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

Loris Bennett loris.bennett at fu-berlin.de
Wed Oct 30 09:03:55 EDT 2024


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

> On 2024-10-29, Loris Bennett <loris.bennett at 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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