如何通过最小数字/字符串android studio对recylerview进行排序

Question

我的应用程序中有一个recylerview,它包含两个textview,一个是rankTextView,另一个是nameTextview.像这样的东西;

我想要做的是对这个rec​​ylerview进行排序,首先是最低的数字顺序,如果有两个相同的数字,那么我希望它由字符串整理出来.在上面的截图我有例如,两个人排名1,首先我希望recylerview将这些放在顶部,然后按String排序.

我在网上搜索过,看看我怎样才能做到这一点,但对于android新手我还没有能够将调查结果调整到我的项目中.例如;

如何在Android RecyclerView上对字符串进行排序？

什么是使用RecyclerView.Adapter的SortedList <T>？

我创建了一个自定义适配器,其中包含一个名为rankTextview的textview和另一个名为nameTextview的textview,就像这样;

 rankTextview = (TextView) itemView.findViewById(R.id.ranktextview);
 nameTextview = (TextView) itemView.findViewById(R.id.nametextview);

然后我有一个方法,将这些textview作为参数放入什么值,像这样;

 public addPerson(String rankTextview, String personTextview,) {
        this.rankTextview = rankTextview;
        this.personTextview = personTextview;
    }

然后我在我的主类中调用这个方法来添加数据,就像这样;

person.add(new addPerson
        ("1\nrank", "James Kub"));
person.add(new addPerson
        ("2\nrank", "Peter Hanly"));
person.add(new addPerson
        ("3\nrank", "Josh Penny"));
person.add(new addPerson
        ("1\nrank", "Danny Jackson"));
person.add(new addPerson
        ("3\nrank", "Brad Black"));

现在我想要做的是首先按等级最低数字顺序排序这些数据,例如1,2 ......如果有两个相同的数字,那么我想按名称字母顺序排序.此外,将来我的应用程序将包含点而不是排名,这将是十进制数字.1.1,1.5,1.1,2.1,2.5等等,所以当按等级排序时,可以接收计数十进制数.

在此先感谢,如果我的问题不够明确,请告诉我.此外,由于我有这么多行代码,我不确定要提供哪个部分以及哪些部分不提供,如果有任何我想要的代码,请告诉我.

编辑:

public void animateTo(List<ExampleModel> models) {
        applyAndAnimateRemovals(models);
        applyAndAnimateAdditions(models);
        applyAndAnimateMovedItems(models);
    }

    private void applyAndAnimateRemovals(List<ExampleModel> newModels) {
        for (int i = mModels.size() - 1; i >= 0; i--) {
            final ExampleModel model = mModels.get(i);
            if (!newModels.contains(model)) {
                removeItem(i);
            }
        }
    }

    private void applyAndAnimateAdditions(List<ExampleModel> newModels) {
        for (int i = 0, count = newModels.size(); i < count; i++) {
            final ExampleModel model = newModels.get(i);
            if (!mModels.contains(model)) { // error here, saying cannot resolve method contains
                addItem(i, model);
            }
        }
    }

    private void applyAndAnimateMovedItems(List<ExampleModel> newModels) {
        for (int toPosition = newModels.size() - 1; toPosition >= 0; toPosition--) {
            final ExampleModel model = newModels.get(toPosition);
            final int fromPosition = mModels.indexOf(model);
            if (fromPosition >= 0 && fromPosition != toPosition) {
                moveItem(fromPosition, toPosition);
            }
        }
    }

    public ExampleModel removeItem(int position) {
        final ExampleModel model = mModels.remove(position); // Error here, saying in sortedlist cannot be applied to (int)
        notifyItemRemoved(position);
        return model;
    }

    public void addItem(int position, ExampleModel model) {
        mModels.add(position, model); // Error here, saying add has private access in 'android.support.v7.util.SortedList'
        notifyItemInserted(position);
    }

    public void moveItem(int fromPosition, int toPosition) {
        final ExampleModel model = mModels.remove(fromPosition); // Error here, saying in sortedlist cannot be applied to (int)
        mModels.add(toPosition, model); // Error here, saying add has private access in 'android.support.v7.util.SortedList'
        notifyItemMoved(fromPosition, toPosition);
    }

Answer 1

在a中实现排序有一些选项RecyclerView.当然可以依赖Comparable<T>和Comparator<T>接口,但正如您所提到的,也可以利用SortedList<T>Android SDK中定义的类.

目的SortedList<T>是简化a中元素的排序RecyclerView,允许您拦截重要事件,如"添加新项目","删除项目"等.

在您的情况下,您可以按以下步骤操作:

1. 定义包含排名和名称的Person类.请注意,在这个版本中,我假设有排名的整数值,但很容易移动到十进制值.

   class Person {
   
       private String rank;
       private String name;
   
       public Person(String rank, String name) {
           this.rank = rank;
           this.name = name;
       }
   
       // getters and setters here
   
   }
   

2. 定义Activity在哪里构建RecyclerView和相应的适配器.在这个例子中,我包含了一个FloatingActionButton用于插入新的随机人员.如您所见,在创建新的Person方法时addPerson,在适配器上调用该方法.它的作用是更新RecyclerView,根据适配器本身定义的标准对其进行排序(参见第3点).

   public class SortPersonsActivity extends AppCompatActivity {
   
       private List<Person> mPersons;
   
       private SortPersonsAdapter mPersonsAdapter;
   
       @Override
       protected void onCreate(Bundle savedInstanceState) {
           super.onCreate(savedInstanceState);
           setContentView(R.layout.activity_persons_list);
           Toolbar toolbar = (Toolbar) findViewById(R.id.toolbar);
           setSupportActionBar(toolbar);
   
           mPersons = new ArrayList<>();
           mPersons.add(new Person("1\nrank", "James Kub"));
           mPersons.add(new Person("2\nrank", "Peter Hanly"));
           mPersons.add(new Person("3\nrank", "Josh Penny"));
           mPersons.add(new Person("1\nrank", "Danny Jackson"));
           mPersons.add(new Person("3\nrank", "Brad Black"));
   
           RecyclerView recyclerView = (RecyclerView) findViewById(R.id.lst_items);
           recyclerView.setLayoutManager(getLayoutManager());
           mPersonsAdapter = new SortPersonsAdapter(this, mPersons);
           recyclerView.setAdapter(mPersonsAdapter);
   
           FloatingActionButton fab = (FloatingActionButton) findViewById(R.id.fab);
           fab.setOnClickListener(new View.OnClickListener() {
               @Override
               public void onClick(View v) {
                   // building new fake person
                   Person person = new Person(
                           buildRandomInt(10) + "\nrank",
                           buildRandomName(5) + " " + buildRandomName(5));
                   // let's keep also basic list updated
                   mPersons.add(person);
                   // let's update adapter
                   mPersonsAdapter.addPerson(person);
               }
           });
       }
   
       private RecyclerView.LayoutManager getLayoutManager() {
           LinearLayoutManager llm = new LinearLayoutManager(this);
           llm.setOrientation(LinearLayoutManager.VERTICAL);
           return llm;
       }
   
       // support method for random names and ranks here
   
   }
   

3. 实现依赖的RecyclerView适配器SortedList<Person>.重要的是要注意所有人都被插入到SortedList<Person>.创建a SortedList<T>需要Callback定义拦截事件以及定义排序标准.在我们的例子中,正如您所看到的,compare方法定义了对人员进行排序的条件,而onInserted方法定义了插入新Person时要执行的操作(RecyclerView在这种情况下通知数据集更改以进行更新).请注意addPerson第2点描述的方法的实现.它只是添加了一个Person SortedList,因为更新的逻辑RecyclerView嵌入到前面提到的Callback方法onInserted中.

   class SortPersonsAdapter extends RecyclerView.Adapter<SortPersonsAdapter.PersonViewHolder> {
   
       protected static class PersonViewHolder extends RecyclerView.ViewHolder {
   
           View layout;
           TextView txt_rank;
           TextView txt_full_name;
   
           public PersonViewHolder(View itemView) {
               super(itemView);
               layout = itemView;
               txt_rank = (TextView) itemView.findViewById(R.id.txt_rank);
               txt_full_name = (TextView) itemView.findViewById(R.id.txt_full_name);
           }
   
       }
   
       private Context mContext;
       private LayoutInflater mLayoutInflater;
   
       private SortedList<Person> mPersons;
   
       public SortPersonsAdapter(Context context, List<Person> persons) {
           mContext = context;
           mLayoutInflater = (LayoutInflater) mContext.getSystemService(Context.LAYOUT_INFLATER_SERVICE);
           mPersons = new SortedList<>(Person.class, new PersonListCallback());
           mPersons.addAll(persons);
       }
   
       public void addPerson(Person person) {
           mPersons.add(person);
       }
   
       @Override
       public int getItemCount() {
           return mPersons.size();
       }
   
       @Override
       public SortPersonsAdapter.PersonViewHolder onCreateViewHolder(ViewGroup parent, int viewType) {
           View itemView = mLayoutInflater.inflate(R.layout.view_person_item, parent, false);
           return new PersonViewHolder(itemView);
       }
   
       @Override
       public void onBindViewHolder(final PersonViewHolder viewHolder, final int position) {
           Person person = mPersons.get(position);
           viewHolder.txt_rank.setText(person.getRank());
           viewHolder.txt_full_name.setText(person.getName());
       }
   
       /**
        * Implementation of callback for getting updates on person list changes.
        */
       private class PersonListCallback extends SortedList.Callback<Person> {
   
           @Override
           public int compare(Person p1, Person p2) {
               String[] rank1 = p1.getStringRank().split("\n");
               String[] rank2 = p2.getStringRank().split("\n");
               int diff = Integer.parseInt(rank1[0]) - Integer.parseInt(rank2[0]);
               return (diff == 0) ? p1.getName().compareTo(p2.getName()) : diff;
           }
   
           @Override
           public void onInserted(int position, int count) {
               notifyItemInserted(position);
           }
   
           @Override
           public void onRemoved(int position, int count) {
               notifyItemRemoved(position);
           }
   
           @Override
           public void onMoved(int fromPosition, int toPosition) {
           }
   
           @Override
           public void onChanged(int position, int count) {
           }
   
           @Override
           public boolean areContentsTheSame(Person oldItem, Person newItem) {
               return false;
           }
   
           @Override
           public boolean areItemsTheSame(Person item1, Person item2) {
               return false;
           }
   
       }
   
   }
   

希望这可以帮助.在这里,我为您提供了一个实现RecyclerView,以防您需要有关代码的更多详细信息.

Answer 2

public int compare(Person p1, Person p2) {
   if(p1.getRank() == p2.getRank()){
      return p1.getName().compareTo(p2.getName());
   }else if(p1.getRank() > p2.getRank()){
      return 1;
   }else{
      return -1;
   }
}  

排序取决于返回的比较结果.你需要返回-1,0或1.在代码片段中,我所做的就是先检查排名.如果它们是相同的等级,我比较它们的名字是Strings,每个String都有按compareTo( )字典顺序比较两个字符串.

如果没有,我们只是根据排名对它们进行排序.

您可以compare( )通过创建Person类实现Comparable接口来进一步简化方法.这将允许您Collections开箱即用的框架好东西.