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，当我在open函数中将它用作参数时，出现了一些问题，但是当我使用它加载图片和图标时，一切都很好。

Answer 1

您应该在路径的最后一个“/”中再添加一个“/”，即： open('C:\Python34\book.csv')到open('C:\Python34\\book.csv')。例如：

import csv
with open('C:\Python34\\book.csv', newline='') as csvfile:
    spamreader = csv.reader(csvfile, delimiter='', quotechar='|')
    for row in spamreader:
        print(row)