我有一个复选框,当选中它时,它会显示一个微调器,但我无法加载微调器包含的数据。
llF = (LinearLayout) view.findViewById(R.id.init_tour_foreign_producer_layout);
llF.setVisibility(View.INVISIBLE);
mySpinner = (Spinner)view.findViewById(R.id.spinner_init_tour);
mySpinner.setAdapter(new ArrayAdapter<Tour>(this.getActivity(), android.R.layout.simple_spinner_item, s.getUnstartedTours()));
/*---------- Managing the checkbox ----------*/
cbAllProducer = (CheckBox) view
.findViewById(R.id.enter_collection_check_other_tour);
cbAllProducer
.setOnCheckedChangeListener(new CompoundButton.OnCheckedChangeListener() {
@Override
public void onCheckedChanged(CompoundButton buttonView,
boolean isChecked) {
//checked = isChecked;
if(isChecked){
llF.setVisibility(View.VISIBLE);
//mySpinner.setSelection(1);
}
else{
llF.setVisibility(View.GONE);
}
}
});
答案 0 :(得分:1)
获取selectedItem:
mySpinner.getSelectedItem();
获取某个位置的项目:
mySpinner.getAdapter().getItem(position);
希望这对你有帮助!
答案 1 :(得分:0)
@Override
protected void onCreate(Bundle savedInstanceState) {
// TODO Auto-generated method stub
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);
checkBox = (CheckBox) findViewById(R.id.chkbox);
spinner = (Spinner) findViewById(R.id.spinner);
List<String> list = new ArrayList<String>();
list.add("Value1");
list.add("Value2");
list.add("Value3");
list.add("Value4");
ArrayAdapter<String> dataAdapter = new ArrayAdapter<String>(this, R.layout.sample_list, list);
spinner.setAdapter(dataAdapter);
spinner.setVisibility(View.INVISIBLE);
checkBox.setOnCheckedChangeListener(new OnCheckedChangeListener() {
@Override
public void onCheckedChanged(CompoundButton buttonView, boolean isChecked) {
if (isChecked) {
spinner.setVisibility(View.VISIBLE);
} else {
System.out.println("Not checked!!!!");
spinner.setVisibility(View.INVISIBLE);
}
}
});
}