我遇到了一个相当简单的问题,我无法想出一个优雅的解决方案.
我正在str.format一个函数中创建一个字符串,该函数在一个dict替换中传递以用于格式.我想创建字符串并使用值格式化它们,如果它们被传递,否则将它们留空.
防爆
kwargs = {"name": "mark"}
"My name is {name} and I'm really {adjective}.".format(**kwargs)
应该回来
"My name is mark and I'm really ."
而不是抛出KeyError(如果我们不做任何事情会发生什么).
令人尴尬的是,我甚至无法为这个问题提出一个不优雅的解决方案.我想我可以通过不使用来解决这个问题str.format,但如果可能的话,我宁愿使用内置(主要是我想要的).
注意:我事先并不知道将使用哪些密钥.如果有人包含一把钥匙但是没有把它放在kwargs dict中,我试图优雅地失败.如果我100%准确地知道将要查找哪些键,我只需填充所有键并完成它.
(对不起.标题很不清楚.我无法想出一个好的.)
假设我有一个这样的网址(它是root-relative):
"/forums/support/windows/help_i_deleted_sys32/6/"
我试图将其拆分为这样的类结构:
class Forum_Spot:
def __init__(self, url):
parts = url.strip("/").split("/")
#parts is now ["forums", "support", "windows", "help...", "6"]
self.root = "forums"
self.section = "support"
self.subsection = "windows"
self.thread = "help..."
self.post = "6"
但是说我不知道网址到底有多长(可能是"/ forums/support /","/ forums/support/windows /"等)(但我知道它不会更深比5级).任何人都可以想出一种优雅的方式来分配这些值,让任何未分配的部分None？(即"/ forums/support/windows /",线程和帖子属性为None)
我知道我可以这样做:
class Forum_Spot:
def __init__(self, url):
parts = url.strip("/").split("/")
#parts is now ["forums", "support", "windows", "help...", "6"]
if len(parts) > 0:
self.root = parts[0]
else:
self.root = None
if len(parts) > 1:
self.section = parts[1]
else: …为什么这个查询:
SELECT inv.batch_number_id AS batch_number_id,
inv.expiry_date AS expiry_date,
inv.facility_id AS facility_id,
inv.id AS id,
fac.code AS facility_code,
inv_st.description AS status,
bn.batch_nbr AS batch_number,
bn.expiry_date AS batch_expiry_date,
sum(CASE WHEN al.status_id < 90 THEN al.alloc_qty ELSE 0 END) AS alloc_qty
FROM inventory inv
INNER JOIN facility fac ON inv.facility_id = fac.id
INNER JOIN inventory_status inv_st ON inv.status_id = inv_st.id
LEFT OUTER JOIN batch_number bn ON inv.batch_number_id = bn.id
LEFT OUTER JOIN allocation al ON al.from_inventory_id = inv.id
GROUP BY inv.id,
facility_code,
status, …