Question

我有这个代码，我从stackoverflow中取出它并稍微改了一下。代码检索图库的内容并将每个图像路径放在数组列表中。然后它随机选择ArrayList中的一个路径并作为ImageView的资源。谢谢你的关注。

import android.database.Cursor;
import android.graphics.Bitmap;
import android.graphics.BitmapFactory;
import android.net.Uri;
import android.os.Handler;
import android.provider.MediaStore;
import android.support.v7.app.AppCompatActivity;
import android.os.Bundle;
import android.widget.ImageView;
import java.util.ArrayList;
import java.util.Random;

public class MainActivity extends AppCompatActivity {

    Handler handler = new Handler();
    private ImageView randomPicture;
    private Bitmap currentBitmap = null;

    @Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_main);
        randomPicture = (ImageView)findViewById(R.id.random_picture);
        final ArrayList<String> imagesPath = new ArrayList<>();

        String[] projection = new String[]{
                MediaStore.Images.Media.DATA,
        };

        Uri images = MediaStore.Images.Media.EXTERNAL_CONTENT_URI;
        Cursor cur = getContentResolver().query(images, projection, null, null, null)

        if(cur == null) {
            randomPicture.setImageResource(R.drawable.picture);
        }
        else {
            if (cur.moveToFirst()) {

                int dataColumn = cur.getColumnIndex(
                        MediaStore.Images.Media.DATA);
                do {
                    imagesPath.add(cur.getString(dataColumn));
                } while (cur.moveToNext());
            }
            cur.close();
            final Random random = new Random();
            final int count = imagesPath.size();
            handler.post(new Runnable() {
                @Override
                public void run() {
                    int number = random.nextInt(count);
                    String path = imagesPath.get(number);
                    if (currentBitmap != null)
                        currentBitmap.recycle();
                    currentBitmap = BitmapFactory.decodeFile(path);
                    randomPicture.setImageBitmap(currentBitmap);
                    handler.postDelayed(this, 1000);
                }
            });
        }
    }
}

它在Cursor的行上崩溃并出现错误：

E/AndroidRuntime: FATAL EXCEPTION: main
Process: com.example.postcards, PID: 2424
java.lang.IllegalArgumentException: n <= 0: 0
at java.util.Random.nextInt(Random.java:182)
at com.example.postcards.MainActivity$1.run(MainActivity.java:53)
at android.os.Handler.handleCallback(Handler.java:739)
at android.os.Handler.dispatchMessage(Handler.java:95)
at android.os.Looper.loop(Looper.java:135)
at android.app.ActivityThread.main(ActivityThread.java:5254)
at java.lang.reflect.Method.invoke(Native Method)
at java.lang.reflect.Method.invoke(Method.java:372)
at com.android.internal.os.ZygoteInit$MethodAndArgsCaller.run(ZygoteInit.java:903)
at com.android.internal.os.ZygoteInit.main(ZygoteInit.java:698)

Answer 1

错误明确指出您无法向Random.nextInt()方法传递小于或等于0的值。

java.lang.IllegalArgumentException: n <= 0: 0

Answer 2

这一行：

int number = random.nextInt(count);

给出错误，因为您将0作为计数传递。使用nextInt(int)需要输入大于0.输入不能为负数或0

Answer 3

问题是随机的nextInt。您的计数可能为零？

Answer 4

Random#nextInt(int)期望这个论点是积极的。这不是这种情况。

Source

为什么我会得到＆＃34; java.lang.IllegalArgumentException＆＃34;错误？

4 个答案: