Re: Using 'with open(...) as ...' together with configparser.ConfigParser.read

Jon Ribbens <jon+usenet@unequivocal.eu> writes:

On 2024-10-29, Loris Bennett <loris.bennett@fu-berlin.de> wrote:

Hi,
>
With Python 3.9.18, if I do
>
    try:
        with open(args.config_file, 'r') as config_file:
            config = configparser.ConfigParser()
            config.read(config_file)
            print(config.sections())
>
i.e try to read the configuration with the variable defined via 'with
... as', I get
>
   []
>
whereas if I use the file name directly
>
    try:
        with open(args.config_file, 'r') as config_file:
            config = configparser.ConfigParser()
            config.read(args.config_file)
            print(config.sections())
I get
>
  ['loggers', 'handlers', 'formatters', 'logger_root', 'handler_fileHandler', 'handler_consoleHandler', 'formatter_defaultFormatter']
>
which is what I expect.
>
If I print type of 'config_file' I get
>
 <class '_io.TextIOWrapper'>
>
whereas 'args.config_file' is just
>
 <class 'str'>
>
Should I be able to use the '_io.TextIOWrapper' object variable here?  If so how?
>
Here
>
  https://docs.python.org/3.9/library/configparser.html
>
there are examples which use the 'with open ... as' variable for writing
a configuration file, but not for reading one.

>
As per the docs you link to, the read() method only takes filename(s)
as arguments, if you have an already-open file you want to read then
you should use the read_file() method instead.

As you and others have pointed out, this is indeed covered in the docs,
so mea culpa.

However, whereas I can see why you might want to read the config from a
dict or a string, what would be a use case in which I would want to
read from an open file rather than just reading from a file(name)?

Cheers,

Loris

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