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

MRAB python at mrabarnett.plus.com
Tue Oct 29 12:10:47 EDT 2024


On 2024-10-29 13:56, Loris Bennett via Python-list 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.
> 
> Cheers,
> 
> Loris
> 
'config.read' expects a path or paths. If you give it a file handle, it 
treats it as an iterable. (It might be reading the line as paths of 
files, but I haven't tested it).

If you want to read from an open file, use 'config.read_file' instead.



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