Question

我试图让我的按钮改为不同的图片和一套不同的法则它是一个字母应用程序，我目前已经工作，它在市场上....我想让它随机改变图片，然后跟随alphebet通过整个程序，但我无法弄清楚如何让按钮随机化所以这是我正在使用的基本代码

package com.alphafree;

import java.util.Random;
import android.app.Activity;
import android.app.AlertDialog;
import android.content.Intent;
import android.media.MediaPlayer;
import android.os.Bundle;
import android.view.View;
import android.view.View.OnClickListener;

public class letterA extends Activity implements OnClickListener {
    /** Called when the activity is first created. */
    private MediaPlayer mp;

    @Override
    public void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);

        int[] _26Num = new int[6];
        int i, number;

        Random r1 = new Random(); // Random generator
        for (i = 1; i < _26Num.length; i++) { // Pick 6 Random Numbers
            do {
                number = (int) (r1.nextDouble() * 26);
                _26Num[1] = 1;
            } while (!(number != 0 && number != _26Num[1]
                    && number != _26Num[2] && number != _26Num[3]
                    && number != _26Num[4] && number != _26Num[5]));
            _26Num[i] = number;
        }

        // defines the buttons
        View but1 = findViewById(R.id.let1);
        View but2 = findViewById(R.id.let2);
        View but3 = findViewById(R.id.let3);
        View but4 = findViewById(R.id.let4);
        View but5 = findViewById(R.id.let5);

        but1.setOnClickListener(this);
        but2.setOnClickListener(this);
        but3.setOnClickListener(this);
        but4.setOnClickListener(this);
        but5.setOnClickListener(this);

        // set the random content view
        int random = (int) Math.ceil(Math.random() * 5);
        if (random == 1) {
            setContentView(R.layout.letter1a);

        } else if (random == 2) {
            setContentView(R.layout.letter1a);

        } else if (random == 3) {
            setContentView(R.layout.letter1a);

        } else if (random == 4) {
            setContentView(R.layout.letter1a);

        } else if (random == 5) {
            setContentView(R.layout.letter1a);

        }

        // here is were i need to set up the buttons
        if (_26Num[1] == 1) {

            let1.setBackgroundResource(R.drawable.let_a);

        } else if (_26Num[1] == 2) {
            setContentView(R.layout.letter1a);

        } 


    }

    public void onClick(View v) {

        if (mp != null)
            mp.release();

        switch (v.getId()) {

        case R.id.let1:

            mp = MediaPlayer.create(this, R.raw.correct);
            {
                new AlertDialog.Builder(this)
                        .setTitle("Y A Y!!!!   Y O U R    R I G H T")
                        .setMessage(
                                "Thats the letter      'A'                  You Chose Correct")
                        .setNeutralButton("Done", null).show();
                startActivity(new Intent("com.alphafree.LETTERB"));

            }
            break;

        }
        mp.seekTo(0);
        mp.start();
    }

}

Answer 1

你可以这样做：

private static final int[] num2DrawableMap = new int[] {
    R.drawable.letterA,
    R.drawable.letterB,
    // etc
};

private static final int[] num2LayoutMap = new int[] {
    R.layout.layoutA,
    R.layout.layoutB,
    // etc
};

@Override
public void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);

    int[] _26Num = new int[6];
    int i, number;

    Random r1 = new Random(); // Random generator
    for (i = 1; i < _26Num.length; i++) { // Pick 6 Random Numbers
        do {
            // don't you like r1.nextInt(26) ?
            number = (int) (r1.nextDouble() * 26);
            // always 1? why in that loop?
            _26Num[1] = 1;
        } while (!(number != 0 && number != _26Num[1]
                && number != _26Num[2] && number != _26Num[3]
                && number != _26Num[4] && number != _26Num[5]));
        _26Num[i] = number;
    }

    // set the random content view
    int random = r1.nextInt(num2LayoutMap.size);
    setContentView(num2LayoutMap[random]);

    // find the buttons
    View[] buts = new View[5];
    buts[0] = findViewById(R.id.let1);
    buts[1] = findViewById(R.id.let2);
    buts[2] = findViewById(R.id.let3);
    buts[3] = findViewById(R.id.let4);
    buts[4] = findViewById(R.id.let5);

    for (View but : buts) {
        but.setOnClickListener(this);
        int drawableId = num2DrawableMap[_26Num[i]];
        but.setBackgroundResource(drawableId);
    }
}

但是：Buttons必须在每个layout.xml文件中，并且它们需要具有相同的id。否则你找不到它们。您还需要setContentView()才能使用findViewById()

Answer 2

每次除了setContentView之外，您是否可以一次添加所有内容并保持其对GONE的可见性。你想要显示的按钮只是让它可见到VISIBLE。如果你想改变使上一个为GONE并使新的可见。

无法用java随机设置一个新按钮

2 个答案: