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

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.