删除重复从ListView的机器人 [英] Removing Duplicates From ListView android

问题描述

我张贴我的code，我无法从列表视图中删除重复值？ 是否有人可以帮助我吗？提前致谢！ 我贴我的code在这里，我已经使用 BaseAdapter 。

---

  @覆盖
            公共无效onCompleted（最终名单，其中，值得推荐＆GT;的结果）{
                android.util.Log.w（suggestionview＆GT;＆GT;＆GT;＆GT;＆gt;中，建议+ result.size（））;
                （（活动）mContext）.runOnUiThread（新的Runnable（）{
                    公共无效的run（）{
                        迭代器＆LT;值得推荐的＆GT; ITR = result.iterator（）;
                        而（itr.hasNext（））{
                            值得推荐的元素= itr.next（）;
                            suggestions.add（元）;
                            android.util.Log.w（suggestionview，添加元素::＆GT;+ suggestions.add（元））;
                        }
                        suggestionListView.setAdapter（新Suggestiondapter（mContext））;
                        android.util.Log.w（suggestionview，建议适配器价值观::＆GT;+ suggestionListView）;
                    }
                }）;
 

以及code中的第二个

 公共类Suggestiondapter扩展了BaseAdapter {


    //私人LayoutInflater mInflater = NULL;
    私人语境mContext;

    公共Suggestiondapter（上下文mContext）{
        this.mContext = mContext;
        android.util.Log.w（建议适配器，vlues都正在添加......）;
    }

    @覆盖
    公众诠释getCount将（）{
        android.util.Log.w（suugestion适配器，suggstion尺寸::＆gt;中+ suggestions.size（））;
        返回suggestions.size（）;
    }

    @覆盖
    公共对象的getItem（INT位置）{
        返回的位置;
    }

    @覆盖
    众长getItemId（INT位置）{
        返回的位置;
    }

    @覆盖
    公共查看getView（INT位置，查看convertView，ViewGroup中父）{
        值得推荐推荐= suggestions.get（位置）;
        如果（convertView == NULL）{
            android.util.Log.w（convertView，adaptervalues​​ ........ ::＆gt;中+ recommendable.Program）;
            android.util.Log.w（系列conveter，计划值系列::＆GT;+ recommendable.Series）;
            convertView =新HeaderView（mContext，recommendable.Program，recommendable.Series，SuggestionView.class）;
        }其他{
            Toast.makeText（mContext，为helloView，Toast.LENGTH_SHORT）;
        }
        返回convertView;
    }
};
 

解决方案

如果您不希望允许重复到您的收藏对象，然后使用Set对象，现在将允许重复的值。

 设置建议=新的HashSet（）;
 

也可以添加您的列表对象这个对象，现在将增加重复值

  ArrayList的人=新的ArrayList（）;
//添加元素人，包括重复
HashSet的HS =新的HashSet（）;
hs.addAll（人）;
al.clear（）;
al.addAll（HS）;
 

更新你的情况：

使用您的建议对象，而不是上面的人对象只是它。

I am posting my code, I am unable to remove the duplicate values from the listview? Can someone please help me? Thanks in advance! I am pasting my code here and I have used BaseAdapter.

---

@Override
            public void onCompleted(final List<Recommendable> result) {
                android.util.Log.w("suggestionview>>>>>", "suggestion"+ result.size());
                ((Activity) mContext).runOnUiThread(new Runnable() {
                    public void run() {
                        Iterator<Recommendable> itr = result.iterator();
                        while (itr.hasNext()) {
                            Recommendable element = itr.next();
                            suggestions.add(element);
                            android.util.Log.w("suggestionview", "Adding elements::>"+suggestions.add(element));
                        }
                        suggestionListView.setAdapter(new Suggestiondapter(mContext));
                        android.util.Log.w("suggestionview","suggestion adapter Values::>"+suggestionListView);
                    }
                });

And the second of the code

public class Suggestiondapter extends BaseAdapter {


    // private LayoutInflater mInflater = null;
    private Context mContext;

    public Suggestiondapter(Context mContext) {
        this.mContext=mContext;
        android.util.Log.w("Suggestion Adapter","vlues are comming.....");
    }

    @Override
    public int getCount() {
        android.util.Log.w("suugestion adapter","suggstion size::>"+suggestions.size());
        return suggestions.size();
    }

    @Override
    public Object getItem(int position) {
        return position;
    }

    @Override
    public long getItemId(int position) {
        return position;
    }

    @Override
    public View getView(int position, View convertView, ViewGroup parent) {
        Recommendable recommendable = suggestions.get(position);
        if(convertView==null){
            android.util.Log.w("convertView", "adaptervalues........::>"+ recommendable.Program);
            android.util.Log.w("series conveter", "program series values::>"+recommendable.Series);
            convertView = new HeaderView(mContext, recommendable.Program,recommendable.Series, SuggestionView.class);
        }else{
            Toast.makeText(mContext, "HelloView", Toast.LENGTH_SHORT);
        }
        return convertView;
    }
};

解决方案

If you don't want to allow duplicate into collection of your object then use the Set object it will now allow the duplicate value.

Set suggestion = new HashSet();

or you can add your list object to this object it will now add duplicate value

ArrayList al = new ArrayList();
// add elements to al, including duplicates
HashSet hs = new HashSet();
hs.addAll(al);
al.clear();
al.addAll(hs);

updated for your case:

use your suggestion object instead of the above al object just it.

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