java - 数组列表。 ArrayList 到 int 和 double

Question

我遇到的问题是我无法从 arraylist 中取出一个数字并将其变成我拥有的 int 或 double使用体重和高度来计算 BMI。请看!
作业是输入客人的体重、身长和姓名，然后将长高比不佳的客人挑出来。主要是我创建了一个有几个客人的数组，当我运行它时说:

"Exception in thread "main" java.lang.NumberFormatException: For input string "name"

和

java.lang.NumberFormatException.forInputString(NumberFormatException.java:65)
at java.lang.Integer.parseInt(Integer.java:492)
at java.lang.Integer.parseInt(Integer.java:527)
at Diet.putOnDiet(Diet.java:12)
at TestDiet.main(TestDiet.java:7)

Diet类如下:

public class Diet{

    public static ArrayList<Guest> putOnDiet(ArrayList <Guest> list){
        ArrayList<Guest> namn = new ArrayList<Guest>();
        ArrayList<Guest> hej = new ArrayList<Guest>();
        for(int i = 0; i<=list.size()/3; i = i+3){

            int langd = Integer.parseInt(list.get(i+1).toString()); //I dont know how to make this work
            double vikt = Double.parseDouble(list.get(i).toString());
            String name = list.get(i+2).toString();

            if ((vikt) > 1.08*(0.9*(langd -100))){
                namn.add(new Guest(vikt, langd, name));
            }
        }

        return namn;
    }
}

Guest 类:

public class Guest { 

   private double weight; private double length; private String name; 

   public Guest(double weight, double length, String name){ 
      this.name = name; this.weight = weight; this.length = length; // Maybe the problem is here. How do you modify the class to make it work?
   } 
   public String getName() {
      return name; 
   } 
   public double getWeight() 
   { 
      return weight; 
   } 
   public double getLength() { 
      return length; 
   } 
   public void setName(String name) { 
      this.name = name; 
   } 
   public void setWeight(double weight) { 
      this.weight = weight; 
   } 
   public void setLength(double length) 
   { this.length = length; 
   } 
   public boolean isSlim() { 
      if (weight >= 1.08 * 0.9 * (length - 100)) { 
         return false; 
      }
      else 
         return true; 
   }
   public String toString() { 
      return name + "\n" + weight + "\n" + length; 
   } 
}

Answer 1

你确定你正在解析一个整数吗？

井号解析失败时抛出异常。当字符串不是像“somthing#$%^&”这样的数字时。所以尝试替换这一行

int langd = Integer.parseInt(list.get(i+1).toString());

有了这个

try {
       int langd = Integer.parseInt(list.get(i+1).toString());
} catch (NumberFormatException e) {
       System.out.println(list.get(i+1).toString() +" : This is not a number");
       System.out.println(e.getMessage());
}

编辑 在阅读了 WOUNDEDStevenJones 的回答后，我还认为您甚至不应该使用 toString() 或解析方法。有关详细信息，请参阅 WOUNDEDEStevenJones 的回答。