在 Python 中使用 open() 时 OSError [Errno 22] 无效参数 [英] OSError [Errno 22] invalid argument when use open() in Python

问题描述

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，当我在 open 函数中使用它作为参数时出现问题，但是当我使用它加载图片和图标时一切正常.

I am using resource files and something is going wrong when i am using it as argument in open function, but when i am using it for loading of pictures and icons everything is fine.

推荐答案

这不是有效的文件路径.您必须使用完整路径

That is not a valid file path. You must either use a full path

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

或者相对路径

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

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