＆QUOT;不能使静态引用非静态方法＆QUOT; Android中 [英] "Cannot make a static reference to a non-static method" in Android

问题描述

我在与老一些问题不能使静态引用非静态方法的错误在我的Andr​​oid程序。我创建一个沙子掉落游戏（类似火药游戏）和我创建了一个名为控件创建一个控制栏，在与画笔大小滑块（即正常工作）屏幕的底部和一个按钮，弹出一个对话框，允许用户选择所选择的元素。然而，当我打电话DemoActivity.showDialog（2）从我的code，它给人的静态引用非静态误差（DemoActivity是我的应用程序的主要活动）。我也试着改变它只是Activity.showDialog（2），但我得到了完全同样的错误！请帮帮忙，我究竟做错了什么？这是我的code和在此先感谢：

I'm having some issues with the old "Cannot make a static reference to a non-static method" error in my Android program. I am creating a sand falling game (similar to the Powder Game) and I created a class called Control to create a Control Bar at the bottom of the screen with a slider for brush size (that works fine) and a button to pop up a Dialog to allow users to pick the selected element. However, when I call DemoActivity.showDialog(2) from my code, it gives the static reference to non-static error (DemoActivity is the main activity of my application). I also tried changing it to just Activity.showDialog(2), but I got exactly the same error! Please help, what am I doing wrong? Here's my code and thanks in advance:

package sand.falling.opengl;

import android.content.Context;
import android.util.AttributeSet;
import android.util.Log;
import android.view.View;
import android.widget.ImageButton;
import android.widget.LinearLayout;
import android.widget.SeekBar;

public class Control extends LinearLayout
{
    private ImageButton control_button;
    private SeekBar brush_size_slider;

    final CharSequence[] elementslist = {"Sand", "Water", "Plant", "Wall", "Fire", "Ice", "Generator", "Oil", "Magma", "Stone", "C4"};

    public Control(Context context, AttributeSet attrs)
    {
        super(context, attrs);
    }

    @Override
    protected void onFinishInflate()
    {
        control_button = (ImageButton) findViewById(R.id.element_picker_button);
        brush_size_slider = (SeekBar) findViewById(R.id.brush_size_slider);

        control_button.setOnClickListener
            (
                    new OnClickListener()
                    {
                        public void onClick(View v)
                        {
                            //THIS DOESN'T WORK!!!!
                            DemoActivity.showDialog(2); //Run the element picker dialog
                        }
                    }
            );
        control_button.setImageResource(R.drawable.palette);

        brush_size_slider.setOnSeekBarChangeListener
            (
                    new SeekBar.OnSeekBarChangeListener()
                    {
                        public void onProgressChanged(SeekBar seekbar, int progress, boolean fromTouch)
                        {
                            int p = 32 * progress/100;
                            DemoActivity.setBrushSize(p);
                            Log.v("DemoActivity", "size:" + p);
                        }
                        public void onStartTrackingTouch(SeekBar seekbar) {}
                        public void onStopTrackingTouch(SeekBar seekbar) {}
                    }
            );
        brush_size_slider.setProgress((int)400/32);
    }
}

编辑：我固定它通过添加以下到我的Control.java code：

I fixed it by adding the following to my Control.java code:

public class Control extends LinearLayout
{
    private DemoActivity activity;
        ...
    public void setActivity(DemoActivity act)
    {
        activity = act;
    }
        ...
        //Set a click listener for the button which should pop up element picker dialog when clicked
        control_button.setOnClickListener
            (
                    new OnClickListener()
                    {
                        public void onClick(View v)
                        {
                            activity.showDialog(2); //Run the element picker dialog
                        }
                    }
            );
}

，然后调用 control.setActivity（本）; 从我的onResume DemoActivity.java的部分！希望它可以帮助那些有类似问题！

And then calling control.setActivity(this); from my onResume section of DemoActivity.java! Hope it helps those of you with similar issues!!

推荐答案

您必须调用的ShowDialog 在 DemoActivity 实例，而不是类本身。你可以调用的唯一时间 ClassName.methodName（）是，如果该方法被定义为静态的。 的ShowDialog 不是一个静态方法。

You have to call showDialog on a DemoActivity instance, NOT on the class itself. The only time you can call ClassName.methodName() is if the method is defined as static. showDialog is not a static method.

要解决这个问题，您可能需要实例化一个新的 DemoActivity 或让现有的，然后调用的ShowDialog 在这一点。

To fix this, you either need to instantiate a new DemoActivity or get an existing one, then call showDialog on that.

编辑：如果您已经有一个 DemoActivity 例如，当你实例化这个控制对象，或许下面的修改将工作：

If you already have a DemoActivity instance when you instantiate this Control object, perhaps the following modification will work:

public class Control extends LinearLayout
{

    ...

    // add an Activity instance
    private Activity activity;

    // set the Activity in your constructor
    public Control(Context context, AttributeSet attrs, Activity activity)
    {
        super(context, attrs);
        this.activity = activity;
    }

    @Override
    protected void onFinishInflate()
    {
        ...

           // Use the instance activity here
           activity.showDialog(2);
        ...
    }
}

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