Question

是否可以从OS X中的程序模拟鼠标的动作？具体来说，简短的版本是我试图使用两个网络摄像头模拟触摸屏。假设我可以获得X，Y位置，我可以将信息作为鼠标移动或点击发送给操作系统吗？

编辑 - 或者如果在另一个操作系统中特别容易，我愿意考虑这一点。

Answer 1

是的，有可能。您可以使用Quartz Event Services来模拟输入事件。

假设C，我写了这个简单的例子：

#include <ApplicationServices/ApplicationServices.h>
#include <unistd.h>

int main() {
    // Move to 200x200
    CGEventRef move1 = CGEventCreateMouseEvent(
        NULL, kCGEventMouseMoved,
        CGPointMake(200, 200),
        kCGMouseButtonLeft // ignored
    );
    // Move to 250x250
    CGEventRef move2 = CGEventCreateMouseEvent(
        NULL, kCGEventMouseMoved,
        CGPointMake(250, 250),
        kCGMouseButtonLeft // ignored
    );
    // Left button down at 250x250
    CGEventRef click1_down = CGEventCreateMouseEvent(
        NULL, kCGEventLeftMouseDown,
        CGPointMake(250, 250),
        kCGMouseButtonLeft
    );
    // Left button up at 250x250
    CGEventRef click1_up = CGEventCreateMouseEvent(
        NULL, kCGEventLeftMouseUp,
        CGPointMake(250, 250),
        kCGMouseButtonLeft
    );

    // Now, execute these events with an interval to make them noticeable
    CGEventPost(kCGHIDEventTap, move1);
    sleep(1);
    CGEventPost(kCGHIDEventTap, move2);
    sleep(1);
    CGEventPost(kCGHIDEventTap, click1_down);
    CGEventPost(kCGHIDEventTap, click1_up);

    // Release the events
    CFRelease(click1_up);
    CFRelease(click1_down);
    CFRelease(move2);
    CFRelease(move1);

    return 0;
}

假设是GCC，请编译：

gcc -o program program.c -Wall -framework ApplicationServices

享受魔力。

Answer 2

Swift鼠标移动并点击示例：

func mouseMoveAndClick(onPoint point: CGPoint) { 
    guard let moveEvent = CGEvent(mouseEventSource: nil, mouseType: .mouseMoved, mouseCursorPosition: point, mouseButton: .left) else {
        return
    }
    guard let downEvent = CGEvent(mouseEventSource: nil, mouseType: .leftMouseDown, mouseCursorPosition: point, mouseButton: .left) else {
        return
    }
    guard let upEvent = CGEvent(mouseEventSource: nil, mouseType: .leftMouseUp, mouseCursorPosition: point, mouseButton: .left) else {
        return
    }
    moveEvent.post(tap: CGEventTapLocation.cghidEventTap)
    downEvent.post(tap: CGEventTapLocation.cghidEventTap)
    upEvent.post(tap: CGEventTapLocation.cghidEventTap)
}

Answer 3

如果您不想编译并且正在寻找基于shell的工具，Cliclick可能是解决方案。

Answer 4

这是一个基于jweyrick的答案的工作C程序：

// Compile instructions:
//
// gcc -o click click.c -Wall -framework ApplicationServices

#include <ApplicationServices/ApplicationServices.h>
#include <unistd.h>

int main(int argc, char *argv[]) {
  int x = 0, y = 0, n = 1;
  float duration = 0.1;

  if (argc < 3) {
    printf("USAGE: click X Y [N] [DURATION]\n");
    exit(1);
  }

  x = atoi(argv[1]);
  y = atoi(argv[2]);

  if (argc >= 4) {
    n = atoi(argv[3]);
  }

  if (argc >= 5) {
    duration = atof(argv[4]);
  }

  CGEventRef click_down = CGEventCreateMouseEvent(
    NULL, kCGEventLeftMouseDown,
    CGPointMake(x, y),
    kCGMouseButtonLeft
  );

  CGEventRef click_up = CGEventCreateMouseEvent(
    NULL, kCGEventLeftMouseUp,
    CGPointMake(x, y),
    kCGMouseButtonLeft
  );

  // Now, execute these events with an interval to make them noticeable
  for (int i = 0; i < n; i++) {
    CGEventPost(kCGHIDEventTap, click_down);
    sleep(duration);
    CGEventPost(kCGHIDEventTap, click_up);
    sleep(duration);
  }

  // Release the events
  CFRelease(click_down);
  CFRelease(click_up);

  return 0;
}

托管于https://gist.github.com/Dorian/5ae010cd70f02adf2107

在OS X中以编程方式模拟鼠标输入

4 个答案: