Reading text file with wierd file extension?

[Lionel]

On Feb 2, 1:07 pm, "Diez B. Roggisch" <de... at nospam.web.de> wrote:

This is written very slowly, so you can read it better:

Please post without sarcasm.


This is the output from my Python shell:

>>> DatafilePath = "C:\\C8Example1.slc"
>>> ResourcefilePath = DatafilePath + ".rsc"
>>> DatafileFH = open(DatafilePath)
>>> ResourceFh = open(ResourcefilePath)
>>> DatafilePath
'C:\\C8Example1.slc'
>>> ResourcefilePath
'C:\\C8Example1.slc.rsc'

It seems to run without trouble. However, here is the offending code
in my class (followed by console output):

class C8DataType:

    def __init__(self, DataFilepath):

        try:
            DataFH = open(DataFilepath, "rb")

        except IOError, message:
            # Error opening file.
            print(message)
            return None

        ResourceFilepath = DataFilepath + ".src"

        print(DataFilepath)
        print(ResourceFilepath)

        # Try to open resource file, catch exception:
        try:
            ResourceFH = open(ResourceFilepath)

        except IOError, message:
            # Error opening file.
            print(message)
            print("Error opening " + ResourceFilepath)
            DataFH.close()
            return None

Console output when invoking as "someObject = C8DataType("C:\
\C8Example1.slc")" :

C:\C8Example1.slc
C:\C8Example1.slc.src
[Errno 2] No such file or directory: 'C:\\C8Example1.slc.src'
Error opening C:\C8Example1.slc.src


Thank you


> Lionel schrieb:
>
>
>
>
>
> > On Feb 2, 12:10 pm, Mike Driscoll <kyoso... at gmail.com> wrote:
> >> On Feb 2, 1:20 pm, Lionel <lionel.ke... at gmail.com> wrote:
>
> >>> On Feb 2, 10:41 am, Mike Driscoll <kyoso... at gmail.com> wrote:
> >>>> On Feb 2, 12:36 pm, Lionel <lionel.ke... at gmail.com> wrote:
> >>>>> Hi Folks, Python newbie here.
> >>>>> I'm trying to open (for reading) a text file with the following
> >>>>> filenaming convension:
> >>>>> "MyTextFile.slc.rsc"
> >>>>> My code is as follows:
> >>>>> Filepath = "C:\\MyTextFile.slc.rsc"
> >>>>> FileH = open(Filepath)
> >>>>> The above throws an IOError exception. On a hunch I changed the
> >>>>> filename (only the filename) and tried again:
> >>>>> Filepath = "C:\\MyTextFile.txt"
> >>>>> FileH = open(Filepath)
> >>>>> The above works well. I am able to open the file and read it's
> >>>>> contents. I assume to read a file in text file "mode" the parameter is
> >>>>> scanned for a ".txt" extension, otherwise the Python runtime doesn't
> >>>>> know what version of "open(...)" to invoke. How do I pass the original
> >>>>> filename (MyTextFile.slc.rsc) and get Python to open it as a text
> >>>>> file? Thanks in advance everyone!
> >>>> The extension shouldn't matter. I tried creating a file with the same
> >>>> extension as yours and Python 2.5.2 opened it and read it no problem.
> >>>> I tried it in IDLE and with Wing on Windows XP. What are you using?
> >>>> What's the complete traceback?
> >>>> Mike- Hide quoted text -
> >>>> - Show quoted text -
> >>> Hi Mike,
> >>> maybe it's not a "true" text file? Opening it in Microsoft Notepad
> >>> gives an unformatted view of the file (text with no line wrapping,
> >>> just the end-of-line square box character followed by more text, end-
> >>> of-line character, etc). Wordpad opens it properly i.e. respects the
> >>> end-of-line wrapping. I'm unsure of how these files are being
> >>> generated, I was just given them and told they wanted to be able to
> >>> read them.
> >>> How do I collect the traceback to post it?
> >> The traceback should look something like this fake one:
>
> >> Traceback (most recent call last):
> >>   File "<pyshell#3>", line 1, in <module>
> >>     raise IOError
> >> IOError
>
> >> Just copy and paste it in your next message. The other guys are
> >> probably right in that it is a line ending issue, but as they and I
> >> have said, Python shouldn't care (and doesn't on my machine).
>
> >> Mike- Hide quoted text -
>
> >> - Show quoted text -
>
> > Okay, I think I see what's going on. My class takes a single parameter
> > when it is instantiated...the file path of the data file the user
> > wants to open. This is of the form "someFile.slc". In the same
> > directory of "someFile.slc" is a resource file that (like a file
> > header) contains a host of parameters associated with the data file.
> > The resource file is of the form "someFile.slc.rsc". So, when the user
> > creates an instance of my class which, it invokes the __init__ method
> > where I add the ".rsc" extension to the original filename/path
> > parameter that was passed to the class "constructor". For example:
>
> > Note: try-catch blocks ommitted.
>
> > class MyUtilityClass:
> >     def __init__(self, DataFilepath):
> >         Resourcepath  = DataFilepath + ".rsc"
> >         DataFileH     = open(DataFilepath)
> >         ResourceFileH = open(Resourcepath)
>
> > Invoking this from the Python shell explicitly is no problem i.e.
> > "TestH = open("C:\\TestResourceFile.slc.rsc") works. BUT...something
> > is lost when I append the ".rsc" extension to the DataFilePath
> > parameter as above. When the __init__ method is invoked, Python will
> > open the data file but generates the exception with the "open
> > (Resourcepath)" instruction. I think it is somehow related to the
> > backslashes but I'm not entirely sure of this. Any ideas?
>
> This is written very slowly, so you can read it better:
>
> Please post the traceback.
>
> Diez- Hide quoted text -
>
> - Show quoted text -