python - OSError [Errno 22] 在 Python 中使用 open() 时参数无效

Question

def choose_option(self):
        if self.option_picker.currentRow() == 0:
            description = open(":/description_files/program_description.txt","r")
            self.information_shower.setText(description.read())
        elif self.option_picker.currentRow() == 1:
            requirements = open(":/description_files/requirements_for_client_data.txt", "r")
            self.information_shower.setText(requirements.read())
        elif self.option_picker.currentRow() == 2:
            menus = open(":/description_files/menus.txt", "r")
            self.information_shower.setText(menus.read())

我正在使用资源文件，当我将它用作 open 函数的参数时出了点问题，但是当我将它用于加载图片和图标时一切正常。

Answer 1

这不是一个有效的文件路径。您必须使用完整路径

open(r"C:\description_files\program_description.txt","r")

或者相对路径

open("program_description.txt","r")