How Python can display data on Label by Access (SQL)?

Question

I want Python to display data from Access on Label. I tried to do smth like this but it doesn't work. There is no error but Label displays pypyodbc.Cursor object at 0x05E02990. What should I do? Thank you

from tkinter import *
import pypyodbc
import ctypes

#Create connection
con = pypyodbc.connect('DRIVER={Microsoft Access Driver (*.mdb)};UID=admin;UserCommitSync=Yes;Threads=3;SafeTransactions=0;PageTimeout=5;MaxScanRows=8;MaxBufferSize=2048;FIL={MS Access};DriverId=25;DefaultDir=C:/Users/HP/Desktop/PITL;DBQ=C:/Users/HP/Desktop/PITL/PITL.mdb;')
cursor = con.cursor ()

form = Tk ()
form.title ("Main")
form.geometry ('400x400')

form2 = Tk ()
form2.title ("Main")
form2.geometry ('400x400')
form2.withdraw()

def Show():
    cursor.execute ("SELECT Law_ID FROM Laws WHERE Fine=1")
    a=cursor.execute
    for a in cursor:
        print ("Law ID where Fine is 1 is", a)
        Label(form2, text=cursor).pack() 
    form.withdraw()
    form2.deiconify()

Button(form, text = 'PUSH ME', command = Show).pack()

form.mainloop ()

con.commit ()
cursor.close ()
con.close ()

What happens when you replace print ("Law ID where Fine is 1 is", a) with print ("Law ID where Fine is 1 is", list(a))? — Deadlock
– Deadlock, Commented Feb 18, 2018 at 8:21
Possible duplicate of Python cannot display data on Label twice — Reblochon Masque
– Reblochon Masque, Commented Feb 18, 2018 at 13:11

Shubhitgarg · Accepted Answer · 2018-02-18 10:20:35Z

2

The error is because you are printing the value of cursor object which always be a location, in your case "0x05E02990". To get the value retrieved by the cursor you have to write:

variable_name = cursor.fetchall()

In your code:

def Show():
    cursor.execute ("SELECT Law_ID FROM Laws WHERE Fine=1")
    a=cursor.fetchall()
    ##print as you like

It will fetch all the data retrieved by cursor

answered Feb 18, 2018 at 10:20

Shubhitgarg

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Comments

Parfait · Accepted Answer · 2023-11-22 03:50:29Z

0

Try this one on your function

def Show():

    cursor.execute ("SELECT Law_ID FROM Laws WHERE Fine=1")
    a=cursor.fetchall()

    for i in range(len(a)):
        print ("Law ID where Fine is 1 is", i)
        Label(form2, text=cursor).grid(row = i, column = 0)

edited Nov 22, 2023 at 3:50

Parfait

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answered Nov 22, 2023 at 3:00

Gines Frank Ellis

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1 Comment

Parfait Over a year ago

Your solution still passes cursor object in Label().

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