强智教务系统验证码识别 java

之前没有想过要通过爬取页面的方式来获取数据，想想毕竟api得到的数据有限，国内使用强智系统的不少，也算是做了点贡献吧。
识别的过程倒是蛮顺利。肉眼验证似乎没有不正确的
话不多说，我们看到的登录界面的验证码是这样的：

如果你的目标和这个一毛一样，那么你便可以拿来用咯
思路很清晰，首先我们知道图像实际上是个二维数组，我们将图片二值化，即处理成只有黑白两色的图像，与训练好的的二维数组逐一相比较，最接近哪个便是哪个字符。
如字母a的二维数组：

//a{{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},{0,0,0,1,1,1,1,1,1,0,0,0,0,0,0,0,0,0,0,0},{0,0,1,1,1,1,1,1,1,1,0,0,0,0,0,0,0,0,0,0},{0,0,1,0,0,0,0,1,1,1,1,0,0,0,0,0,0,0,0,0},{0,0,0,0,0,0,0,0,1,1,1,0,0,0,0,0,0,0,0,0},{0,0,0,0,0,0,0,0,1,1,1,0,0,0,0,0,0,0,0,0},{0,0,0,0,1,1,1,1,1,1,1,0,0,0,0,0,0,0,0,0},{0,0,1,1,1,1,1,1,1,1,1,0,0,0,0,0,0,0,0,0},{0,1,1,1,1,0,0,0,1,1,1,0,0,0,0,0,0,0,0,0},{0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,0,0,0,0,0},{0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,0,0,0,0,0},{0,1,1,1,0,0,0,1,1,1,1,0,0,0,0,0,0,0,0,0},{0,0,1,1,1,1,1,1,1,1,1,1,0,0,0,0,0,0,0,0},{0,0,0,1,1,1,1,0,0,1,1,1,0,0,0,0,0,0,0,0},{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0}}

图片分析

验证码形如，经过爬取大量验证码可以发现，验证码图片为80*40px，周围有1px黑边，实际上黑边我忽略了，因为再后面的切割中被剪掉了。接着说图片，图片只含有小写字母和数字，没有字母o和数字0，即一共34个字符，后期要至少准备34个二维数组用于比较

进一步分析

打开ps

找一个图片有重合的例子，发现每个字符占水平20px，垂直24px,可以初步处理下，裁剪掉上下的空白：

BufferedImage image = ImageIO.read(f);
image = image.getSubimage(0, 9, 80, 24);

二值化

 public static BufferedImage grayImage(BufferedImage bufferedImage) throws Exception {int width = bufferedImage.getWidth();int height = bufferedImage.getHeight();BufferedImage grayBufferedImage = new BufferedImage(width, height, bufferedImage.getType());for (int i = 0; i < bufferedImage.getWidth(); i++) {for (int j = 0; j < bufferedImage.getHeight(); j++) {final int color = bufferedImage.getRGB(i, j);final int r = (color >> 16) & 0xff;final int g = (color >> 8) & 0xff;final int b = color & 0xff;int gray = (int) (0.3 * r + 0.59 * g + 0.11 * b);int newPixel = colorToRGB(255, gray, gray, gray);grayBufferedImage.setRGB(i, j, newPixel);}}return grayBufferedImage;}/*** 颜色分量转换为RGB值* * @param alpha* @param red* @param green* @param blue* @return*/private static int colorToRGB(int alpha, int red, int green, int blue) {int newPixel = 0;newPixel += alpha;newPixel = newPixel << 8;newPixel += red;newPixel = newPixel << 8;newPixel += green;newPixel = newPixel << 8;newPixel += blue;return newPixel;}public static BufferedImage binaryImage(BufferedImage image) throws Exception {int w = image.getWidth();int h = image.getHeight();float[] rgb = new float[3];double[][] coordinates = new double[w][h];int black = new Color(0, 0, 0).getRGB();int white = new Color(255, 255, 255).getRGB();BufferedImage bi = new BufferedImage(w, h, BufferedImage.TYPE_BYTE_BINARY);;for (int x = 0; x < w; x++) {for (int y = 0; y < h; y++) {int pixel = image.getRGB(x, y);rgb[0] = (pixel & 0xff0000) >> 16;rgb[1] = (pixel & 0xff00) >> 8;rgb[2] = (pixel & 0xff);float avg = (rgb[0] + rgb[1] + rgb[2]) / 3;coordinates[x][y] = avg;}}// 这里是阈值，白底黑字还是黑底白字，大多数情况下建议白底黑字，后面都以白底黑字为例double SW = 192;for (int x = 0; x < w; x++) {for (int y = 0; y < h; y++) {if (coordinates[x][y] < SW) {bi.setRGB(x, y, black);} else {bi.setRGB(x, y, white);}}}return bi;}

按照前面的分析将图片分割成4部分，分别处理之：

     int width = image.getWidth();int height = image.getHeight();//System.out.println("width\t" + width + "\theight\t" + height);//int subWidth = width / 4;newim[0] = binaryImage(image.getSubimage(4, 0, 20, height));newim[1] = binaryImage(image.getSubimage(22, 0, 20, height));newim[2] = binaryImage(image.getSubimage(40, 0, 20, height));newim[3] = binaryImage(image.getSubimage(58, 0, 20, height));

实际上现在每个分割的图片便可以用0,1表示，如果用██填充的话。。。

    ████████████                        ██████████████████                    ██          ████████                  ██████                  ██████                  ██████                  ██████                    ██████                      ████████████                        ████████████████                    ████████                  ████████                ██████                ██████                ██████                ████          ██████                  ██████████████████                    ██████████████                      {{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1},{0,0,1,1,1,1,1,1,0,0,0,0,0,0,0,0,0,0,0,0},{0,1,1,1,1,1,1,1,1,1,0,0,0,0,0,0,0,0,0,0},{0,1,0,0,0,0,0,1,1,1,1,0,0,0,0,0,0,0,0,0},{0,0,0,0,0,0,0,0,1,1,1,0,0,0,0,0,0,0,0,0},{0,0,0,0,0,0,0,0,1,1,1,0,0,0,0,0,0,0,0,0},{0,0,0,0,0,0,0,0,1,1,1,0,0,0,0,0,0,0,0,0},{0,0,0,0,0,0,0,1,1,1,0,0,0,0,0,0,0,0,0,0},{0,0,0,0,0,0,1,1,1,0,0,0,0,0,0,0,0,0,0,0},{0,0,1,1,1,1,1,1,0,0,0,0,0,0,0,0,0,0,0,0},{0,0,1,1,1,1,1,1,1,0,0,0,0,0,0,0,0,0,0,0},{0,0,0,0,0,0,0,1,1,1,1,0,0,0,0,0,0,0,0,0},{0,0,0,0,0,0,0,0,1,1,1,1,0,0,0,0,0,0,0,0},{0,0,0,0,0,0,0,0,0,1,1,1,0,0,0,0,0,0,0,0},{0,0,0,0,0,0,0,0,0,1,1,1,0,0,0,0,0,0,0,0},{0,0,0,0,0,0,0,0,0,1,1,1,0,0,0,0,0,0,0,0},{0,1,1,0,0,0,0,0,1,1,1,0,0,0,0,0,0,0,0,0},{0,1,0,1,1,1,1,1,1,1,0,0,0,0,0,0,0,0,0,0},{0,0,1,1,1,1,1,1,1,0,0,0,0,0,0,0,0,0,0,0},{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0}},

之后我们用训练好的数组比较选出结果即可

     StringBuilder sb = new StringBuilder();int count = -1;double total = 1;int hit = 0;while (++count < 4) {int chCount = 0;int target = 0;//最接近的指针double temp = -1.0;while (chCount < MyCharacter.g.length) {total = 1;hit = 0;for (int i = 0; i < g[count].length; i++) {for (int j = 0; j < g[count][i].length; j++) {if (MyCharacter.g[chCount][i][j] == 1) {//System.out.print("-");++total;if(g[count][i][j] == MyCharacter.g[chCount][i][j]) {//System.out.print("+");++hit;}}}}//System.out.println(hit);//System.out.println(total);//System.out.println(MyCharacter.ch[chCount]+" "+hit/total);if(hit/total > temp) {target = chCount;temp = hit/total;}else {}++chCount;}//比例最大的给他sb.append(MyCharacter.ch[target]);}return sb.toString();

效果不错

完整代码：github

参考: java图像处理