将图像与实际屏幕进行比较 [英] Compare image to actual screen

问题描述

我想让我的Java程序将实际屏幕与图片进行比较（截图）。

I'd like to make my Java program compare the actual screen with a picture (screenshot).

我不知道是否可能，但我有在Jitbit（宏录制器）中看到它，我想自己实现它。 （也许用这个例子你明白我的意思）。

I don't know if it's possible, but I have seen it in Jitbit (a macro recorder) and I would like to implement it myself. (Maybe with that example you understand what I mean).

谢谢

----编辑 - ---
换句话说，是否可以检查图像是否显示？要查找并比较屏幕中的像素？

----edit----- In other words, is it possible to check if an image is showing in? To find and compare that pixels in the screen?

推荐答案

好的，所以几天后我找到了答案。

Ok then, so I found an answer after a few days.

此方法采用屏幕截图：

public static void takeScreenshot() {
        try {
            BufferedImage image = new Robot().createScreenCapture(new Rectangle(490,490,30,30));
/* this two first parameters are the initial X and Y coordinates. And the last ones are the increment of each axis*/
            ImageIO.write(image, "png", new File("C:\\Example\\Folder\\capture.png"));

        } catch (IOException e) {
            e.printStackTrace();
        } catch (HeadlessException e) {
            e.printStackTrace();
        } catch (AWTException e) {
            e.printStackTrace();
        }
    }

另一个将比较图像

public static String compareImage() throws Exception {

// savedImage is the image we want to look for in the new screenshot.
// Both must have the same width and height

    String c1 = "savedImage";
    String c2 = "capture";

    BufferedInputStream in = new BufferedInputStream(new FileInputStream(c1
            + ".png"));
    BufferedInputStream in1 = new BufferedInputStream(new FileInputStream(
            c2 + ".png"));
    int i, j;
    int k = 1;
    while (((i = in.read()) != -1) && ((j = in1.read()) != -1)) {
        if (i != j) {
            k = 0;
            break;
        }
    }

    in.close();
    in1.close();
    if (k == 1) {

        System.out.println("Ok...");
        return "Ok";            

    } else {
        System.out.println("Fail ...");
        return "Fail";
    }
}

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