Python: Alter a given SQLite table

Last update on December 21 2024 07:46:06 (UTC/GMT +8 hours)

Write a Python program to alter a given SQLite table.

Sample Solution:

Python Code :

import sqlite3
from sqlite3 import Error
def sql_connection():
   try:
     conn = sqlite3.connect('mydatabase.db')
     return conn
   except Error:
     print(Error)
def sql_table(conn):
   cursorObj = conn.cursor()
   cursorObj.execute("CREATE TABLE agent_master(agent_code char(6),agent_name char(40),working_area char(35),commission decimal(10,2),phone_no char(15) NULL);")
   print("\nagent_master file has created.")
   print("\nTable structure of agent_master:")
   a= conn.execute("PRAGMA table_info('agent_master')")
   for i in a:
     print(i)
   # adding a new column in the agent_master table
   cursorObj.execute("""
   ALTER TABLE agent_master
   ADD COLUMN FLAG BOOLEAN;
   """)
   print("\nagent_master file altered.")
   a= conn.execute("PRAGMA table_info('agent_master')")
   print("\nTable structure of agent_master:")
   for i in a:
     print(i)
   conn.commit()
sqllite_conn = sql_connection()
sql_table(sqllite_conn)
if (sqllite_conn):
 sqllite_conn.close()
 print("\nThe SQLite connection is closed.")

Sample Output:

agent_master file has created.

Table structure of agent_master:
(0, 'agent_code', 'char(6)', 0, None, 0)
(1, 'agent_name', 'char(40)', 0, None, 0)
(2, 'working_area', 'char(35)', 0, None, 0)
(3, 'commission', 'decimal(10,2)', 0, None, 0)
(4, 'phone_no', 'char(15)', 0, None, 0)

agent_master file altered.

Table structure of agent_master:
(0, 'agent_code', 'char(6)', 0, None, 0)
(1, 'agent_name', 'char(40)', 0, None, 0)
(2, 'working_area', 'char(35)', 0, None, 0)
(3, 'commission', 'decimal(10,2)', 0, None, 0)
(4, 'phone_no', 'char(15)', 0, None, 0)
(5, 'FLAG', 'BOOLEAN', 0, None, 0)

The SQLite connection is closed.

Python Code Editor:

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