行为参数化-白红宇

假设有如下业务：有一堆有颜色和重量的苹果，我需要通过颜色和重量取出相应苹果定义苹果

public class Apple {        private int weight = 0;        private String color = "";        public Apple(int weight, String color){            this.weight = weight;            this.color = color;        }        public Integer getWeight() {            return weight;        }        public void setWeight(Integer weight) {            this.weight = weight;        }        public String getColor() {            return color;        }        public void setColor(String color) {            this.color = color;        }        public String toString() {            return "Apple{" +                    "color='" + color + '\'' +                    ", weight=" + weight +                    '}';        }    }复制代码

假设

inventory = Arrays.asList(new Apple(80,"green"), new Apple(155, "green"), new Apple(120, "red"));复制代码

解决方案1:

List
    
      result = new ArrayList<>();        for(Apple apple: inventory){            if("green".equals(apple.getColor())){                result.add(apple);            }        }复制代码

这是最常见的方法。但是这样的结构很难复用。比如我颜色不确定呢？

解决方案2:

List
    
      result = new ArrayList<>();        for(Apple apple: inventory){            if(apple.getColor().equals(color)){                result.add(apple);            }        }复制代码

如果我需要100g以上的且红色的苹果我就需要

List
    
      result = new ArrayList<>();        for(Apple apple: inventory){            if(apple.getColor().equals(color)                             && apple.getWeight() > weight){                result.add(apple);            }        }复制代码

如果我需要100g以上或者红色的苹果

List
    
      result = new ArrayList<>();        for(Apple apple: inventory){            if(apple.getColor().equals(color)                             || apple.getWeight() > weight){                result.add(apple);            }        }复制代码

是不是变得没完没了了？解决方案3:

public static List
    
      filterApples(List
     
       inventory, ApplePredicate p){        List
      
        result = new ArrayList<>();        for(Apple apple : inventory){            if(p.test(apple)){                result.add(apple);            }        }        return result;    }复制代码

interface ApplePredicate{        boolean test(Apple a);    }     class AppleWeightPredicate implements ApplePredicate{        public boolean test(Apple apple){            return apple.getWeight() > 150;        }    }     class AppleColorPredicate implements ApplePredicate{        public boolean test(Apple apple){            return "green".equals(apple.getColor());        }    }     class AppleRedAndHeavyPredicate implements ApplePredicate{        public boolean test(Apple apple){            return "red".equals(apple.getColor())                    && apple.getWeight() > 150;        }    }复制代码

List
    
      greenApples2 = filterApples(inventory, new AppleColorPredicate());复制代码

这种方法和合适。不过如果规则也是不确定的呢？

解决方案4:

List
    
      redApples2 = filterApples(inventory, new ApplePredicate() {            public boolean test(Apple a){                return a.getColor().equals("red");            }        });复制代码

Good!这样就能做到定制化了。不过通过lambda写起来更加优美

解决方案5:

List
    
      redApples2 = filterApples(inventory, (Apple a)-> a.getColor().equals("red"));        复制代码

如果我们要推广。不只是苹果而是所有的判断规则？

解决方案6:

interface Predicate
    
     {        boolean test(T t);    }    public static 
     
       List
      
        filter(List
       
         inventory, Predicate
        
          p){        List
         
           result = new ArrayList<>(); for(T apple : inventory){ if(p.test(apple)){ result.add(apple); } } return result; } 复制代码

List
    
      redApples2 = filter(inventory, (Apple a)-> a.getColor().equals("red"));复制代码

其实java 8 的思路也是这样的解决方案7:

List
    
      redApples2 = inventory                .stream()                .filter((Apple a)-> a.getColor().equals("red"))                .collect(Collectors.toList());复制代码