Querying a persisted SQLite using Python 3.7

Python uses the sqlite3 module written by Gerhard Häring. It provides a SQL interface compliant with the DB-API 2.0 specification described by PEP 249.

I launch the Python Shell in my All Programs.

PythonSQLite_00

Python 3.7.0b2 (v3.7.0b2:b0ef5c979b, Feb 28 2018, 01:32:48) [MSC v.1912 32 bit (Intel)] on win32

Type "copyright", "credits" or "license()" for more information.

Connecting to an existing SQLite database. UsersOwnerApr25.sqlite has a table called 'friends' with just one row of data.

 First you need to import sqlite3 and then establish a connection by providing the path to the database (note the direction of slashes).

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>>> import sqlite3

>>> conn=sqlite3.connect("C:/Users/Owner/Desktop/Blog2017/MSSS2017/SQLite3_DBS/UsersOwnerApr25.sqlite")

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After connecting create a cursor and the subsequently use the Execute() method of the cursor.

>>> c=conn.cursor()

>>> conn.commit()

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For example, for inserting data into 'friends' table use the following statement. I already had a table with the name 'friends'

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>>> c.execute("Insert Into friends values('Jay', 'Kris',45)")

--This is something I need to figure out---

I try to print the results but I get the same response after the print() statement.

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>>> y= c.execute('Select * from friends')

>>> print (y)

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Now I modify the print() statement and I succeed in printing the first row using the fetchone()method.

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>>> print(c.fetchone())

('John', 'Chiavetta', 35)

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I print all the rows using the fetchall().

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>>> for row in c.execute('Select * from friends'):

print(row)

('John', 'Chiavetta', 35)

('Jay', 'Kris', 45)

>>>

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I have not figured out why I get this:

The IDLE interface is very useful as mentioned here. You get a lot of help.