繁体   English   中英

在if语句中调用函数

[英]Calling function inside if statement

我试图在 if 语句中调用函数,但它不起作用。 这是我使用 Python 的第一次尝试。 我做错了什么?

#!/usr/bin/python


menu = raw_input ("Hello, please choose form following options (1,2,3) and press enter:\n"
    "Option 1\n"
    "Option 2\n"
    "Option 3\n")

if menu == str("1"):
    savinginfile = raw_input ("Please, state your name: ")
    option1()
elif menu == str("2"):
    print ("Option 2")
elif menu == str("3"):
    print ("Option 3")

def option1():
    test = open ("test.txt", "rw")
    test.write(savinginfile)
    print ("Option 1 used")
    test.close()

建议您将savinginfile作为参数传递:

def option1(savinginfile):
    test = open ("test.txt", "rw")
    test.write(savinginfile)
    print ("Option 1 used")
    test.close()

您需要在调用之前定义option1 Python 从上到下解释。

在尝试调用它之前,您需要定义您的函数。 只需将def option1(): #and all that code below it放在 if 语句上方即可。

抛出太多全局变量也是不好的做法。 您不应该savinginfile现在这样使用savinginfile —— 而是将它作为参数传递给函数,并让函数在它自己的范围内运行。 您需要将要使用的文件名传递给函数,然后才能使用savinginfile 试试吧:

def option1(whattosaveinfile):
  test = open("test.txt","a+") #probably better to use a with statement -- I'll comment below.
  test.write(whattosaveinfile) #note that you use the parameter name, not the var you pass to it
  print("Option 1 used")
  test.close()

#that with statement works better for file-like objects because it automatically
#catches and handles any errors that occur, leaving you with a closed object.
#it's also a little prettier :) Use it like this:
#
# with open("test.txt","a+") as f:
#   f.write(whattosaveinfile)
# print("Option 1 used")
#
#note that you didn't have to call f.close(), because the with block does that for you
#if you'd like to know more, look up the docs for contextlib

if menu == "1": #no reason to turn this to a string -- you've already defined it by such by enclosing it in quotes
  savinginfile = raw_input("Please state your name: ")
  option1(savinginfile) #putting the var in the parens will pass it to the function as a parameter.

elif menu == "2": #etc
#etc
#etc

暂无
暂无

声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM