Python3 _io.TextIOWrapper error when opening a file with notepad - python-3.x

Question

I am stuck from a couple of days on an issue in my micro Address Book project. I have a function that writes all records from a SQLite3 Db on file in order to open in via OS module, but as soon as I try to open the file, Python gives me the following error:
Error while opening tempfile. Error:startfile: filepath should be string, bytes or os.PathLike, not _io.TextIOWrapper
This is the code that I have to write records on file and to open it:
source_file_name = open("C:\\workdir\\temp.txt","w")
#Fetching results from database and storing in result variable
self.cur.execute("SELECT id, first_name, last_name, address1, address2, zipcode, city, country, nation, phone1, phone2, email FROM contacts")
result = self.cur.fetchall()
#Writing results into tempfile
source_file_name.write("Stampa Elenco Contatti\n")
for element in result:
source_file_name.write(str(element[0]) + "|" + str(element[1]) + "|" + str(element[2]) + "|" + str(element[3]) + "|" + str(element[4]) + "|" + str(element[5]) + "|" + \
str(element[6]) + "|" + str(element[7]) + "|" + str(element[8]) + "|" + str(element[9]) + "|" + str(element[10]) + "|" + str(element[11]) + "\n")
#TODO: Before exiting printing function you MUST:
# 1. filename.close()
# 2. exit to main() function
source_file_name.close()
try:
os.startfile(source_file_name,"open")
except Exception as generic_error:
print("Error while opening tempfile. Error:" + str(generic_error))
finally:
main()
Frankly I don't understand what this error means, in my previous code snippets I've always handled text files without issues, but I realize this time it's different because I am picking my stream from a database. Any ideas how to fix it?
Thanks in advance, and sorry for my english...

Answer 1

Your problem ultimately stems from poor variable naming. Here
source_file_name = open("C:\\workdir\\temp.txt","w")
source_file_name does not contain the source file name. It contains the source file itself (i.e., a file handle). You can't give that to os.startfile(), which expects a file path (as the error also says).
What you meant to do is
source_file_name = "C:\\workdir\\temp.txt"
source_file = open(source_file_name,"w")
But in fact, it's much better to use a with block in Python, as this will handle closing the file for you.
It's also better to use a CSV writer instead of creating the CSV manually, and it's highly advisable to set the file encoding explicitly.
import csv
# ...
source_file_name = "C:\\workdir\\temp.txt"
with open(source_file_name, "w", encoding="utf8", newline="") as source_file:
writer = csv.writer(source_file, delimiter='|')
source_file.write("Stampa Elenco Contatti\n")
for record in self.cur.fetchall():
writer.writerow(record)
# alternative to the above for loop on one line
# writer.writerows(self.cur.fetchall())