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android - ListView 始终只显示最后一个元素

转载 作者:搜寻专家 更新时间:2023-11-01 08:37:08 26 4
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我在一个 Activity 中有一个 ListView ,我已经为它创建了自定义 ListView 。在自定义列表中,我有一个 ImageView 和两个 TextView 。当我向列表中添加值时,它始终只显示最后一个元素。

public class ContactUs extends Activity {
ImageView imageView;
TextView tvDetails,tv1;
private static String details[] = {"details1","details2","details3","details4","details5","details6"};
private static String desc[] = {"Name","Call","Email",
"Skype","Head Office","Website"};

private static int icons[] = {R.drawable.user, R.drawable.smartphone,
R.drawable.e_mail_envelope, R.drawable.facebook_logo, R.drawable.map_localization,
R.drawable.grid_world};
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);

setContentView(R.layout.activity_contact_us);
imageView=(ImageView)findViewById(R.id.imageView);
tvDetails=(TextView)findViewById(R.id.tvDetails);
tv1=(TextView)findViewById(R.id.tv11);
//final ListView lv1 = (ListView) findViewById(R.id.lvContact);


ArrayList<addValuesContact> searchResults = GetSearchResults();

final ListView lv1 = (ListView) findViewById(R.id.lvContact);
lv1.setAdapter(new MyCustomAdapterContact(this, searchResults));

lv1.setOnItemClickListener(new AdapterView.OnItemClickListener() {
@Override
public void onItemClick(AdapterView<?> a, View v, int position, long id) {
Object o = lv1.getItemAtPosition(position);
addValuesContact fullObject = (addValuesContact) o;
//Toast.makeText(getBaseContext(), "You have chosen: " + " " + fullObject.getDetails(), Toast.LENGTH_LONG).show();
}
});

}

private ArrayList<addValuesContact> GetSearchResults() {
ArrayList<addValuesContact> results = new ArrayList<>();
addValuesContact sr1 = new addValuesContact();

for (int i=0;i<6;i++) {
sr1.setImage(icons[i]);
sr1.setTv1(desc[i]);
sr1.setDetails(details[i]);
results.add(sr1);
}

return results;
}}

下面给出了我的自定义适配器类的详细信息

public class MyCustomAdapterContact extends BaseAdapter {
private static ArrayList<addValuesContact> searchArrayList;
private LayoutInflater mInflater;

public MyCustomAdapterContact(Contextcontext,ArrayList<addValuesContact>results) {
searchArrayList = results;
mInflater = LayoutInflater.from(context);
}

public int getCount() {
return searchArrayList.size();
}

public Object getItem(int position) {
return searchArrayList.get(position);
}

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

public View getView(int position, View convertView, ViewGroup parent) {
ViewHolder holder;
if (convertView == null) {
convertView = mInflater.inflate(R.layout.contact_list, null);
holder = new ViewHolder();
holder.imageView(ImageView)convertView.findViewById(R.id.imageView);
holder.tvDetails (TextView)convertView.findViewById(R.id.tvDetails);
holder.tv1 = (TextView) convertView.findViewById(R.id.tv1);


convertView.setTag(holder);
} else {
holder = (ViewHolder) convertView.getTag();
}
addValuesContact bean = (addValuesContact)searchArrayList.get(position);

holder.imageView.setImageResource(bean.getImage());
holder.tvDetails.setText(searchArrayList.get(position).getDetails());
holder.tv1.setText(searchArrayList.get(position).getTv1());


return convertView;
}

static class ViewHolder {
ImageView imageView;
TextView tvDetails;
TextView tv1;

}}

下面给出了Pojo类

public class addValuesContact {
private String details = "";
private String tv1 = "";
int image;
public int getImage() {
return image;
}
public void setImage(int image) {
this.image = image;
}
public void setDetails(String details) {
this.details = details;
}

public String getTv1() {
return tv1;
}

public void setTv1(String tv1) {
this.tv1 = tv1;
}

public String getDetails() {
return details;
}}

最佳答案

更新以下方法:

private ArrayList<addValuesContact> GetSearchResults() {
ArrayList<addValuesContact> results = new ArrayList<>();

for (int i=0;i<6;i++) {
addValuesContact sr1 = new addValuesContact();
sr1.setImage(icons[i]);
sr1.setTv1(desc[i]);
sr1.setDetails(details[i]);
results.add(sr1);
}

return results;
}

原因:您每次都需要创建模型类的新对象。

完成

关于android - ListView 始终只显示最后一个元素,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/35860835/

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