Os Error Code 22 Invalid Argument
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Oserror: [errno 22] Invalid Argument Pickle
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Ioerror Errno 22 Invalid Argument Python
Tags Users Badges Ask Question x Dismiss Join the Stack Overflow Community Stack Overflow is a community of 6.2 million programmers, just like you, helping each other. Join them; it errno 22 invalid argument socket only takes a minute: Sign up OSError [Errno 22] invalid argument when use open() in Python up vote 0 down vote favorite 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 oserror errno 22 invalid argument read i am using it as argument in open function, but when i am using it for loading of pictures and icons everything is fine. python share|improve this question edited Aug 30 '14 at 15:45 CoryKramer 52.2k93983 asked Aug 30 '14 at 15:45 eugene 1111 add a comment| 2 Answers 2 active oldest votes up vote 3 down vote 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") share|improve this answer answered Aug 30 '14 at 15:47 CoryKramer 52.2k93983 add a comment| up vote 0 down vote you should add one more "/" in the last "/" of path for example: open('C:\Python34\book.csv') to open('C:\Python34\book.csv') import csv with open('C:\Python34\\book.csv', newline='') as csvfile: spamreader = csv.reader(csvfile, delimiter='', quotechar='|') for row in spamreader: print(row) share|improve this answer edited Jan 5 at 18:45 Tomasz Jakub Rup 5,06471436 answered Jan 5 at 15:58 Hiep Tran 512 1 Welcome to Stack Overflow! Whilst this may theoretically answer the question, it would be preferable to include the essential parts of the answer here, and prov
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Python 3 Oserror Errno 22 Invalid Argument
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Python Open Errno 22 Invalid Argument
ads with us Stack Overflow Questions Jobs Documentation Tags Users Badges Ask Question x Dismiss Join the Stack Overflow Community Stack Overflow is a community of 6.2 http://stackoverflow.com/questions/25584124/oserror-errno-22-invalid-argument-when-use-open-in-python million programmers, just like you, helping each other. Join them; it only takes a minute: Sign up Python Error: OSError: [Errno 22] Invalid argument up vote 2 down vote favorite 2 I am using an automation framework and I am getting random error after many iterations which is as follows. Can someone help me understand http://stackoverflow.com/questions/17392568/python-error-oserror-errno-22-invalid-argument what this could correspond to !! _os.environ['PATH'] = r'C:\DAL;' + _os.environ['PATH'] File "c:\Python26\lib\os.py", line 420, in __setitem__ putenv(key, item) OSError: [Errno 22] Invalid argument Function Call where it fails: function: plugin_xml_file_name = plugin_name else: plugin_xml_file_name = plugin_path + "\\" + plugin_name # _os.environ['PATH'] = r'C:\Intel\DAL;' + _os.environ['PATH'] _os.environ['PATH'] = r'C:\intel\dal;' + _os.environ['PATH'] _os.environ['PATH'] = _lakemore_path + ';' + _os.environ['PATH'] _os.environ['PATH'] = plugin_path + ';' + _os.environ['PATH'] python error-handling share|improve this question edited Jun 30 '13 at 16:56 asked Jun 30 '13 at 16:40 user2510612 1516 1 How long is os.environ['PATH'] by this stage? –Martijn Pieters♦ Jun 30 '13 at 16:44 I am not sure . Is there any way to dump out that information? Or clear if it exceeds some threshold ? –user2510612 Jun 30 '13 at 16:47 What is the rest of the traceback? It is impossible to tell what is extending the path here. –Martijn Pieters♦ Jun 30 '13 at 16:48 It traces b
Post #1 of 6 (2229 views) Permalink which is the right file path http://www.gossamer-threads.com/lists/python/python/1173217 format in python3.4 ? My system is :win7+python3.4 . I want to write a new file "named names.txt" in disk f: >>> ff=open(r"F:\names.txt","w") Traceback (most recent call last): File "