操作栏的弹出窗口

Question

I'm trying to create a pop-up window in Android by clicking button in action bar. Like this:

http://pix.am/yo2E.jpg

In my idea I realised it with two fragments in one container, where 1 (pop-up) is in View.GONE state and becomes visible when I click button.

Is there an easier way to solve my problem?

Answer 1

Basically you can do that with very small amount of code and you will end up like this

but if you want to customize you have to design a custom layout

To Achieve that, create a xml menu file like below

<?xml version="1.0" encoding="utf-8"?>
<menu xmlns:android="http://schemas.android.com/apk/res/android" >

    <item
        android:id="@+id/search"
        android:icon="@android:drawable/ic_menu_search"
        android:title="Search"/>
    <item
        android:id="@+id/add"
        android:icon="@android:drawable/ic_menu_add"
        android:title="Add"/>
    <item
        android:id="@+id/edit"
        android:icon="@android:drawable/ic_menu_edit"
        android:title="Edit">
        <menu>
            <item
                android:id="@+id/share"
                android:icon="@android:drawable/ic_menu_share"
                android:title="Share"/>
        </menu>
    </item>

</menu>

Now, write PopupMenu1 Activity.java file

package com.example.popuptest;

import android.app.Activity;
import android.os.Bundle;
import android.view.MenuItem;
import android.view.View;
import android.widget.PopupMenu;
import android.widget.Toast;

public class PopupMenu1 extends Activity {
    @Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.popup_menu_1);
    }

    public void onPopupButtonClick(View button) {
        PopupMenu popup = new PopupMenu(this, button);
        popup.getMenuInflater().inflate(R.menu.popup, popup.getMenu());

        popup.setOnMenuItemClickListener(new PopupMenu.OnMenuItemClickListener() {
            public boolean onMenuItemClick(MenuItem item) {
                Toast.makeText(PopupMenu1.this,
                        "Clicked popup menu item " + item.getTitle(),
                        Toast.LENGTH_SHORT).show();
                return true;
            }
        });

        popup.show();
    }
}

source

Answer 2

EDIT:Now you Can use to show as simple dropdown for that actionbar button.

I used because you can display in any location of the screen but with small bug which I will talk about later, this is kinda good cause it takes as one of the parameter along with and positions,so use this to place where ever you want on the screen

Oh and the action button wont work yet after click action,so use these values to make popup window react to actions outside the screen
popupWindow.setBackgroundDrawable(new BitmapDrawable()); popupWindow.setOutsideTouchable(true); now launch the window from OnOptionsItemSelected() inside your activity

The catch here is with , your window works fine for actionbar clicks , but makes the window also to react for touches outside the window area,meaning it dismisses the window if u touch outside, there is a method called which would have helped in that case but it's buggy

Answer 3

Take a look at official PopupMenu support from android. You can launch a popup window on button click and inflate your own UI for that window.

Link: http://developer.android.com/reference/android/widget/PopupMenu.html

If you want to support older versions of android, you can use PopupMenuCompat

Link: http://developer.android.com/reference/android/support/v7/widget/PopupMenu.html