Question

我正在尝试编写一个程序，该程序可以将文件的内容存储到用户选择的变量中。例如，用户将选择一个位于当前目录中的文件，然后选择他们想要存储在其中的变量。这是到目前为止我的代码的一部分。

print("What .antonio file do you want to load?")
loadfile = input("")
open(loadfile, "r")

print("Which variable do you want to load it for? - list_1, list_2, list_3")
whichvariable = input("") 

if whichvariable == "list_1":
    list_1 = loadfile.read()
elif whichvariable == "list_2":
    list_2 = loadfile.read()
elif whichvariable == "list_3":
    list_3 = loadfile.read()
else:
    print("ERROR")

当我输入那个loadfile = list1.antonio（它是一个现有文件）和whichvariable = list_1时，它会引发以下错误：

Traceback (most recent call last):
  File "D:\Antonio\Projetos\Python\hello.py", line 29, in <module>
    list_1 = loadfile.read()
AttributeError: 'str' object has no attribute 'read'

我尝试了各种各样的方法，但没有找到解决方案。

Answer 1

您需要将open的结果存储到变量中，并从该变量中执行read方法。

这是您的代码修复程序：

print("What .antonio file do you want to load?")
loadfile = input("")
loadfile = open(loadfile, "r") # you forget to store the result of open into loadfile

print("Which variable do you want to load it for? - list_1, list_2, list_3")
whichvariable = input("") 

if whichvariable == "list_1":
    list_1 = loadfile.read()
elif whichvariable == "list_2":
    list_2 = loadfile.read()
elif whichvariable == "list_3":
    list_3 = loadfile.read()
else:
    print("ERROR")

不要忘记关闭打开的loadfile文件。

更好

print("What .antonio file do you want to load?")
loadfile = input("")
with open(loadfile, "r") as openedfile:

    print("Which variable do you want to load it for? - list_1, list_2, list_3")
    whichvariable = input("") 

    if whichvariable == "list_1":
        list_1 = loadfile.read()
    elif whichvariable == "list_2":
        list_2 = loadfile.read()
    elif whichvariable == "list_3":
        list_3 = loadfile.read()
   else:
       print("ERROR")

如何读取打开的文件？

1 个答案: