In a database table, if a specific row has been previously modified or updated, I want that row to not be changed or cannot be modified for let's say I set in 5 seconds even if the program is trying to modify that row, then after that the row can be modified or update again. Is that possible?
I didn't try anything yet because I dont have an idea if it is possible or not
Yes this is possible. To implement this kind of behaviour you can add a new column to your table let's call it last_modified
. Which get's a default value of the current timestamp
CREATE TABLE your_table (
...
last_modified TIMESTAMP DEFAULT CURRENT_TIMESTAMP
);
or
ALTER TABLE your_table ADD COLUMN last_modified TIMESTAMP DEFAULT CURRENT_TIMESTAMP;
Now each time you want to update your table you check if the affected row has been updated in the last 5 seconds if not update the row if yes throw an error.
import sqlite3
import time
# Connect to the SQLite database
conn = sqlite3.connect('your_database.db')
cursor = conn.cursor()
# Create the table if it doesn't exist
cursor.execute('''CREATE TABLE IF NOT EXISTS your_table
(id INTEGER PRIMARY KEY AUTOINCREMENT,
column1 TEXT,
column2 TEXT,
last_modified TIMESTAMP DEFAULT CURRENT_TIMESTAMP)''')
# Function to check if a row can be modified
def can_modify_row(row_id, interval):
cursor.execute("SELECT ROUND((JULIANDAY('now') - JULIANDAY(last_modified)) * 86400) FROM your_table WHERE id = ?", (row_id,))
elapsed_time = cursor.fetchone()[0]
return elapsed_time > interval
# Update a row, setting the last_modified timestamp to the current time
def update_row(row_id, column1_value, column2_value):
cursor.execute("UPDATE your_table SET column1 = ?, column2 = ?, last_modified = CURRENT_TIMESTAMP WHERE id = ?", (column1_value, column2_value, row_id))
conn.commit()
# Example usage
row_id = 1
interval = 5 # seconds
if can_modify_row(row_id, interval):
print("Modifying the row...")
update_row(row_id, "new_value1", "new_value2")
else:
print("Cannot modify the row yet. Please try again later.")
# Close the database connection
conn.close()
The code is currently untested, but I hope you get the idea