try... except... except... ：如何避免重复代码

• 我想避免写作errorCount += 1在不止一处。
• 我正在寻找比更好的方法

    success = False
    try:
        ...
    else:
        success = True
    finally:
        if success:
            storage.store.commit()
        else:
            storage.store.rollback()
  

• 我试图避免store.rollback()在每个 except 子句中。

关于如何做到这一点有什么想法吗？

count = 0
successCount = 0
errorCount = 0
for row in rows:
    success = False
    count += 1
    newOrder = storage.RepeatedOrder()
    storage.store.add(newOrder)
    try:
        try:
            newOrder.customer = customers[row.customer_id]
        except KeyError:
            raise CustomerNotFoundError, (row.customer_id,)
        newOrder.nextDate = dates[row.weekday]
        _fillOrder(newOrder, row.id)
    except CustomerNotFoundError as e:
        errorCount += 1
        print u"Error: Customer not found. order_id: {0}, customer_id: {1}".format(row.id, e.id)
    except ProductNotFoundError as e:
        errorCount += 1
        print u"Error: Product not found. order_id: {0}, product_id: {1}".format(row.id, e.id)
    else:
        success = True
        successCount += 1
    finally:
        if success:
            storage.store.commit()
        else:
            storage.store.rollback()
print u"{0} of {1} repeated orders imported. {2} error(s).".format(successCount, count, errorCount)

这看起来像是 Python 新功能的一个可能应用with陈述。无论代码块的结果如何，它都允许安全地展开操作并释放资源。

阅读有关它的内容PEP 343