Python变量替换

Question

我有一个脚本，它调用我正在整理的linux来宾列表。以下是代码：

#!/usr/bin/python


guests = ['guest1','guest2','guest3','guest*']
def serverCheck(guestList)
    for g in guestList:
        server = AdminControl.completeObjectName('cell=tstenvironment,node=guest1,name=uatenvironment,type=Server,*')
        try:
            status = AdminControl.getAttribute(server, 'state')
            print g + status
        except:
            print "Error %s is down." % g
serverCheck(guests)

问题就在这一行上：

server = AdminControl.completeObjectName('cell=Afcutst,node=%s,name=afcuuat1,type=Server,*') % g

如何使用我的列表填充节点变量，同时仍然能够将括号内的信息传递给AdminControl函数？

Answer 1

参数字符串本身是%操作符的参数，而不是函数调用的返回值。

server = AdminControl.completeObjectName(
    'cell=Afcutst,node=%s,name=afcuuat1,type=Server,*' % (g,)
)

通过窥视水晶球，Python 3.6将允许您编写

server = AdminControl.completeObjectName(
    f'cell=Afcutst,node={g},name=afcuuat1,type=Server,*'
)

将变量直接嵌入到特殊格式字符串文本中。

Answer 2

你能这样试试吗？

AdminControl.completeObjectName('cell=tstenvironment,node=%s,name=uatenvironment,type=Server,*'%g)

Answer 3

为了获得更多的可读性，我建议这样做，并使用相同的方式从变量格式化字符串(在这里，我选择了str.format)

guests = ['guest1','guest2','guest3','guest*']

def serverCheck(guestList)
    name_tpl = 'cell=tstenvironment,node={},name=uatenvironment,type=Server,*'

    for g in guestList:
        obj_name = name_tpl.format(g)
        server = AdminControl.completeObjectName(obj_name)
        try:
            status = AdminControl.getAttribute(server, 'state')
            print '{}: {}'.format(g, status)
        except:
            print 'Error {} is down'.format(g)

serverCheck(guests)