我写了一段非常简单的代码，但一直有这种唠叨的感觉，有一种更简单的方法来编写它。特别是其他两个；我觉得一定有办法避免在这个循环中写两个相同的 else 子句。我是不是很笨？
parentid = ca.get_parent_id(spacename, parentname)
if parentid is None:
raise Exception("Unable to find parent page")
for subpage in subpages:
## check to see if subpage is a child of the previous page
childreninfo = ca.get_page_children(parentid)
if childreninfo['children']['page']['results']:
for pageinfo in childreninfo['children']['page']['results']:
if pageinfo['title'] == subpage:
break
else:
# if we find the page somewhere in the space but it's not a child, stop!
info = ca.get_page_info_by_title(subpage, spacename)
if len(info['results']) == 1:
raise Exception("Found page but not in right location")
else:
# if we find the page somewhere in the space but it's not a child, stop!
info = ca.get_page_info_by_title(subpage, spacename)
if len(info['results']) == 1:
raise Exception("Found page but not in right location")
## DO STUFF with subpage here
请您参考如下方法:
在这里你根本不需要外层的if:..else:。直接进入 for 循环:
for subpage in subpages:
## check to see if subpage is a child of the previous page
childreninfo = ca.get_page_children(parentid)
for pageinfo in childreninfo['children']['page']['results']:
if pageinfo['title'] == subpage:
break
else:
# if we find the page somewhere in the space but it's not a child, stop!
info = ca.get_page_info_by_title(subpage, spacename)
if len(info['results']) == 1:
raise Exception("Found page but not in right location")
当遍历空序列时，立即执行 for 语句的 else: 组。这与您的 if ...: ... else: ... 使用代码重复测试的效果完全相同。
如果 childreninfo['children']['page']['results'] 可以是无法迭代的不同假值(例如 None)，在for 循环可迭代表达式中添加或()以强制对空序列进行迭代:
for pageinfo in childreninfo['children']['page']['results'] or ():
