OSError Errno 22 invalid argument when use open in Python

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      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

User · Answer

In my case the error was due to lack of permissions to the folder path  I entered and saved the credentials the issue was solved

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In Windows-Pycharm  If File Location Path contains any string like  t then need to escape that with additional   like   t

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That is not a valid file path  You must either use a full path  open r C  description files program description txt   r     Or a relative path  open  program description txt   r

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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   filename   filename replace  special character to replace    -

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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

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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   import csv with open  C  Python34  book csv   newline     as csvfile      spamreader   csv reader csvfile  delimiter     quotechar          for row in spamreader          print row

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Just replace with     for file path       open  description files program description txt   r

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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

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In my case  I was using an invalid string prefix   Wrong   path   f D  Folder file txt    Right   path   r D  Folder file txt

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import pandas as pd df   pd read excel   C  Users yourlogin new folder file xlsx   print  df

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I got this error because old server instance was running and using log file  hence new instance was not able to write to log file  Post deleting log file this issue got resolved

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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

User · Answer

Add  r  in starting of path   path   r D  Folder file txt    That works for me

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I also ran into this fault when I used open file path   My reason for this fault was that my file path had a special character like     or   lt

[python] OSError [Errno 22] invalid argument when use open() in Python

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