我正在编写这个 Python(2.7) 脚本,它是一系列菜单。对于每个子菜单,都有一个继承 BaseMenu 的子类。有一个名为 SubMenu 的子类,我想在子菜单名称的标题下方打印 server_name 。在 BaseMenu() 中,我尝试放置一个 if 语句来检查属性是否存在,如果存在则打印它:
def display(self):
header = "FooBar YO"
term = getTerminalSize()
#sys.stdout.write("\x1b[2J\x1b[H")
print header.center(term, '*')
print self.menu_name.center(term, '+')
###Below is the check
if self.modify_server_class:
print self.modify_server_class.center(term, '+')
当我尝试运行不具有该属性的其他类时,它不起作用并提示 AttributeError: 'Servers' object has no attribute 'modify_server_class'
。
我应该如何让 BaseClass 检查此属性并打印它(如果存在)?我不想将该属性放在每个类中并将其设置为 false。似乎可能有一种更干净的方法。
class BaseMenu(object):
__metaclass__ = abc.ABCMeta
@abc.abstractproperty
def options(self):
pass
@abc.abstractproperty
def menu_name(self):
pass
def display(self):
header = "FooBar YO"
term = getTerminalSize()
#sys.stdout.write("\x1b[2J\x1b[H")
print header.center(term, '*')
print self.menu_name.center(term, '+')
if self.modify_server_class:
print self.modify_server_class.center(term, '+')
print "Please choose which option:"
for i in self.options:
print(
str(self.options.index(i)+1) + ") "
+ i.__name__
)
while True:
value = int(raw_input("Please Choose[ENTER to exit]: ")) - 1
try:
if value == "":
break
else:
#with Redirect(self.file_object):
self.options[value](self)
self.display()
except IndexError:
print "Out of range!"
class ModifyServer(BaseMenu):
def __init__(self, a):
self.servers_object = a
menu_name = "Modify Server"
server_chosen = "None"
modify_server_class = "hi"
def choose_server(self):
if not self.servers_object.servers:
cs = pyrax.cloudservers
self.servers_object.servers = cs.servers.list()
temp = self.servers_object.servers
for index, item in enumerate(temp):
print (
"%s) %s" % (index+1, item.name)
)
def jj(self):
pass
options = (
choose_server,
jj
)
最佳答案
您可以为您创建一个设置此属性的BaseMenu
构造函数,或者使用hasattr()
关于Python 父类检查子类中要在类函数中使用的属性,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/17384804/