Python error [WinError 123] (File name, directory name or volume label syntax incorrect) when using __file__
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
add a comment |
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
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
add a comment |
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
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
python
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
add a comment |
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
add a comment |
1 Answer
1
active
oldest
votes
Perhaps try this?
currentpath = os.path.dirname(os.path.abspath(__file__))
Thank you so much, you're a life saver
– EMarshall
Nov 11 '18 at 17:34
add a comment |
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1 Answer
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Perhaps try this?
currentpath = os.path.dirname(os.path.abspath(__file__))
Thank you so much, you're a life saver
– EMarshall
Nov 11 '18 at 17:34
add a comment |
Perhaps try this?
currentpath = os.path.dirname(os.path.abspath(__file__))
Thank you so much, you're a life saver
– EMarshall
Nov 11 '18 at 17:34
add a comment |
Perhaps try this?
currentpath = os.path.dirname(os.path.abspath(__file__))
Perhaps try this?
currentpath = os.path.dirname(os.path.abspath(__file__))
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
add a comment |
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
add a comment |
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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