Python error [WinError 123] (File name, directory name or volume label syntax incorrect) when using __file__










2















I've been writing a program which writes to text files in the directory it's placed in. So, to find the path to the directory it is placed in, I used this statement:



currentpath = os.path.dirname(__file__)


But whenever I call the program outside of Idle it gives the error:



OSError: [WinError 123] The filename, directory name or volume label syntax is incorrect



I have no idea why this is happening, and even less why it happens outside of Idle and not inside.



So please can someone help because I have little hope of solving this on my own.



Oh, and PS. The name of the file is "File sprayer.py" and the directory name is "File sprayer test". I'm not sure if that helps.










share|improve this question






















  • What do you mean by "outside of Idle"?

    – user1394
    Nov 11 '18 at 11:48











  • I mean when I run from the command line or from double-clicking in Windows Explorer rather than using the Run>Run Module feature of Idle.

    – EMarshall
    Nov 11 '18 at 17:31















2















I've been writing a program which writes to text files in the directory it's placed in. So, to find the path to the directory it is placed in, I used this statement:



currentpath = os.path.dirname(__file__)


But whenever I call the program outside of Idle it gives the error:



OSError: [WinError 123] The filename, directory name or volume label syntax is incorrect



I have no idea why this is happening, and even less why it happens outside of Idle and not inside.



So please can someone help because I have little hope of solving this on my own.



Oh, and PS. The name of the file is "File sprayer.py" and the directory name is "File sprayer test". I'm not sure if that helps.










share|improve this question






















  • What do you mean by "outside of Idle"?

    – user1394
    Nov 11 '18 at 11:48











  • I mean when I run from the command line or from double-clicking in Windows Explorer rather than using the Run>Run Module feature of Idle.

    – EMarshall
    Nov 11 '18 at 17:31













2












2








2








I've been writing a program which writes to text files in the directory it's placed in. So, to find the path to the directory it is placed in, I used this statement:



currentpath = os.path.dirname(__file__)


But whenever I call the program outside of Idle it gives the error:



OSError: [WinError 123] The filename, directory name or volume label syntax is incorrect



I have no idea why this is happening, and even less why it happens outside of Idle and not inside.



So please can someone help because I have little hope of solving this on my own.



Oh, and PS. The name of the file is "File sprayer.py" and the directory name is "File sprayer test". I'm not sure if that helps.










share|improve this question














I've been writing a program which writes to text files in the directory it's placed in. So, to find the path to the directory it is placed in, I used this statement:



currentpath = os.path.dirname(__file__)


But whenever I call the program outside of Idle it gives the error:



OSError: [WinError 123] The filename, directory name or volume label syntax is incorrect



I have no idea why this is happening, and even less why it happens outside of Idle and not inside.



So please can someone help because I have little hope of solving this on my own.



Oh, and PS. The name of the file is "File sprayer.py" and the directory name is "File sprayer test". I'm not sure if that helps.







python






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share|improve this question











share|improve this question




share|improve this question










asked Nov 11 '18 at 11:25









EMarshallEMarshall

135




135












  • What do you mean by "outside of Idle"?

    – user1394
    Nov 11 '18 at 11:48











  • I mean when I run from the command line or from double-clicking in Windows Explorer rather than using the Run>Run Module feature of Idle.

    – EMarshall
    Nov 11 '18 at 17:31

















  • What do you mean by "outside of Idle"?

    – user1394
    Nov 11 '18 at 11:48











  • I mean when I run from the command line or from double-clicking in Windows Explorer rather than using the Run>Run Module feature of Idle.

    – EMarshall
    Nov 11 '18 at 17:31
















What do you mean by "outside of Idle"?

– user1394
Nov 11 '18 at 11:48





What do you mean by "outside of Idle"?

– user1394
Nov 11 '18 at 11:48













I mean when I run from the command line or from double-clicking in Windows Explorer rather than using the Run>Run Module feature of Idle.

– EMarshall
Nov 11 '18 at 17:31





I mean when I run from the command line or from double-clicking in Windows Explorer rather than using the Run>Run Module feature of Idle.

– EMarshall
Nov 11 '18 at 17:31












1 Answer
1






active

oldest

votes


















0














Perhaps try this?



currentpath = os.path.dirname(os.path.abspath(__file__))





share|improve this answer























  • Thank you so much, you're a life saver

    – EMarshall
    Nov 11 '18 at 17:34










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1 Answer
1






active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0














Perhaps try this?



currentpath = os.path.dirname(os.path.abspath(__file__))





share|improve this answer























  • Thank you so much, you're a life saver

    – EMarshall
    Nov 11 '18 at 17:34















0














Perhaps try this?



currentpath = os.path.dirname(os.path.abspath(__file__))





share|improve this answer























  • Thank you so much, you're a life saver

    – EMarshall
    Nov 11 '18 at 17:34













0












0








0







Perhaps try this?



currentpath = os.path.dirname(os.path.abspath(__file__))





share|improve this answer













Perhaps try this?



currentpath = os.path.dirname(os.path.abspath(__file__))






share|improve this answer












share|improve this answer



share|improve this answer










answered Nov 11 '18 at 11:50









user1394user1394

3121313




3121313












  • Thank you so much, you're a life saver

    – EMarshall
    Nov 11 '18 at 17:34

















  • Thank you so much, you're a life saver

    – EMarshall
    Nov 11 '18 at 17:34
















Thank you so much, you're a life saver

– EMarshall
Nov 11 '18 at 17:34





Thank you so much, you're a life saver

– EMarshall
Nov 11 '18 at 17:34

















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