如何解决缺少多个 ao.lock 的问题？

Question

我正在编写一个简单的 pyS60 应用程序，之前没有真正用 python 做过任何事情或使用多线程，所以这对我来说有点新鲜。 为了保持应用程序打开，我在应用程序主体初始化后将 e32.Ao_lock 设置为 wait()，然后向 exit_key_handler 发出锁定信号。

该程序可能执行的任务之一是打开第三方应用程序 UpCode。 这会扫描条形码并将条形码字符串复制到剪贴板。 当我关闭 UpCode 时，我的应用程序应该恢复并粘贴剪贴板中的输入。 我知道这可以使用 Ao.lock 来完成，但我已经调用了它的一个实例。 理想情况下，我的应用程序会在注意到某些内容已粘贴到剪贴板后重新获得焦点。 我可以使用睡眠或定时器功能之一来完成我需要的功能吗？

您可以在此处找到完整的脚本，我已将其缩写为必要的部分如下：

lock=e32.Ao_lock()

# Quit the script
def quit():
    lock.signal()

# Callback function will be called when the requested service is complete. 
def launch_app_callback(trans_id, event_id, input_params):
    if trans_id != appmanager_id and event_id != scriptext.EventCompleted:
        print "Error in servicing the request"
        print "Error code is: " + str(input_params["ReturnValue"]["ErrorCode"])
        if "ErrorMessage" in input_params["ReturnValue"]:
            print "Error message is: " + input_params["ReturnValue"]["ErrorMessage"]
    else:
        print "\nWaiting for UpCode to close"
    #lock.signal()

# launch UpCode to scan barcode and get barcode from clipboard
def scan_barcode():
    msg('Launching UpCode to scan barcode.\nPlease exit UpCode after the barcode has been copied to the clipboard')
    # Load appmanage service
    appmanager_handle = scriptext.load('Service.AppManager', 'IAppManager')
    # Make a request to query the required information in asynchronous mode
    appmanager_id = appmanager_handle.call('LaunchApp', {'ApplicationID': u's60uid://0x2000c83e'}, callback=launch_app_callback)
    #lock.wait()
    #print "Request complete!"
    barcode = clipboard.Get()
    return barcode

# handle the selection made from the main body listbox
def handle_selection():
    if (lb.current() == 0):
        barcode = scan_barcode()
    elif (lb.current() ==1):
        barcode = clipboard.Get()
    elif (lb.current() ==2):
        barcode = input_barcode()

    found = False
    if is_barcode(barcode):
        found, mbid, album, artist = identify_release(barcode)
    else:
        msg('Valid barcode not found. Please try again/ another method/ another CD')
        return

    if found:
        go = appuifw.query(unicode('Found: ' + artist + ' - ' + album + '\nScrobble it?'), 'query')
        if (go == 1):
            now = datetime.datetime.utcnow()
            scrobble_tracks(mbid, album, artist, now)
        else:
            appuifw.note(u'Scrobbling cancelled', 'info')
    else:
        appuifw.note(u'No match found for this barcode.', 'info')

# Set the application body up
appuifw.app.exit_key_handler = quit
appuifw.app.title = u"ScanScrobbler"
entries = [(u"Scan a barcode", u"Opens UpCode for scanning"),
           (u"Submit barcode from clipboard", u"If you've already copied a barcode there"),
           (u"Enter barcode by hand", u"Using numeric keypad")
          ]

lb = appuifw.Listbox(entries, handle_selection)
appuifw.app.body = lb

lock.wait()

感谢所有帮助。

Answer 1

我通过定义一个单独的第二个锁来解决这个问题，并确保一次只有一个在等待。 它似乎工作没有任何问题。 当前代码可以托管在 Google 代码上找到