我应该扩展ArrayList中添加属性不为null？ [英] Should I extend ArrayList to add attributes that isn't null?

问题描述

我想对象的集合添加到一个ArrayList中，只有在特定的属性不为null。

I would like to add a collection of objects to an arrayList ,only if the particular attribute is not null.

我想延长ArrayList和落实儿童类中的检查。

I am thinking of extending the ArrayList and implementing the check inside the child class.

一另一种方法是把它在一个ArrayList中之前检查的属性，但是这意味着，我将不得不驱散，如果检查每一个地方，如果我需要的对象添加到基于逻辑的数组列表。

One alternate way is to check for the the attribute before putting it in a Arraylist, but that would mean , i will have to scatter the if checks every where if i need to add the objects to the arraylist based on the logic.

我想知道你的就可以了的想法...再一想它是一个矫枉过正？

I would like to know your thoughts on it ... on a second thought is it a overkill ?

推荐答案

我真的建议包装的ArrayList 使用详细记录的的装饰的格局。您只需换你的ArrayList 另一个列表的实现，代表大多数的方法，但增加了验证逻辑：

Decorator pattern

I would actually recommend wrapping ArrayList using well-documented Decorator pattern. You simply wrap your ArrayList with another List implementation that delegates most of the methods but adds validation logic:

public class ValidatingListDecorator extends AbstractList<MyBusinessObject>
{

    private final List<MyBusinessObject> target;

    public ValidatingListDecorator(List<MyBusinessObject> target) {
        this.target = target;
    }

    @Override
    public MyBusinessObject set(int index, MyBusinessObject element)
    {
        validate(element);
        return target.set(index, element);
    }

    @Override
    public boolean add(MyBusinessObject o)
    {
        validate(o);
        return target.add(o);
    }

    //few more to implement

}

优点：

• 您仍可以访问原始列表未经验证，如果你想要的（但你可以限制此）


• 更易于堆叠不同的验证，打开和关闭可选择开启他们。


• 在继承组成通过的 @helios 的
  
• 提高可测试性


• 不配合你到一个特定的列表实现，可以添加验证的LinkedList 或休眠 -backed持续名单。你甚至可以考虑一下通用收藏装饰来验证任何集合。



Advantages:

• You can still access raw list without validation if you want (but you can restrict this)
• Easier to stack different validations, turn them on and off selectively.
• Promotes composition over inheritance as noted by @helios
• Improves testability
• Does not tie you to a specific List implementation, you can add validation to LinkedList or Hibernate-backed persistent lists. You can even think about generic Collection decorator to validate any collection.

尽管实施记得有相当多的方法，你必须要记住，同时覆盖：添加（），的addAll（），设置（），子列表（）（？）等。

Despite the implementation remember there are quite a lot of methods you have to remember about while overriding: add(), addAll(), set(), subList() (?), etc.

此外，您的对象必须是不可变的，否则用户可以添加/设置有效的对象并对其进行修改后违反合同。

Also your object must be immutable, otherwise the user can add/set valid object and modify it afterwards to violate the contract.

Finaly我写的：

Finaly I wrote:

validate(element)

但考虑：

element.validate()

这是一个更好的设计。

which is a better design.

正如前面提到的，如果你想堆栈验证，在一个单一的，独立的类验证每个proprty / apsect，考虑下面的语句：

As noted before if you want to stack validations, validating each proprty/apsect in a single, separate class, consider the following idiom:

public abstract class ValidatingListDecorator extends AbstractList<MyBusinessObject>
{

    private final List<MyBusinessObject> target;

    public ValidatingListDecorator(List<MyBusinessObject> target) {
        this.target = target;
    }

    @Override
    public MyBusinessObject set(int index, MyBusinessObject element)
    {
        validate(element);
        return target.set(index, element);
    }

    protected abstract void validate(MyBusinessObject element);

}

...很少实现：

...and few implementations:

class FooValidatingDecorator extends ValidatingListDecorator {

    public FooValidatingDecorator(List<MyBusinessObject> target)
    {
        super(target);
    }

    @Override
    protected void validate(MyBusinessObject element)
    {
        //throw if "foo" not met
    }
}

class BarValidatingDecorator extends ValidatingListDecorator {

    public BarValidatingDecorator(List<MyBusinessObject> target)
    {
        super(target);
    }

    @Override
    protected void validate(MyBusinessObject element)
    {
        //throw if "bar" not met
    }
}

要只验证的富的？

List<MyBusinessObject> list = new FooValidatingDecorator(rawArrayList);

要验证这两个富的和的栏的？

List<MyBusinessObject> list = 
  new BarValidatingDecorator(new FooValidatingDecorator(rawArrayList));

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