Updating tkinter label text

I made a little script that takes some values from a web page and displays them as tkinter labels. The values are updated every 5 minutes. How can I make the labels update every time the values change?

url = 'https://myurl.something'

def check():

    sleep(300)

    value1 = str(requests.get(url))



root = Tk()

label1 = label(root, text=value1)

label1.grid()



while True:

    check()

root.mainloop()

I know that the sleep loop could never work, but that's how I did this before adding a GUI

edited Nov 15 '18 at 23:36

asked Nov 15 '18 at 23:19

Stefano

add your code please

– makaros
Nov 15 '18 at 23:22

Here's some docmentation that describes the universal after() method. Note you shouldn't call time.sleep() in a tkinter program, it will make it hang—but after() won't. Lastly, your while True: loop will prevent the root.mainloop() from ever being executed since it's an infinite loop.

– martineau
Nov 16 '18 at 0:36

add a comment |

url = 'https://myurl.something'

def check():

    sleep(300)

    value1 = str(requests.get(url))



root = Tk()

label1 = label(root, text=value1)

label1.grid()



while True:

    check()

root.mainloop()

I know that the sleep loop could never work, but that's how I did this before adding a GUI

edited Nov 15 '18 at 23:36

asked Nov 15 '18 at 23:19

Stefano

add your code please

– makaros
Nov 15 '18 at 23:22

Here's some docmentation that describes the universal after() method. Note you shouldn't call time.sleep() in a tkinter program, it will make it hang—but after() won't. Lastly, your while True: loop will prevent the root.mainloop() from ever being executed since it's an infinite loop.

– martineau
Nov 16 '18 at 0:36

add a comment |

url = 'https://myurl.something'

def check():

    sleep(300)

    value1 = str(requests.get(url))



root = Tk()

label1 = label(root, text=value1)

label1.grid()



while True:

    check()

root.mainloop()

I know that the sleep loop could never work, but that's how I did this before adding a GUI

edited Nov 15 '18 at 23:36

asked Nov 15 '18 at 23:19

Stefano

url = 'https://myurl.something'

def check():

    sleep(300)

    value1 = str(requests.get(url))



root = Tk()

label1 = label(root, text=value1)

label1.grid()



while True:

    check()

root.mainloop()

I know that the sleep loop could never work, but that's how I did this before adding a GUI

python tkinter

edited Nov 15 '18 at 23:36

asked Nov 15 '18 at 23:19

Stefano

edited Nov 15 '18 at 23:36

asked Nov 15 '18 at 23:19

Stefano

edited Nov 15 '18 at 23:36

asked Nov 15 '18 at 23:19

Stefano

asked Nov 15 '18 at 23:19

Stefano

asked Nov 15 '18 at 23:19

Stefano

add your code please

– makaros
Nov 15 '18 at 23:22

Here's some docmentation that describes the universal after() method. Note you shouldn't call time.sleep() in a tkinter program, it will make it hang—but after() won't. Lastly, your while True: loop will prevent the root.mainloop() from ever being executed since it's an infinite loop.

– martineau
Nov 16 '18 at 0:36

add a comment |

add your code please

– makaros
Nov 15 '18 at 23:22

Here's some docmentation that describes the universal after() method. Note you shouldn't call time.sleep() in a tkinter program, it will make it hang—but after() won't. Lastly, your while True: loop will prevent the root.mainloop() from ever being executed since it's an infinite loop.

– martineau
Nov 16 '18 at 0:36

add your code please

– makaros
Nov 15 '18 at 23:22

Here's some docmentation that describes the universal after() method. Note you shouldn't call time.sleep() in a tkinter program, it will make it hang—but after() won't. Lastly, your while True: loop will prevent the root.mainloop() from ever being executed since it's an infinite loop.

– martineau
Nov 16 '18 at 0:36

add a comment |

1 Answer
1

active

oldest

votes

Here is some basic code to get you started. I took out the requests stuff, as it is not really relevant to your problem.

The key things are

The StringVar, which allows you to update your text.

The use of after, to shift your periodic task away from the gui
thread. No need for sleep.

code:

from tkinter import Tk, Label, StringVar



def check(val):

    # do your request.get here

    sv.set("dummy text %d" % val)

    root.after(1000, lambda: check(val + 1))



root = Tk()



sv = StringVar()

sv.set("waiting...")

label1 = Label(root, textvariable=sv)

label1.grid()



root.after(1000, lambda: check(1))

root.mainloop()

Given this code, hopefully you can fit your http request back in and it will do what you want.

edited Nov 16 '18 at 0:35

answered Nov 16 '18 at 0:07

Paul Rooney

12.7k72845

add a comment |

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

Here is some basic code to get you started. I took out the requests stuff, as it is not really relevant to your problem.

The key things are

The StringVar, which allows you to update your text.

The use of after, to shift your periodic task away from the gui
thread. No need for sleep.

code:

from tkinter import Tk, Label, StringVar



def check(val):

    # do your request.get here

    sv.set("dummy text %d" % val)

    root.after(1000, lambda: check(val + 1))



root = Tk()



sv = StringVar()

sv.set("waiting...")

label1 = Label(root, textvariable=sv)

label1.grid()



root.after(1000, lambda: check(1))

root.mainloop()

Given this code, hopefully you can fit your http request back in and it will do what you want.

edited Nov 16 '18 at 0:35

answered Nov 16 '18 at 0:07

Paul Rooney

12.7k72845

add a comment |

Here is some basic code to get you started. I took out the requests stuff, as it is not really relevant to your problem.

The key things are

The StringVar, which allows you to update your text.

The use of after, to shift your periodic task away from the gui
thread. No need for sleep.

code:

from tkinter import Tk, Label, StringVar



def check(val):

    # do your request.get here

    sv.set("dummy text %d" % val)

    root.after(1000, lambda: check(val + 1))



root = Tk()



sv = StringVar()

sv.set("waiting...")

label1 = Label(root, textvariable=sv)

label1.grid()



root.after(1000, lambda: check(1))

root.mainloop()

Given this code, hopefully you can fit your http request back in and it will do what you want.

edited Nov 16 '18 at 0:35

answered Nov 16 '18 at 0:07

Paul Rooney

12.7k72845

add a comment |

Here is some basic code to get you started. I took out the requests stuff, as it is not really relevant to your problem.

The key things are

The StringVar, which allows you to update your text.

The use of after, to shift your periodic task away from the gui
thread. No need for sleep.

code:

from tkinter import Tk, Label, StringVar



def check(val):

    # do your request.get here

    sv.set("dummy text %d" % val)

    root.after(1000, lambda: check(val + 1))



root = Tk()



sv = StringVar()

sv.set("waiting...")

label1 = Label(root, textvariable=sv)

label1.grid()



root.after(1000, lambda: check(1))

root.mainloop()

Given this code, hopefully you can fit your http request back in and it will do what you want.

edited Nov 16 '18 at 0:35

answered Nov 16 '18 at 0:07

Paul Rooney

12.7k72845

Here is some basic code to get you started. I took out the requests stuff, as it is not really relevant to your problem.

The key things are

The StringVar, which allows you to update your text.

The use of after, to shift your periodic task away from the gui
thread. No need for sleep.

code:

from tkinter import Tk, Label, StringVar



def check(val):

    # do your request.get here

    sv.set("dummy text %d" % val)

    root.after(1000, lambda: check(val + 1))



root = Tk()



sv = StringVar()

sv.set("waiting...")

label1 = Label(root, textvariable=sv)

label1.grid()



root.after(1000, lambda: check(1))

root.mainloop()

Given this code, hopefully you can fit your http request back in and it will do what you want.

edited Nov 16 '18 at 0:35

answered Nov 16 '18 at 0:07

Paul Rooney

12.7k72845

edited Nov 16 '18 at 0:35

answered Nov 16 '18 at 0:07

Paul Rooney

12.7k72845

answered Nov 16 '18 at 0:07

Paul Rooney

12.7k72845

answered Nov 16 '18 at 0:07

Paul Rooney

12.7k72845

add a comment |

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