当我从 Spinner 类中选择选项时,我想打开新表单
我试试这个,但我有语法错误
public void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.tf);
Spinner spinner2 = (Spinner) findViewById(R.id.spinner2);
ArrayAdapter<CharSequence> adapter2 = ArrayAdapter.createFromResource(
this, R.array.tfoptions,android.R.layout.simple_spinner_item);
adapter2.setDropDownViewResource(
android.R.layout.simple_spinner_dropdown_item);
spinner2.setAdapter(adapter2);
spinner2.setOnItemSelectedListener(new MyOnItemSelectedListener());
public class MyOnItemSelectedListener implements OnItemSelectedListener {
public void onItemSelected(AdapterView<?> parent,
View view, int pos, long id) {
switch (view.isClickable()) { <---------------- syntax error
Spinner spinner2;
case spinner2.setSelection(0):
startActivity(new Intent(this,To.class));
break;
case spinner2.setSelection(1):
startActivity(new Intent(this,out.class));
default:
break;
}
}
public void onNothingSelected(AdapterView parent) {
// Do nothing.
}
}
}
最佳答案
您试图在第一个 case
块之前声明一个变量。你不可以做这个。将变量声明移至 switch
上方。
关于java - token "{"上的语法错误,此 token 后预期的 SwitchLabels,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/5601040/