Question

我已成功从compareHexaRGB方法中检索RGB像素值。输出似乎并不令人信服，因为输出也包含空字符。我怎样才能消除它们？

其次，当我尝试组合String [] [] char 1 + char 2时，我得到一个错误，声明它是类型“+”的坏操作数。但我想结合2个字符串，以便我能够将其转换为ASCII字符。

以下是compareHexaRGB方法的代码：

public class compareHexaRGB
{
 private static int w;
 private static int h;
 private static BufferedImage img;
 private static BufferedImage img2;
 private static String[][] check_hex2;
 private static String[][] check_hex4;
 private static String[][] message;

 public static void compareHexaRGB(BufferedImage image, BufferedImage image2, int width, int height) throws IOException
{
w = width;
h = height;
img = image;
img2 = image2;

}

public void check() throws IOException
{
    getPixelRGB1 pixel = new getPixelRGB1();
    getPixelData1 newPD = new getPixelData1();

    int[] rgb;
    int count = 0;

    int[][] pixelData = new int[w * h][3];
    check_hex2 = new String[w][h];
    check_hex4 = new String[w][h];

    for(int i = 0; i < w; i++) 
    {
        for(int j = 0; j < h; j++)
        {
            rgb = newPD.getPixelData(img, i, j);

            for(int k = 0; k < rgb.length; k++)
            {
                pixelData[count][k] = rgb[k];
            }

                if(pixel.display_imgHex2()[i][j].equals(pixel.display_img2Hex2()[i][j]))
                {
                    System.out.println("\nPixel values at position 2 are the same." + "\n" + pixel.display_imgHex2()[i][j] + "  " + pixel.display_img2Hex2()[i][j]);
                }
                if(pixel.display_imgHex4()[i][j].equals(pixel.display_img2Hex4()[i][j]))
                {
                    System.out.println("\nPixel values at position 4 are the same." + "\n" + pixel.display_imgHex4()[i][j] + "  " + pixel.display_img2Hex4()[i][j]);
                }
                if(!pixel.display_imgHex2()[i][j].equals(pixel.display_img2Hex2()[i][j]))
                {
                    System.out.println("\nPixel values at position 2 are not the same." + "\n" + pixel.display_imgHex2()[i][j] + "  " + pixel.display_img2Hex2()[i][j]);
                    check_hex2[i][j] = pixel.display_img2Hex2()[i][j];
                    System.out.println("\nOutput Hex 2: " + check_hex2[i][j]);
                }
                if(!pixel.display_imgHex4()[i][j].equals(pixel.display_img2Hex4()[i][j]))
                {
                    System.out.println("\nPixel values at position 4 are not the same." + "\n" + pixel.display_imgHex4()[i][j] + "  " + pixel.display_img2Hex4()[i][j]);
                    check_hex4[i][j] = pixel.display_img2Hex4()[i][j];
                    System.out.println("\nOutput Hex 4: " + check_hex4[i][j]);
                }
                if(!pixel.display_imgHex2()[i][j].equals(pixel.display_img2Hex2()[i][j]) || (!pixel.display_imgHex4()[i][j].equals(pixel.display_img2Hex4()[i][j])))
                {
                    System.out.println("\nOne of the pixel values at position 2 and 4 are not the same." + "\n" + pixel.display_imgHex2()[i][j] + "  " + pixel.display_img2Hex2()[i][j] + "\n" + pixel.display_imgHex4()[i][j] + "  " + pixel.display_img2Hex4()[i][j]);

                    if(!pixel.display_imgHex2()[i][j].equals(pixel.display_img2Hex2()[i][j]) || (pixel.display_imgHex2()[i][j].equals(pixel.display_img2Hex2()[i][j])))
                    {
                        check_hex2[i][j] = pixel.display_img2Hex2()[i][j];
                        System.out.println("\nOutput Hex 2: " + check_hex2[i][j]);
                    }

                    if(!pixel.display_imgHex4()[i][j].equals(pixel.display_img2Hex4()[i][j]) || (pixel.display_imgHex4()[i][j].equals(pixel.display_img2Hex4()[i][j])))
                    {
                        check_hex4[i][j] = pixel.display_img2Hex4()[i][j];
                        System.out.println("\nOutput Hex 4: " + check_hex4[i][j]);
                    }

                }
            count++;
            System.out.println("\nOutput Count: " + count);
        }

    }

}

public String[][] getCheck_hex2()
{
    return check_hex2;
}

public String[][] getCheck_hex4()
{
    return check_hex4;
}
}

extractMessage方法的代码：

public class extractMessage
{
private static String character;
private static String[][] char1;
private static String[][] char2;
private static int w;
private static int h;
private static String[][] in;

public static void extractMessage(int width, int height, String[][] inn)
{
    character = "";
    w = width;
    h = height;
    in = inn;
}

public static void printString2DArray(String[][] inn) 
{
    for (int i = 0; i < inn.length; i++) 
    {
        for(int j = 0; j < inn[i].length; j++)
        {
            if (i != 0 && j == 0)
            {
                System.out.print(" ");
            }
            System.out.println(inn[i][j]);
        }
    }
}

public static void charExtract()
{
    compareHexaRGB hexRGB = new compareHexaRGB();

    char1 = hexRGB.getCheck_hex2();
    char2 = hexRGB.getCheck_hex4();
    String[][] combine = char1 + char2;

    System.out.println("Char 1: ");
    printString2DArray(char1);
    System.out.println("Char 2: ");
    printString2DArray(char2);

}
}

输出：

Char 1: 
null
null
null
null
null
null
 null
null
6
6
null
null
 6
null
null
null
null
null
Char 2: 
null
null
null
null
null
null
 null
null
2
3
null
null
 1
null
null
null
null
null

我非常感谢任何建议或更正！

Answer 1

由于这些位置的像素相同，因此获得大量空值。在check()方法中，当像素相同时，您不会在该位置向数组添加任何内容 - 因此它保持为空。在这些点添加一个持有角色的地方，例如-或其他。
您的打印方法也会在一行上打印每个元素，如果您像打印矩阵一样打印2D数组，它将更具可读性。您可以通过将方法修改为此
来完成此操作
```
for (int i = 0; i < inn.length; i++) 
{
    for(int j = 0; j < inn[i].length; j++)
    {
       System.out.print(inn[i][j] + " ");
    }
    System.out.println();
}  
```
您可能感兴趣的另一个有用的方法是Arrays.deepToString(Array)，但这会创建一个字符串，并且需要操作才能在多行上打印更容易执行以下操作
```
for(String[] arr : inn)
     System.out.println(Arrays.toString(arr));
```

尝试添加数组时出现错误，因为您根本无法做到这一点 - 将两个数组一起添加而不循环内容。您将需要一个类似于上面的循环

char1 = hexRGB.getCheck_hex2();
char2 = hexRGB.getCheck_hex4();
//assuming char1 and char2 are the same size and symmetrical
String[][] combine = new String[char1.length][char1[0].length]

for (int i = 0; i < char1.length; i++) 
{
    for(int j = 0; j < char1[i].length; j++)
    {
         // This will only concatenate the strings though,
         // If you want the numerical addition you will have to 
         // convert them to ints first
         combine[i][j] = char1[i][j] + char2[i][j];
    }
}

检索到的值具有空值。如何消除它们？

1 个答案: