why is this type of variable not compatible with string?

Question

I am trying to ask the user what they want to name a file that is about to be created on my desktop. When I try to add the variable to the string, it gives me this error:

appendFile = open('%s.txt', 'a') % cusername
TypeError: unsupported operand type(s) for %: '_io.TextIOWrapper' and 'str'

Here is my program:

def CNA():
    cusername = input("Create username\n>>")
    filehandler = open("C:/Users/CJ Peine/Desktop/%s.txt", "w") % cusername
    filehandler.close()
    cpassword = input("Create password\n>>")
    appendFile = open('%s.txt', 'a') % cusername
    appendFile.write(cpassword)
    appendFile.close()
    print ("Account Created")

How do I make the variable compatible to the string?

The `% cusername` needs to be with `'%s.txt'` and not with `open()` — ZdaR, Oct 24 '17 at 03:56
`open(...)` supposed to return a file-handler. what's the meaning of `file_handler % cusername` ? — Nir Alfasi, Oct 24 '17 at 03:59
when i do that a text file shows up on my desktop with % cusername instead of the variable i put in. — CJ Peine, Oct 24 '17 at 04:01
@CJPeine: what variable you are referring to? ("C:/Users/CJ Peine/Desktop/%s.txt" % cusername ... should work as well. — Sameer Mahajan, Oct 24 '17 at 04:11

Nae · Accepted Answer · 2017-10-31T19:55:31.420

1

Try doing

cusername = input("Create username\n>>")

filehandler = open("C:/Users/CJ Peine/Desktop/" + cusername + ".txt", "w")

instead. Or you're just trying use modulus operator % on open function.

edited Oct 31 '17 at 19:55

answered Oct 24 '17 at 04:02

Nae

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errantlinguist · Answer 2 · 2017-10-24T04:30:05.680

The % operator for formatting strings should take a string str as a first argument but instead you're passing the object returned from open(...). You can use this expression instead:

open("C:/Users/CJ Peine/Desktop/%s.txt" % cusername, "w")

Alternatively, Python 3 supports using the str.format(str) method ("C:/Users/CJ Peine/Desktop/{}.txt".format(cusername)), which IMHO is much more readable, and Python 3 is much better than Python 2 anyway. Python 2 has been out of active development for ages and is scheduled to no longer be supported; please don't use it unless you absolutely have to.

why is this type of variable not compatible with string?

2 Answers2