的Python：如何忽略一個例外，同時還處理其他

Question

我與一些代碼搞亂，並面臨着一個特殊問題：

def find_available_slug(object, instance, slug) 
    try: 
     sender_node = object.objects.get(slug=slug) 
    except object.DoesNotExist: 
     instance.slug = slug 
    else: 
     slug = '%s_' % slug 
     find_available_slug(object, instance, slug) 
    return 

我遇到的問題是，有時objects.get(slug=slug)拋出一個MultipleObjectsReturned異常，因爲這字段在我的數據庫中不是唯一的。我不知道我怎麼可以乾淨地趕上MultipleObjectsReturned而「else」語句仍然會被執行。

Answer 1

另外，不要使用else條款都：

def find_available_slug(object, instance, slug) 
    try: 
     sender_node = object.objects.get(slug=slug) 
    except object.DoesNotExist: 
     instance.slug = slug 
     return 
    except object.MultipleObjectsReturned: 
     pass 

    slug = '%s_' % slug 
    find_available_slug(object, instance, slug) 

Answer 2

簡單的解決方案：訣竅是在調用get方法時，將MultipleObjectsReturned置於第二個try語句中。這樣，不會引發異常並且正常執行繼續。

作品：

def find_available_slug(object, instance, slug) 
    try: 
     try: 
      sender_node = object.objects.get(slug=slug) 
     except object.MultipleObjectsReturned: 
      pass 
    except object.DoesNotExist: 
     instance.slug = slug 
    else: 
     slug = '%s_' % slug 
     find_available_slug(object, instance, slug) 
    return 

不起作用：

def find_available_slug(object, instance, slug) 
    try: 
     sender_node = object.objects.get(slug=slug) 
    except object.MultipleObjectsReturned: 
     pass 
    except object.DoesNotExist: 
     instance.slug = slug 
    else: 
     slug = '%s_' % slug 
     find_available_slug(object, instance, slug) 
    return 

第二個「天真」的方法不起作用的原因是，如果一個異常被捕獲，解釋器將不經過else:條款。它會寧靜地return。