Django，不序列化的坚持

Question

我正在做一个Django项目，它使用平板扫描仪。连接到扫描仪需要很长时间。所以我在寻找一种重复使用扫描仪实例的方法。

序列化似乎是解决这个问题的方法。不幸的是，我不能序列化或挑选扫描器实例。我经常遇到错误，告诉我序列化失败了。

是否有另一种方法来重复使用同一个扫描器实例进行多个扫描？后端魔术还是前端魔法？(注意，我对前端开发一无所知。)

我们可以作弊！

该项目将在本地计算机上脱机运行，根本没有internet或网络连接。这可能会提供其他不安全的选项。

我用来扫描的东西

• 神智，用于访问扫描器的包
• 丙酮-3-理智，一个用于理智的Python包装器
• 图像扫描，扫描仪驱动程序

Answer 1

环顾四周后，我发现了RPyC。本教程提供了一个工作示例，我让它在几分钟内运行。

服务启动并连接到扫描仪。多个客户端可以立即调用该服务并进行扫描。

import rpyc
import sane


class MyService(rpyc.Service):

    def __init__(self):
        self.scanner = Scanner()

    def on_connect(self, conn):
        pass

    def on_disconnect(self, conn):
        pass

    def exposed_scan_image(self):
        self.scanner.low_dpi_scan('scanner_service')


class Scanner(object):
    """
    The Scanner class is used to interact with flatbed scanners.
    """

    def __init__(self):
        sane.init()
        self.device_name = None
        self.error_message = None
        self.device = self.get_device()

    def get_device(self):
        """
        Return the first detected scanner and set the name.

        @return: sane.SaneDev
        """
        devices = sane.get_devices()
        print('Available devices:', devices)

        # Empty list means no scanner is connect or the connected scanner is
        # already being used
        if not devices:
            self.error_message = 'Scanner disconnect or already being used.'
            print(self.error_message)
            return None

        # open the first scanner you see
        try:
            device = sane.open(devices[0][0])
        except Exception as e:
            self.error_message = e
            print(e)
            return None

        brand = devices[0][1]
        model = devices[0][2]
        self.device_name = "{brand}_{model}".format(
            brand=brand,
            model=model
        )

        print("Connected to:", self.device_name)

        # set to color scanning mode, this is not always the default mode
        device.mode = 'color'

        return device

    def low_dpi_scan(self, file_name):
        """
        Scan at 300 dpi and store as jpeg.

        @param file_name: string
        """
        image = self.scan_image(300)
        image.save(file_name+'.jpeg')

    def scan_image(self, dpi):
        """
        Scan an image.

        @param dpi: integer
        @return: image file
        """
        self.device.resolution = dpi
        self.device.start()
        image = self.device.snap()

        return image


if __name__ == "__main__":
    from rpyc.utils.server import ThreadedServer
    t = ThreadedServer(MyService(), port=18861)
    t.start()