如何通過傳遞包含文件所在目錄的變量名來打開文件 python？

Question

我正在處理 function，我需要通過傳遞包含文件所在目錄而不是文件名的變量名來打開文件，例如“hello.txt”。 下面是我為此設計的虛擬代碼。 簡而言之，我需要能夠通過傳遞目錄名稱來打開位於該目錄中的文件，如您在更新中看到的那樣。

def folder_things():
path = 'C:\\Users\\blazh\\Documents\\Vladyslav\\City-Project\\Python\\'
folder_code = "518Z%"
updated = path + folder_code
cwd = os.listdir(updated)
print("You have:", cwd)

st = ""
for x in cwd:
    st += x

print(st)

# BECOMES A STRING
str(st)
print(type(st))
print(st)
final = ("'"+st+"'")
f = open(st, "r")
print(f)

Answer 1

我希望這回答了你的問題。 不確定為什么您的代碼如此重復。

def open_file(path:str) -> None:
    try:
        with open(path, "r") as file:
            #do stuff with file
            for line in file:
                print(line)
    except FileNotFoundError:
        print(f"File was not found in path: {path}.")

if __name__ == "__main__":
    path = "C:\\Users\\dejon\\Desktop\\Python Training\\demo.txt"
    open_file(path)