'displaying data of table from database using python
I am newbie to python ,trying with simple programs given below is a program which i tried to get the data from table and displaying it. Installed Python3.4 and mysql.connector 2.0.4, runningin localhost as http://localhost:804/cgi-bin/ex7.py
It's connecting to database but not fetching the data from table
#!"C:\python34\python.exe"
import sys
import mysql.connector
print("Content-Type: text/html;charset=utf-8")
print()
conn = mysql.connector.connect(host='localhost',port='8051',
database='example',
user='root',
password='tiger')
cursor = conn.cursor()
if conn.is_connected():
print('Connected to MySQL database')
cursor.execute(""" SELECT * FROM album """)
for row in cursor:
print (row[1])
It's giving output as : Connected to MySQL database
not printing data from table Please suggest where went wrong
Solution 1:[1]
you missed this part i think
conn = mysql.connector.MySQLConnection(host='localhost',port='8051',
database='example',
user='root',
password='tiger')
cursor = conn.cursor()
if conn.is_connected():
print('Connected to MySQL database')
cursor.execute(""" SELECT * FROM album """)
# fetch all of the rows from the query
data = cursor.fetchall ()
# print the rows
for row in data :
print row[1]
Solution 2:[2]
import mysql.connector
from mysql.connector import Error
try:
connection = mysql.connector.connect(host='localhost', database='example', user='root', password='tiger')
sql_select_Query = "SELECT * FROM album"
cursor = connection.cursor()
cursor.execute(sql_select_Query)
records = cursor.fetchall()
print("Total number of rows in album is: ", cursor.rowcount)
print("\nPrinting each album record")
for row in records:
print("Id = ", row[0], )
print("Name = ", row[1], "\n")
# print("Price = ", row[2])
# print("Purchase date = ", row[3], "\n")
except Error as e:
print("Error reading data from MySQL table", e)
finally:
if connection.is_connected():
connection.close()
cursor.close()
print("MySQL connection is closed")
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
| Solution | Source |
|---|---|
| Solution 1 | |
| Solution 2 | mahesh chougule |
