open a single file with a wildcard

I know the format of a file that will be dropped into a folder in windows.

The contains the filename and then the date with a timestamp such that the file looks like this:

temp_path = 'H:\\Temp\\file_name_' + yyyymmddhhmmss + '.txt'

Given that i know the date yyyymmdd but i do not know the time hhmmss, i replace the time part with a wild card using the below code.

import datetime as dt

# todays date in yyyymmdd format
today = dt.datetime.today()
today_yyyymmdd = today.strftime('%Y%m%d')

# now the file
temp_path = 'H:\\Temp\\test_file_' + today_yyyymmdd + '*.txt'

print(temp_path)

with open(temp_path, 'r') as f:
    data = f.read()

print(data)

The code works if i remove the wildcard * from in front of the .txt, but fails with it in place.

However, how can open the file with a wildcard ?

>Solution :

You need to use glob module

import glob
temp_path = glob.glob('file_' + today_yyyymmdd + '*.txt')[0]

https://docs.python.org/fr/3.6/library/glob.html