If statement not returning the desired result - mysql

Question

I'm new to Python and I believe the issue with my code is being caused by the fact that I'm a newbie and there's some theory or something that I must not be familiar with yet.
Yes, this question was asked before but, is different from mine. Believe me I tried everything that I thought that needs to be done.
Everything worked until I added everything in "if five in silos" statement.
After I enter the values for the 6 input functions, the program just finishes with exit code 0. Nothing else happens. The for loop is not initiated.
I want for the code to accept either 103 or 106 when prompting to enter something for the "five" variable.
I'm using PyCharm and Python 3.7.
import mysql.connector
try:
db = mysql.connector.connect(
host="",
user="",
passwd="",
database=""
)
one = int(input("Number of requested telephone numbers: "))
two = input("Enter the prefix (4 characters) with a leading 0: ")[:4]
three = int(input("Enter the ccid: "))
four = int(input("Enter the cid: "))
six = input("Enter case number: ")
five = int(input("Enter silo (103, 106 only): "))
cursor = db.cursor()
cursor.execute(f"SELECT * FROM n1 WHERE ddi LIKE '{two}%' AND silo = 1 AND ccid = 0 LIMIT {one}")
cursor.fetchall()
silos = (103, 106)
if five in silos:
if cursor.rowcount > 0:
for row in cursor:
seven = input(f"{row[1]} has been found on our system. Do you want to continue? Type either Y or N.")
if seven == "Y":
cursor.execute(f"INSERT INTO n{five} (ddi, silo, ccid, campaign, assigned, allocated, "
f"internal_notes, client_notes, agentid, carrier, alias) VALUES "
f"('{row[1]}', 1, {three}, {four}, NOW(), NOW(), 'This is a test.', '', 0, "
f"'{row[13]}', '') "
f"ON DUPLICATE KEY UPDATE "
f"silo = VALUES (silo), "
f"ccid = VALUES (ccid), "
f"campaign = VALUES (campaign);")
cursor.execute(f"UPDATE n1 SET silo = {five}, internal_notes = '{six}', allocated = NOW() WHERE "
f"ddi = '{row[1]}'")
else:
print("The operation has been canceled.")
db.commit()
else:
print(f"No results for prefix {two}.")
else:
print("Enter either silo 103 or 106.")
cursor.close()
db.close()
except (ValueError, NameError):
print("Please, enter an integer for all questions, except case number.")

Answer 1

Because it must be:
for row in cursor.fetchall():
// do something
In your code cursor returns a Python Class defined by db.cursor() but you need to call the fetchall() function to read the rows contained in it.
You're actually calling cursor.fetchall() without doing nothing with it, you can assign the call to a variable and than do this:
result = cursor.fetchall()
for row in result:
//do something

Answer 2

I found the problem: I had to store cursor.fetchall() into a variable.
After I put: eight = cursor.fetchall() before the "silos" tuple, everything worked perfectly.