Python request SyntaxError: invalid syntax
Question:
I have written a Python3.7.1 code, and I need to run it in Python3.5 (because I have to run it from Raspberry), an error shows up:
teams = requests.get(f'http://api.football-data.org/v2/competitions/{league}/standings', headers=headers, timeout = 10).json()
SyntaxError: invalid syntax
The error appears on the first comma (…standings’, headers…)
I suppose this is because of python3.5 don’t support the same things that 3.7.
I have tried to install Python 3.7 in raspberry but have no good results
teams = requests.get(f'http://api.football-data.org/v2/competitions/{league}/standings', headers=headers, timeout = 10).json()
Thanks for the help
Answers:
The issue here is that the f string is supported since python 3.6. As you are using python 3.5, you can use an equivalent:
url = 'http://api.football-data.org/v2/competitions/{}/standings'.format(league)
Here you can find some information about string formatting.
I have written a Python3.7.1 code, and I need to run it in Python3.5 (because I have to run it from Raspberry), an error shows up:
teams = requests.get(f'http://api.football-data.org/v2/competitions/{league}/standings', headers=headers, timeout = 10).json()
SyntaxError: invalid syntax
The error appears on the first comma (…standings’, headers…)
I suppose this is because of python3.5 don’t support the same things that 3.7.
I have tried to install Python 3.7 in raspberry but have no good results
teams = requests.get(f'http://api.football-data.org/v2/competitions/{league}/standings', headers=headers, timeout = 10).json()
Thanks for the help
The issue here is that the f string is supported since python 3.6. As you are using python 3.5, you can use an equivalent:
url = 'http://api.football-data.org/v2/competitions/{}/standings'.format(league)
Here you can find some information about string formatting.