Sqlite3 Tutorial Query Python Fixed !exclusive! -

import sqlite3 # Data to use in the fixed query user_id = 42 try: # 1. Establish connection (use context manager for automatic commit/rollback) with sqlite3.connect('my_database.db') as conn: cursor = conn.cursor() # 2. Define query with a '?' placeholder for safe execution sql_query = "SELECT * FROM users WHERE id = ?" # 3. Execute with parameters passed as a TUPLE cursor.execute(sql_query, (user_id,)) # 4. Fetch the result result = cursor.fetchone() if result: print(f"User Found: result") else: print("No user found with that ID.") except sqlite3.Error as e: print(f"Database error: e") Use code with caution. Copied to clipboard Essential Steps for Fixed Queries A Python sqlite3 context manager gotcha - Robin's Blog

cursor.execute('SELECT * FROM users WHERE name = ?', ('John Doe',)) row = cursor.fetchone() sqlite3 tutorial query python fixed

or use a with block to prevent locking.

cursor.execute('SELECT * FROM inventory WHERE quantity > 0') rows = cursor.fetchall() for row in rows: print(row) import sqlite3 # Data to use in the