Oz-in-DFW wrote:
> I'm using Python 3.5.2 (v3.5.2:4def2a2901a5, Jun 25 2016, 22:01:18) [MSC
> v.1900 32 bit (Intel)] on Windows 7
>
> I'm trying to write some file processing that looks at file size,
> extensions, and several other things and I'm having trouble getting a
> reliably usable path to files.
>
> The problem *seems* to be doubled backslashes in the path, but I've read
> elsewhere that this is just an artifact of the way the interpreter
> displays the strings.
>
> I'm getting an error message on an os.path.getsize call;
>
> Path: -
> "C:\Users\Rich\Desktop\2B_Proc\2307e60da6451986dd8d23635b845386.jpg" -
> Traceback (most recent call last):
> File "C:\Users\Rich\workspace\PyTest\test.py", line 19, in <module>
> if os.path.getsize(path)>10000:
> File "C:\Python32\lib\genericpath.py", line 49, in getsize
> return os.stat(filename).st_size
> WindowsError: [Error 123] The filename, directory name, or volume
> label syntax is incorrect:
> '"C:
\\Users\\Rich\\Desktop\\2B_Proc\\2307e60da6451986dd8d23635b845386.jpg"'
>
> From (snippet)
>
> path = '"'+dirpath+name+'"'
> path = os.path.normpath(path)
> print("Path: -",path,"-")
> if os.path.getsize(path)>10000:
> print("Path: ",path," Size:
> ",os.path.getsize(dirpath+name))
>
> but if I manually use a console window and cut and paste the path I
> print, it works;
>
> C:\>dir
> "C:\Users\Rich\Desktop\2B_Proc\2307e60da6451986dd8d23635b845386.jpg"
> Volume in drive C is Windows7_OS
>
> Directory of C:\Users\Rich\Desktop\2B_Proc
>
> 10/03/2016 08:35 AM 59,200
> 2307e60da6451986dd8d23635b845386.jpg
> 1 File(s) 59,200 bytes
> 0 Dir(s) 115,857,260,544 bytes free
>
> So the file is there and the path is correct. I'm adding quotes to the
> path to deal with directories and filenames that have spaces in them.
> If I drop the quotes, everything works as I expect *until* I encounter
> the first file/path with spaces.
You have to omit the extra quotes. That the non-working path has spaces in
it is probably not the cause of the problem. If a string *literal* is used,
e. g. "C:\Path\test file.txt" there may be combinations of the backslash and
the character that follows that are interpreted specially -- in the example
\P is just \ followed by a P whereas \t is a TAB (chr(9)):
>>> print("C:\Path\test file.txt")
C:\Path est file.txt
To fix the problem either use forward slashes (which are understood by
Windows, too) or or duplicate all backslashes,
>>> print("C:\\Path\\test file.txt")
C:\Path\test file.txt
or try the r (for "raw string") prefix:
>>> print(r"C:\Path\test file.txt")
C:\Path\test file.txt
This applies only to string literals in Python source code; if you read a
filename from a file or get it as user input there is no need to process the
string:
>>> input("Enter a file name: ")
Enter a file name: C:\Path\test file.txt
'C:\\Path\\test file.txt'
> I'll happily RTFM, but I need some hints as to which FM to R
>
https://docs.python.org/3.5/tutorial/introduction.html#strings
https://docs.python.org/3.5/reference/lexical_analysis.html#string-and-bytes-literals
PS: Unrelated, but a portable way to combine paths is os.path.join() which
will also insert the platform specific directory separator if necessary:
>>> print(os.path.join("foo", "bar\\", "baz"))
foo\bar\baz
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