在 Python 中使用 open() 时 OSError [Errno 22] 无效参数
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OSError [Errno 22] invalid argument when use open() in Python
提问by eugene
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())
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.
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，一切都很好。
回答by Cory Kramer
That is not a valid file path. You must either use a full path
这不是有效的文件路径。您必须使用完整路径
open(r"C:\description_files\program_description.txt","r")
Or a relative path
或者相对路径
open("program_description.txt","r")
回答by Hiep Tran
you should add one more "/" in the last "/" of path, that is:
open('C:\Python34\book.csv')to open('C:\Python34\\book.csv'). For example:
您应该在路径的最后一个“/”中再添加一个“/”，即：
open('C:\Python34\book.csv')to 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)
回答by Abhijeet
In Windows-Pycharm: If File Location|Path contains any string like \tthen need to escape that with additional \like \\t
在Windows的Pycharm：如果文件位置|路径中包含任何字符串如\t再需要逃避与另外\像\\t
回答by P113305A009D8M
Just replace with "/" for file path :
只需用“/”替换文件路径：
open("description_files/program_description.txt","r")
回答by duhaime
I received the same error when trying to print an absolutely enormous dictionary. When I attempted to print just the keys of the dictionary, all was well!
在尝试打印绝对巨大的字典时，我收到了同样的错误。当我试图只打印字典的键时，一切都很好！
回答by wolfog
I also ran into this fault when I used open(file_path). My reason for this fault was that my file_pathhad a special character like "?"or "<".
我在使用的时候也遇到了这个故障open(file_path)。我这个错误的原因是我file_path有一个特殊的字符，比如"?"or "<"。
回答by Prostak
for folder, subs, files in os.walk(unicode(docs_dir, 'utf-8')):
for filename in files:
if not filename.startswith('.'):
file_path = os.path.join(folder, filename)
回答by Mohsen Hrt
In my case,the problem exists beacause I have not set permission for drive "C:\" and when I change my path to other drive like "F:\" my problem resolved.
就我而言，问题存在是因为我没有为驱动器“C:\”设置权限，当我将路径更改为“F:\”等其他驱动器时，我的问题解决了。
回答by Lio Ak
I had the same problem It happens because files can't contain special characters like ":", "?", ">" and etc. You should replace these files by using replace() function:
我遇到了同样的问题 这是因为文件不能包含特殊字符，如“:”、“?”、“>”等。您应该使用 replace() 函数替换这些文件：
filename = filename.replace("special character to replace", "-")
回答by I Iv
import pandas as pd
df = pd.read_excel ('C:/Users/yourlogin/new folder/file.xlsx')
print (df)
