Question

What will the following code do?
import mysql.connector
db = mysql.connector.connect(...)
cursor = db.cursor()
sali = "update category set name = "% WHERE ID=%"% ('CSS,2)
cursor.execute(sqli)
db.commit()
print ("Rows affected", cursor.rowcount)
db.close()​

Answer 1

1. It will open a database connection to a database.
2. Creates a statement to update category name to CSS where is is 2
3. It executes this statement & commits it permanently.
4. It prints the number of affected rows which in this case is 1
5. Then closes the correction to the database.

$\small\mathsf\color{lightgreen}useful?\: \color{white}\longrightarrow\: \color{orange}brainliest!$

Answer 2

Answer:

really do you love me mate