微調一個有2項:毒藥non_poisonAndroid的微調,如果選擇了一些項目,其他微調了
微調b具有2項:箭毒蛙和毒蛇
微調c具有導2項:雞肉和雞蛋
長話短說,我要讓 「IF」微調上一=毒藥 「然後」微調b =可見,微調c中選擇項目=不見了
繼承人我的代碼,使用setOnItemSelectedListener試過了,沒有錯誤顯示,但它不按照它應該的方式工作,我在這裏錯過了什麼?請幫我....
public class eatable extends Fragment {
public String spinner1x;
@Nullable
@Override
public View onCreateView(final LayoutInflater inflater, final ViewGroup container, final Bundle savedInstanceState) {
final FrameLayout mRelative = (FrameLayout) inflater.inflate(R.layout.berita_layout, container, false);
final TextView text1 = (TextView) mRelative.findViewById(R.id.text_test1);
final TextView text2 = (TextView) mRelative.findViewById(R.id.text_test2);
final TextView text3 = (TextView) mRelative.findViewById(R.id.text_test3);
final Spinner spinner1 = (Spinner) mRelative.findViewById(R.id.spinner_test1);
final Spinner spinner2 = (Spinner) mRelative.findViewById(R.id.spinner_test2);
final Spinner spinner3 = (Spinner) mRelative.findViewById(R.id.spinner_test3);
spinner1x = spinner1.getSelectedItem().toString();
spinner1.setOnItemSelectedListener(new AdapterView.OnItemSelectedListener() {
@Override
public void onItemSelected(AdapterView<?> mRelative, View selectedItemView, int position, long id) {
// your code here
if (spinner1x.equals("poison")){
spinner2.setVisibility(View.VISIBLE);
spinner3.setVisibility(View.GONE);
}
}
});
return mRelative;
}
}
編寫如下spinner1x = spinner1.getSelectedItem()的toString()。在onItemSelected方法 – Sanjeev
嘗試從spinner1裏面的onim選擇 – Shekhar
把spinner1x放到onItemSelected方法裏面,仍然不工作 –