可以在Android应用程序中以编程方式打开Spinner吗？

Question

如果你有一个SpinnerAndroid对象的句柄Spinner,你可以通过编程方式弹出微调器选项 - 从而迫使用户选择一个选项,即使他们没有点击Spinner自己？

Answer 1

要打开Spinner,您只需要调用它的performClick()方法.

请记住,您只能从UI线程调用此方法.如果你需要从一个单独的线程打开Spinner,你应该在UI线程中创建一个Handler,然后从你的第二个线程发送一个runInable对象,调用performClick()给Handler.

package com.example.SpinnerDemo;

import android.app.Activity;
import android.os.Bundle;
import android.widget.ArrayAdapter;
import android.widget.Spinner;
import android.os.Handler;

public class SpinnerDemo extends Activity {

    private Handler h;
    private Spinner s;

    @Override
    public void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.main);

        h = new Handler();

        s = (Spinner) findViewById(R.id.spinner);
        ArrayAdapter adapter = ArrayAdapter.createFromResource(this,
                R.array.planets, android.R.layout.simple_spinner_item);
        adapter.setDropDownViewResource(
                android.R.layout.simple_spinner_dropdown_item);
        s.setAdapter(adapter);

        // Open the Spinner...
        s.performClick();

        // Spawn a thread that triggers the Spinner to open after 5 seconds...
        new Thread(new Runnable() {
            public void run() {
                // DO NOT ATTEMPT TO DIRECTLY UPDATE THE UI HERE, IT WON'T WORK!
                // YOU MUST POST THE WORK TO THE UI THREAD'S HANDLER
                h.postDelayed(new Runnable() {
                    public void run() {
                        // Open the Spinner...
                        s.performClick();
                    }
                }, 5000);
            }
        }).start();
    }
}

可以在此处找到此示例使用的资源.

Answer 2

要显示Spinner项目,您只需要调用它的performClick()方法.

Spinner spDeviceType = (Spinner) findViewById(R.id.spDeviceType);
spDeviceType.performClick();

Answer 3

为什么前面的例子中有2个runnable？

这还不够:

h.postDelayed(new Runnable() {
    public void run() {
        s.performClick();
    }
}, 5000);

Answer 4

只需使用它

yourspinner.performClick();