2009-07-30 39 views
1

我正在編程一個簡單的pyS60應用程序,並沒有真正做過任何事情與python或之前使用多個線程,所以這對我來說有點新。 爲了使應用程序保持打開狀態,我在應用程序主體初始化後設置了e32.Ao_lock wait(),然後在exit_key_handler上發出鎖定信號。如何解決缺少多個ao.lock的問題?

程序可能執行的任務之一是打開第三方應用程序UpCode。這將掃描條形碼並將條形碼字符串複製到剪貼板。當我關閉UpCode時,我的應用程序應該繼續並粘貼來自剪貼板的輸入。 我知道這可以使用Ao.lock來完成,但我已經調用了這個實例。理想情況下,我的應用程序會在注意到某些內容已粘貼到剪貼板後重新獲得焦點。 我可以通過睡眠或定時器功能完成所需的任務嗎?

你可以找到完整的腳本here,我已經將它簡稱爲下面的必要部分:

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() 

任何和讚賞所有幫助。

回答

0

我通過定義一個單獨的第二個鎖來解決這個問題,並且確保一次只有一個正在等待。它似乎沒有任何問題的工作。當前代碼可以找到hosted on google code

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