Question

希望你们一切顺利。

尝试创建一个充当字典的Python程序，但现在在创建elif语句时遇到了一些问题。我是我的空闲我一直有迹象表明我的语法对于elif是错误的，但我不是我做错了什么呢？我想这是一个缩进错误，但究竟是什么呢？

if choice == "0":
   print "good bye"

elif choice == "1":
  name = raw_input("Which philosopher do you want to get")
if name in philosopher:
 country = philosopher [name]
 print name, "comes from" , country
else:
 print "No such term"
***elif choice == "2" :*** ***<<I am being told that I have syntax error in this elif element, what am I doing wrong)**
  name = raw_input(" What name would you like to enter")
if name not in philosopher:
    country = raw_input( "Which country do you want to put your philosopher in")
    philosopher [name] = country
    print name, "has now been added and he is from", country
else:
      print "We already have that name"

Answer 1

假设您修复了缩进，if语句将按此顺序排列：

if x:
  #do something
elif x:
  #do something

if x:
  #do something
else:
  #do something

elif x:#CAUSES ERROR
  #do something

if x:
  #do something
else:
  #do something

您的elif在else声明之后出现。你不能这样做。 elif 必须介于if和else之间。否则编译器永远不会捕获elif（因为它刚刚运行并执行了else语句）。换句话说，您必须按如下顺序排列if语句：

if x:
  #do something
elif x:
  #do something
else:
  #do something

Answer 2

我认为你对缩进问题是正确的。以下是我认为您要做的事情：

if choice == "0":
    print "good bye"
elif choice == "1":
    name = raw_input("Which philosopher do you want to get")
    if name in philosopher:
        country = philosopher [name]
        print name, "comes from" , country
    else:
        print "No such term"
elif choice == "2" :
    name = raw_input(" What name would you like to enter")
    if name not in philosopher:
        country = raw_input( "Which country do you want to put your philosopher in")
        philosopher [name] = country
        print name, "has now been added and he is from", country
    else:
        print "We already have that name"

关键问题是缩进不一致，这使得Python难以确定您想要的内容。在你开发自己的风格并且有充分理由做其他事情之前，每个级别一致的四个缩进空间是一个好习惯。让编辑帮助您持续缩进。哦，并确保在缩进时不要混合标签和空格：这有点似乎有效，然后再回来咬你。

Answer 3

您希望将if name in philosopher ... No such term"部分放在以elif choice == "1":开头的块中。如果是这样，您需要再缩进一次，以便Python正确地对您的if，elif和else语句进行分组。

if choice == "0":
    # code
elif choice == "1":
    if name in philospher: # indented twice; can only run if choice == "1"
        # code
    else:
        # code
elif choice == "2":
    # code
# rest of code

创建简单的python程序时出现语法错误

3 个答案: