Android:带重复项的SortedList [英] Android: SortedList with duplicates

问题描述

我在理解RecyclerView s SortedList时遇到了一些问题.

I have some problems understanding RecyclerViews SortedList.

让我们说我有一个非常简单的类，只有一个非常简单的类拥有数据:

Lets say I have a very simple class only having a very simple class holding data:

public class Pojo {
    public final int id;
    public final char aChar;

    public Pojo(int id, char aChar) {
        this.id = id;
        this.aChar = aChar;
    }

    @Override
    public String toString() {
        return "Pojo[" + "id=" + id
                + ",aChar=" + aChar
                + "]";
    }
}

我的理解是，排序后的列表不会包含任何重复项.

My understanding is that the sorted list won't contain any duplicates.

但是当我有一个带有这样的回调的SortedList时:

But when I have a SortedList with callbacks like this:

....

@Override
public boolean areContentsTheSame(Pojo oldItem, Pojo newItem) {
    return oldItem.aChar == newItem.aChar;
}

@Override
public int compare(Pojo o1, Pojo o2) {
    return Character.compare(o1.aChar, o2.aChar);
}

@Override
public boolean areItemsTheSame(Pojo item1, Pojo item2) {
    return item1.id == item2.id;
}

当我添加多个具有相同ID但不同字符的项目时，我最终会重复.

I end up with duplicates when I add multiple items with the same id but different chars.

sortedList.add(new Pojo(1, 'a'));
sortedList.add(new Pojo(1, 'b'));

我希望列表会更新该项目.相反，即使areItemsTheSame返回了true，我现在还是有多个项目.

I would expect the list to update the item. Instead now I have multiple items even though areItemsTheSame returned true.

推荐答案

SortedList不会按ID保留任何映射(因为API中没有ID). 因此，当排序标准更改时(在您的情况下为a到b)，SortedList无法找到现有元素.

SortedList does not keep any mapping by ids (because there are no ids in the API). So when the sorting criteria changes (a to b in your case), SortedList cannot find the existing element.

您可以自己保留ID映射，然后使用如下的add方法:

You can keep the id mapping yourself, then have your add method as follows:

void add(Item t) {
  Item existing = idMap.get(t.id);
  if (existing == null) {        
     sortedList.add(t);
  } else {
     sortedList.updateItemAt(sortedList.indexOf(existing), t);
  }
  idMap.put(t.id, t);
}

您还需要实现remove方法，以从idMap中删除项目.

You'll also need to implement a remove method to remove the item from the idMap.

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