如何解决 NameError: name 'xx' is not defined？

奥利弗·杨

我在学python，按照书上代码写的逻辑，想看看运行结果，代码如下，但是输出报错NameError: name 'pr' is not defined

代码显示如下：

stack=[]
def pushit():
    stack:append(input(' Enter New String: ').strip())
def popit():
    if len(stack)==0:
        print('Cannot pop from an empty stack!')
    else:
        print ('Removes [','stack.pop()',']')
def viewstack():
    print(stack)
CMDs={'u':pushit,'o':popit,'v':viewstack}
def showmenu():
    pr=''' 
p(U)sh
p(O)p
(V)iew
(Q)uit
Enter choice:'''
while True:
    while True:
        try:
            choice=input(pr).strip()[0].lower()
        except (EOFError,KeyboardInterrupt,IndexError):
            choice='q'
        print('\nYou picked:[%s]'%choice)
        if choice not in 'uovq':
            print('Invalid option,try again')
        else:
            break
        if choice=='q':
            break
        CMDs[choice]()
if _name_=='_main_':
    showmenu()

错误信息如下：

Traceback (most recent call last):
  File "/Users/oliver/Desktop/test.py", line 22, in <module>
    choice=input(pr).strip()[0].lower()
NameError: name 'pr' is not defined

可以罗伯特

您的代码中有两个错误，

方法中的逻辑showmenu需要缩进
_name_=='_main_'->__name__ == "__main__"

stack=[]
def pushit():
    stack:append(input(' Enter New String: ').strip())
def popit():
    if len(stack)==0:
        print('Cannot pop from an empty stack!')
    else:
        print ('Removes [','stack.pop()',']')
def viewstack():
    print(stack)
CMDs={'u':pushit,'o':popit,'v':viewstack}
def showmenu():
    pr=''' 
    p(U)sh
    p(O)p
    (V)iew
    (Q)uit
    Enter choice:'''
    while True:
        while True:
            try:
                choice=input(pr).strip()[0].lower()
            except (EOFError,KeyboardInterrupt,IndexError):
                choice='q'
            print('\nYou picked:[%s]'%choice)
            if choice not in 'uovq':
                print('Invalid option,try again')
            else:
                break
            if choice=='q':
                break
            CMDs[choice]()

if __name__ == "__main__":
    showmenu()

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