Retrieve Multiple images one by one from MySQL database using Python 3

Question

I have one table for inserting images. From that, I just retrieve images using MySQL database and Python 3 code. My code is executed well, but my problem is whenever I retrieve images it is showing me first image which is placed in my table and I don't know how can I retrieve blobs one by one using for loop.

Here is my code:- code.py:

from PIL import Image
import pymysql
from api import mysql

db=pymysql.connect(host="localhost",user="root",passwd="root",db="votelist")
image = Image.open('D:/mine project/face_reg/media/a.jpg')
blob = open('D:/mine project/face_reg/media/a.jpg', 'rb').read()
sql = 'INSERT INTO imo(image) VALUES(%s)'
args = (blob)
cursor=db.cursor()
cursor.execute(sql,args)
sql1='select image from imo'
# i=sql1
# for image in i:
#     print(i)
# else:
#     print("no")
db.commit()
cursor.execute(sql1)
data=cursor.fetchall()
print ('Inserted')
file_like=io.BytesIO(data[0][0])
img=PIL.Image.open(file_like)
img.show()
print('retrieved')
db.close()

I have tried, but I can't get my expected output..Can anyone help me...

args = (blob) is missing a comma. It should be args = (blob,) in order to make it a 1-tuple. — Ilja Everilä
– Ilja Everilä, Commented Nov 11, 2017 at 10:15

Sruthipriyanga · Accepted Answer · 2017-11-11 10:10:43Z

1

I have tried and i got my expected output just now..

import io
from io import BytesIO

import PIL.Image
# import cStringIO
from PIL import Image
import pymysql
from api import mysql

db=pymysql.connect(host="localhost",user="root",passwd="root",db="votelist")
image = Image.open('D:/mine project/face_reg/media/a.jpg')
blob = open('D:/mine project/face_reg/media/a.jpg', 'rb').read()
sql = 'INSERT INTO imo(image) VALUES(%s)'
args = (blob)
cursor=db.cursor()
cursor.execute(sql,args)
sql1='select image from imo'
conn=mysql.connect()
cursor=conn.cursor()
cursor.execute("select * from imo")
data=cursor.fetchall()
for a in data:
    file_like = io.BytesIO(a[0])
    img = PIL.Image.open(file_like)
    img.show()

answered Nov 11, 2017 at 10:10

Sruthipriyanga

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