Question

我尝试在if语句中调用函数但它不起作用。这是我第一次尝试使用Python。我究竟做错了什么？

#!/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()

Answer 1

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()

您需要在致电前定义option1。 Python从上到下解释。

Answer 2

您需要在尝试调用之前定义函数。只需将def option1(): #and all that code below it放在if语句之上。

抛出太多的全局变量也是不好的做法。您不应该按照您的方式使用savinginfile - 而是将其作为参数传递给函数，并让函数在其自己的范围内运行。在能够使用savinginfile之前，您需要将函数传递给要使用的文件的名称。请尝试改为：

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

在if语句中调用函数

2 个答案: