当我选择一个Spiner值然后我想要一个新的活动，这将显示一些消息。但它无法取得价值。 [英] When I Selected A Spiner Value Then I Want To Go A New Activity, That Will Show Some Message. But It Can't Take The Value.

问题描述

public void addListenerOnSpinnerItemSelection1() {
    spinvisitingPlace = (Spinner) findViewById(R.id.spinvisitingPlace);
    spinvisitingPlace.setOnItemSelectedListener(new CustomOnItemSelectedListener());
    if (spinvisitingPlace.equals("Rajbari")) {

        Intent intent = new Intent(MainActivity.this,ActivityTajhat.class);
        startActivity(intent);

  }
    else {
      Toast.makeText(MainActivity.this, "string",1000).show();
  }

当我首先运行它时显示吐司消息但是当我选择我的Rajbari值..这不给我任何答案。请帮帮我

when i run it at first its show the toast message but when i select my "Rajbari" value..it's don't give me any answer. please help me

推荐答案

您发布的代码是错误的。代码if（spinvisitingPlace.equals（Rajbari））是错误的。

只需将此行放在onItemclicklistener（AdapterView

并从中获取值



String string = arg0.getItemAtPosition（arg2）.toString（）。trim（）;

接下来，您将比较此字符串与



if（string.equals（Rajbari））

{

意图意图=新意图（MainActivity.this ，ActivityTajhat.class）;

startActivity（intent）;

}

Hi, The code you posted ,is wrong. the code if(spinvisitingPlace.equals("Rajbari")) is wrong.
Just put this line inside of onItemclicklistener(AdapterView
and get the value from that of using

String string=arg0.getItemAtPosition(arg2).toString().trim();
Next you will compare this string with

if(string.equals("Rajbari))
{
Intent intent = new Intent(MainActivity.this,ActivityTajhat.class);
startActivity(intent);
}

代码中onItemSelected方法的位置

Where's onItemSelected method in your code

public class SpinnerActivity extends Activity implements OnItemSelectedListener {
    ...

    public void onItemSelected(AdapterView<?> parent, View view, int pos, long id) {
        // An item was selected. You can retrieve the selected item using
        // parent.getItemAtPosition(pos)  >> this will solve your problem
    }

    public void onNothingSelected(AdapterView<?> parent) {
        // Another interface callback
    }
}

-KR

-KR

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