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).
---------------------
>>> import sqlite3
>>> conn=sqlite3.connect("C:/Users/Owner/Desktop/Blog2017/MSSS2017/SQLite3_DBS/UsersOwnerApr25.sqlite")
--------------------
After connecting create a cursor and the subsequently use the Execute() method of the cursor.
>>> c=conn.cursor()
>>> conn.commit()
-----------------------------------------------
For example, for inserting data into 'friends' table use the following statement. I already had a table with the name 'friends'
----------------------------------
>>> 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.
--------------------------------------
>>> y= c.execute('Select * from friends')
>>> print (y)
-----------------------------------------
Now I modify the print() statement and I succeed in printing the first row using the fetchone()method.
---------------------------------------
>>> print(c.fetchone())
('John', 'Chiavetta', 35)
-----------------------------
I print all the rows using the fetchall().
-----------------------------
>>> for row in c.execute('Select * from friends'):
print(row)
('John', 'Chiavetta', 35)
('Jay', 'Kris', 45)
>>>
-------
I have not figured out why I get this:
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).
---------------------
>>> import sqlite3
>>> conn=sqlite3.connect("C:/Users/Owner/Desktop/Blog2017/MSSS2017/SQLite3_DBS/UsersOwnerApr25.sqlite")
--------------------
After connecting create a cursor and the subsequently use the Execute() method of the cursor.
>>> c=conn.cursor()
>>> conn.commit()
-----------------------------------------------
For example, for inserting data into 'friends' table use the following statement. I already had a table with the name 'friends'
----------------------------------
>>> 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.
--------------------------------------
>>> y= c.execute('Select * from friends')
>>> print (y)
-----------------------------------------
Now I modify the print() statement and I succeed in printing the first row using the fetchone()method.
---------------------------------------
>>> print(c.fetchone())
('John', 'Chiavetta', 35)
-----------------------------
I print all the rows using the fetchall().
-----------------------------
>>> for row in c.execute('Select * from friends'):
print(row)
('John', 'Chiavetta', 35)
('Jay', 'Kris', 45)
>>>
-------
I have not figured out why I get this:
The IDLE interface is very useful as mentioned here. You get a lot of help.
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