我有一个带有两个开关的列表视图。我希望该功能正常工作,因为一次只能激活一个开关。
我的适配器:
public class NotificationsAdapter extends ArrayAdapter<String> {
private Context context;
private String mTitle[];
private boolean onOff[];
public NotificationsAdapter(Context c, String mTitle[], boolean onOff[]) {
super(c, R.layout.adapter_notifications_layout, R.id.notificationsListTv, mTitle);
this.context = c;
this.mTitle = mTitle;
this.onOff = onOff;
}
@NonNull
@Override
public View getView(int position, @Nullable View convertView, @NonNull ViewGroup parent) {
LayoutInflater layoutInflater = (LayoutInflater) context.getSystemService(Context.LAYOUT_INFLATER_SERVICE);
View v = layoutInflater.inflate(R.layout.adapter_notifications_layout, parent, false);
Switch notificationSwitch = v.findViewById(R.id.switchNotificationsDaily);
TextView myTitle = v.findViewById(R.id.notificationsListTv);
myTitle.setText(mTitle[position]);
notificationSwitch.setChecked(onOff[position]);
return v;
我的班级:
private String[] mTitle = new String[]{"Once Daily", "Twice Daily"};
private Switch notificationSwitch;
private boolean[] onOff = new boolean[] {false, false};
NotificationsAdapter notificationsAdapter = new NotificationsAdapter(getActivity(), mTitle, onOff);
listView = view.findViewById(R.id.notificationsListView);
listView.setAdapter(notificationsAdapter);
listView.setChoiceMode(ListView.CHOICE_MODE_SINGLE);
listView.setOnItemClickListener(new AdapterView.OnItemClickListener() {
@Override
public void onItemClick(AdapterView<?> parent, View view, int position, long id) {
if (onOff[position]) {
notificationSwitch.setChecked(false);
onOff[position] = false;
} else {
onOff[position] = true;
notificationSwitch.setChecked(true);
}
}
当前使用此代码,我可以选择两个开关同时处于活动状态。当我选择多个开关时,我希望另一个开关停用。使该功能正常工作的任何帮助将不胜感激。
listView.setOnItemClickListener(new AdapterView.OnItemClickListener() {
@Override
public void onItemClick(AdapterView<?> parent, View view, int position, long id) {
resetValues(position);
if (onOff[position]) {
notificationSwitch.setChecked(false);
onOff[position] = false;
} else {
onOff[position] = true;
notificationSwitch.setChecked(true);
}
}
}
private void resetValues(int selectedPosition){
int size = onOff.length;
for(int i=0; i < size; i++){
if(i == selectedPosition){
break;
}
onOff[i] = false;
}
}