");
        System.out.println("1. Model, eg Les Paul");
        System.out.println("2. Serial number, eg 101");
        System.out.println("3. Year of manufacture, eg 2001");         

        Scanner input = new Scanner(System.in);
        System.out.println("Please select how you would like to search");        
        int search = input.nextInt(); 
        for(Guitar g : data){
        switch(search){
            case 1:                
                System.out.println("You have chosen to search by model");
                System.out.println("Valid models are Les Paul, Stratocaster or Noisemaker");
                System.out.println("Please enter a model to search");                
                Scanner inputO1 = new Scanner(System.in);
                String modelSearch = input.next();
                if("Les Paul".equals(modelSearch)){
                    g.printDetail();
                }
                break;

            case 2:
                System.out.println("You have chosen to search by Serial number");
                System.out.println("valid serial numbers are between 100 and 105");
                System.out.println("Please enter a serial number to search");
                Scanner inputO2 = new Scanner(System.in);
                int serialSearch = input.nextInt();
                System.out.println(g.getSerialNumber());                
                break;

            case 3:                
                System.out.println("You have chosen to search by year of manufacture");    
                System.out.println("Valid manufacture dates are 1995, 2001 and 2005");
                System.out.println("Please enter a manufacture year to search");
                Scanner input03 = new Scanner(System.in);
                int yearSearch = input.nextInt();
                System.out.println(g.getYearOfManufacture());

                break;
        } 
        break;
        }
    }
}

Answer 1

您正在阅读输入的方式存在问题：当您使用scanner对象并在其上调用nextInt（）时。它只读取数字。同样，当你接下来打电话时，它只读取一个字符串值，即

如果您输入“Les Paul”作为输入，则扫描仪对象next（）方法仅准备Les，因此它与您的条件不符。

所以一旦你读取整数调用nextLine（）方法，那么光标移动到下一行，现在通过再次调用nextLine（）方法从下一行读取输入

 case 1:
   System.out.println("You have chosen to search by model");
   System.out.println("Valid models are Les Paul, Stratocaster or Noisemaker");
   System.out.println("Please enter a model to search");
   Scanner inputO1 = new Scanner(System.in);
   input.nextLine();
   String modelSearch = input.nextLine();
   if ("Les Paul".equals(modelSearch)) {
    g.printDetail();
   }
   break;

如果用户将模型输入为“Les Paul”，则上述代码将起作用。

无法从我的ArrayList返回一个对象的代码

1 个答案: