[英]UnboundLocalError: local variable 'text_in_use' referenced before assignment
I can set the variable and print it directly after if choice == 1. But if choice == 2 i cant print it out the text that was set, I get an error.如果选择 == 1,我可以设置变量并直接打印它。但是如果选择 == 2,我无法将设置的文本打印出来,我会收到错误消息。
(UnboundLocalError: local variable 'text_in_use' referenced before assignment) (UnboundLocalError:赋值前引用了局部变量'text_in_use')
How can i fix this?我怎样才能解决这个问题?
text_in_use = ''
encrypt_key = ''
def menu():
choice = int(input("""1: Input text to work with
2: Print the current text
3: Encrypt the current text
4: Decrypt the current text
5: Exit
Enter Choice: """))
if choice == 1:
text_in_use = str(input("Enter Text: ")).upper()
print("Text to use was set to:", text_in_use)
menu()
elif choice == 2:
print(text_in_use) #this is where i get the error <-----
menu()
elif choice == 3:
print("3")
menu()
elif choice == 4:
#decrypt()
print("4")
menu()
elif choice == 5:
#exit()
print("5")
menu()
menu()
i just want it to print the text that was set.我只想让它打印设置的文本。
-- Hi Linus, ——嗨,莱纳斯,
Your variable你的变量
text_in_use
only is set if your first condition is met.只有在满足您的第一个条件时才设置。 So if your codes skips that condition and moves on to:
因此,如果您的代码跳过该条件并继续执行以下操作:
elif choice == 2
the variable hasn't been set yet.该变量尚未设置。
Since, the function recursively calls itself after every option, you also can't add the variable before the first clause as I initially suggested.由于 function 在每个选项之后递归调用自身,因此您也不能像我最初建议的那样在第一个子句之前添加变量。
So I'm changing my answer to the following:所以我将我的答案更改为以下内容:
At this point I would also like to add that a function without any exit may not be what you ultimately want to use.在这一点上,我还想补充一点,没有任何出口的 function 可能不是您最终想要使用的。 So I commented out the recursive call in option 5.
所以我注释掉了选项 5 中的递归调用。
My suggestion is to use a simple class:我的建议是使用简单的 class:
class Menu:
def __init__(self):
self.text_in_use = ''
self.encrypt_key = ''
self.build_menu()
def build_menu(self):
choice = int(input(
"""
1: Input text to work with
2: Print the current text
3: Encrypt the current text
4: Decrypt the current text
5: Exit
Enter Choice:
"""
))
if choice == 1:
self.text_in_use = str(input("Enter Text: ")).upper()
print("Text to use was set to:", self.text_in_use)
self.build_menu()
elif choice == 2:
print(self.text_in_use)
self.build_menu()
elif choice == 3:
print("3")
self.build_menu()
elif choice == 4:
#decrypt()
print("4")
self.build_menu()
elif choice == 5:
#exit()
print("5")
# self.build_menu() do not call this again so it actually exits.
Menu()
You should mark text_in_use variable as global.您应该将 text_in_use 变量标记为全局变量。 You reference it in function from outer scope
您从外部 scope 在 function 中引用它
def menu():
global text_in_use
choice = int(input("your_text"))
#rest of code
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