Question

这就是我定义微调器的方法

s_province = (Spinner) findViewById(R.id.s_province);
        ArrayAdapter<String> provinceAdapter = new ArrayAdapter<String>(this,
                android.R.layout.simple_spinner_item, Data.provinces);
        provinceAdapter
                .setDropDownViewResource(android.R.layout.simple_spinner_dropdown_item);
        s_province.setAdapter(provinceAdapter);
        s_province.setOnItemSelectedListener(this);

我的类是从OnItemSelectedListener实现的，我重写了这个方法

@Override
    public void onItemSelected(AdapterView<?> arg0, View arg1, int arg2,
            long arg3) {
        // TODO Auto-generated method stub
        switch (arg1.getId()) {
        case R.id.s_province:
            Log.d("here", "there");
            break;

        default:
            break;
        }
    }

    @Override
    public void onNothingSelected(AdapterView<?> arg0) {
        // TODO Auto-generated method stub

    }

但是onItemSelect没有被区分，为什么还请？

Answer 1

两件事：

如果您确实想检查您的方法是否有效，请在Log或switch案例之外添加default语句，以便您可以确定正在调用该方法。
自that represents position以来，您应该使用arg2。使您的开关与位置一起工作，而不是传递View。还要从Eclipse分配的默认名称重命名变量。 arg0,1,2, etc对您或其他任何查看您代码的人都没有帮助。

例如

@Override
public void onItemSelected(AdapterView<?> parent, View view, int position, long id){
  Toast.makeText(view.getContext(),"onItemSelected called", Toast.LENGTH_LONG).show();
  int spinnerId = parent.getId();
  if (spinnerId == R.id.s_province)
  {
    switch (position)
    {
      case 0: 
        //do something if first position was clicked
      break:
      case 1:
        //do something else
        break;
      default: 
        //if for any reason no position matches.
        break;
    }
  }
  else if (spinnerId == R.other_id_in_xml)
  {
    switch (position)
    {
      case 0: 
        //do something if first position was clicked
      break:
      case 1:
        //do something else
        break;
      default: 
        //if for any reason no position matches.
        break;
    }
  }
  //etc
}

android spinner onselect item单击监听器无触发器

1 个答案: