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.
--
https://mail.python.org/mailman/listinfo/python-list
