python - Python：如何在处理其他异常的同时忽略异常

Question

我弄乱了一些代码，并遇到了一个特殊的问题：

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

我遇到的问题是有时会objects.get(slug=slug)引发MultipleObjectsReturned异常，因为该字段在我的数据库中不是唯一的。我想知道如何MultipleObjectsReturned在“else”语句仍会被执行的情况下干净地捕捉到。

Answer 1

或者，根本不要使用该else子句：

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

简单的解决方案：诀窍是在调用方法时捕获MultipleObjectsReturned第二try条语句。get这样，不会引发异常并且继续正常执行。

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

不工作：

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

第二种“幼稚”方法不起作用的原因是，如果捕获到异常，解释器将不会遍历该else:子句。反而会默默无闻return。