I'm trying to create a file file a unique file name, every time my script is ran, it's only intended to be weekly or monthly. so I chose to use the date for the file name.
f = open('%s.csv', 'wb') %name
is where I'm getting this error.
Traceback (most recent call last):
File "C:\Users\User\workspace\new3\stjohnsinvoices\BabblevoiceInvoiceswpath.py", line 143, in <module>
f = open('%s.csv', 'ab') %name
TypeError: unsupported operand type(s) for %: 'file' and 'str'
it works if I use a static filename, is there an issue with the open function, that means you can't pass a string like this?
name is a string and has values such as :
31/1/2013BVI
Many thanks for any help
This question is related to
python
file
python-2.7
f = open('{}.csv'.format(), 'wb')
you can do something like
filename = "%s.csv" % name
f = open(filename , 'wb')
or f = open('%s.csv' % name, 'wb')
Very similar to peixe.
You don't have to mention the number if the variables you add as parameters are in order of appearance
f = open('{}.csv'.format(name), 'wb')
Another option - the f-string formatting (ref):
f = open(f"{name}.csv", 'wb')
Even better are f-strings in python 3!
f = open(f'{name}.csv', 'wb')
And with the new string formatting method...
f = open('{0}.csv'.format(name), 'wb')
Source: Stackoverflow.com