Collections.sort方法引发类java.lang.Integer不能强制转换为类java.lang.Double错误 [英] Collections.sort method is throwing class java.lang.Integer cannot be cast to class java.lang.Double error

问题描述

我从基于空手道的API自动化中的JSON响应中获取了一些运行时值.我将它们存储到ArrayList中，如:

I am getting some runtime values from a JSON response in my Karate based API Automation. I am storing them into an ArrayList like:

   * def ArrayList = Java.type('java.util.ArrayList')
   * def Collections = Java.type('java.util.Collections')
  ArrayList al = new ArrayList();
  * eval for(var i = 0; i < actualPrice.length; i++) expectedPrice.add(actualPrice[i])

当我打印列表时，输出为:

When I printed the list, the output was :

[-150，-210，-100，-112.5]

[-150, -210, -100, -112.5]

在输出上方，当我在Collections.sort(al)的帮助下申请进行排序时，它引发了以下错误:

Above output, When I applied to sort with the help of Collections.sort(al), It threw me the following error:

class java.lang.Integer cannot be cast to class java. lang.Double (java.lang.Integer and java.lang. Double is in module java.base of loader.

我需要在空手道功能文件中使用Asc/Desc排序.有人可以帮我吗?

I need to use Asc/Desc sorting in the Karate feature file. Can someone please help me here?

Java等效代码为:

Java Equivalent Code is:

    ArrayList al = new ArrayList();
    al.add(-150);
    al.add(-210);
    al.add(-100);
    al.add(-112.5);
    System.out.println(al);
    Collections.sort(al);

推荐答案

您要将 Integer s和 Double s都添加到同一 List .当排序方法尝试将 Integer 与 Double 进行比较时，它将尝试将 Integer 转换为 Double ，失败了.

You are adding both Integers and Doubles to the same List. When the sorting method attempts to compare an Integer to a Double it attempts to cast an Integer to a Double, which fails.

不要使用原始类型.如果您需要向 List 中添加双精度值，请使用 List< Double> :

Don't use raw types. If you need to add double values to the List, use a List<Double>:

List<Double> al = new ArrayList<>();
al.add(-150.0);
al.add(-210.0);
al.add(-100.0);
al.add(-112.5);
System.out.println(al);
Collections.sort(al);

这样， List 中的所有元素都将具有相同的类型，并且 Collections.sort()彼此比较不会有问题.

This way all the elements of your List will be of the same type, and Collections.sort() won't have trouble comparing them to each other.

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