We Keep Coding

I try to make filename in upload_photo a global variable so I can use it in the add_client fuction but it throws an error filename is not defined

def upload_photo(self):
global img, filename
global filename
f_types = [('JPG Files','*.jpg'),("PNG File","*.png"),("All Files","*.*")]
filename=filedialog.askopenfilename(filetypes=f_types)
img=Image.open(filename)
img_resized=img.resize((200,200))
img=ImageTk.PhotoImage(img_resized)
b2=Label(self.emp_Manage_frame,image=img,width=250,height=250,bg="red")
b2.place(x=530,y=100,width=199,height=185)
def add_client(self):
global img, filename
global filename
fob=open(filename, 'rb')
fob=fob.read()
data=fob

You will get the error if the add_client() method is called before upload_photo(), but otherwise your code works OK. You could fix that problem by adding a filename = None at the top-level of the script to ensure that the variable is always defined — and the checked its value before attempting to use it.
However it would be even better to get rid of the global variables altogether because they are considered a poor programming practice. The following shows how to implement that:
from tkinter import *
from tkinter import filedialog
import tkinter.messagebox as messagebox
from PIL import Image, ImageTk
class Application:
def __init__(self, parent):
self.emp_Manage_frame = Frame(parent, width=800, height=640)
self.emp_Manage_frame.pack()
self.img = None
self.filename = None
def upload_photo(self):
f_types = [('JPG Files','*.jpg'), ("PNG File","*.png"), ("All Files","*.*")]
self.filename = filedialog.askopenfilename(filetypes=f_types)
img = Image.open(self.filename)
img_resized = img.resize((200, 200))
self.img = ImageTk.PhotoImage(img_resized)
b2 = Label(self.emp_Manage_frame, image=self.img, width=250, height=250, bg="red")
b2.place(x=530, y=100, width=199, height=185)
def add_client(self):
if self.filename is None:
messagebox.showerror('File not defined!', 'Please upload image file first')
else:
with open(self.filename, 'rb') as fob:
data = fob.read()
if __name__ == '__main__':
root = Tk()
app = Application(root)
upload_btn = Button(root, text='Upload Photo', command=app.upload_photo)
upload_btn.pack()
add_client_btn = Button(root, text='Add Client', command=app.add_client)
add_client_btn.pack()
root.mainloop()

Python thread with tkinter (threads can only be started once) - after checking if previous thread isAlive() [duplicate]

but I'm trying to make the GUI for my script, when I click bt_send I start a thread (thread_enviar), that thread also start other thread (core), the problem is that thread_enviar is running for ever so I got this error when I try to click bt_send again:
File "/anaconda3/envs/tensor/lib/python3.6/threading.py", line 842, in start
raise RuntimeError("threads can only be started once")
RuntimeError: threads can only be started once
My code:
import tkinter as tk
from tkinter import filedialog
import pandas as pd
from tkinter import messagebox
from tkinter import ttk
import threading
import rnn10forecasting as rnn10f
filepath = ""
model = ""
'''def change_menu(selection):
global model
selected = selection
print(selected)
model = selected'''
def click():
global filepath
print("click")
filepath = filedialog.askopenfilename(initialdir = "/",title = "Select file",filetypes = (("data files","*.csv"),("all files","*.*")))
print(filepath)
label_filepath.config(text = filepath)
def enviar():
print(filepath)
try:
data = pd.read_csv(filepath)
except:
messagebox.showerror("Error", "Archivo .csv vacio o formato incompatible")
if any(data.columns.values != ['date','close','volume','open','high','low']):
messagebox.showerror("Error", "El archivo .csv no contiene la estructura requerida: [date,close,volume,open,high,low]")
elif len(data) < 300:
print("# registros")
print(len(data))
messagebox.showerror("Error", "El archivo de be contener como minimo 700 registros")
else:
pg_bar.start(500)
core = threading.Thread(target=rnn10f.forecasting, args=(filepath,))
#core.daemon = True
core.start()
core.join()
print("VIVO?")
print(core.isAlive())
pg_bar.stop()
return print(thread_enviar.is_alive())
thread_enviar = threading.Thread(target=enviar, args=())
window = tk.Tk()
window.resizable(width=False, height=False)
window.title("LSTM Module")
window.geometry("600x150")
title = tk.Label(text="StockForecaster", font=("Times New Roman", 30))
title.place(relx=0.1, rely=0.05, relwidth=0.8, relheight=0.25)
bt_select = tk.Button(text="Select File", bg="blue", command= click)
bt_select.place(relx=0.7, rely=0.4, relwidth=0.25, relheight=0.2)
label_filepath = tk.Label(text="Please select a .csv File")
label_filepath.place(relx=0, rely=0.4, relwidth=0.7, relheight=0.15)
options = tk.StringVar()
bt_send = tk.Button(text="Send", bg="blue", command=thread_enviar.start)
bt_send.place(relx=0.70, rely=0.7, relwidth=0.25, relheight=0.20)
pg_bar = ttk.Progressbar(window, orient= tk.HORIZONTAL, mode="indeterminate", )
pg_bar.place(relx=0.10, rely=0.75, relwidth=0.55, relheight=0.05)
window.mainloop()
I don' know if there is any way to kill that thread or If I'm doing something wrong.

Question: RuntimeError: threads can only be started once
As I understand, you don't want to run multiple threads, you only want to do a Task in a thread to avoid freezing the Tk().mainloop().
To inhibit, to start a new thread while the previous thread are still running you have to disable the Button or verify if the previous thread is still .alive().
Try the following approach:
import tkinter as tk
import threading, time
class Task(threading.Thread):
def __init__(self, master, task):
threading.Thread.__init__(self, target=task, args=(master,))
if not hasattr(master, 'thread_enviar') or not master.thread_enviar.is_alive():
master.thread_enviar = self
self.start()
def enviar(master):
# Simulating conditions
if 0:
pass
#if any(...
#elif len(data) < 300:
else:
#master.pg_bar.start(500)
# Simulate long run
time.sleep(10)
#rnn10f.forecasting(filepath)
print("VIVO?")
#master.pg_bar.stop()
class App(tk.Tk):
def __init__(self):
super().__init__()
bt_send = tk.Button(text="Send", bg="blue",
command=lambda :Task(self, enviar))
if __name__ == "__main__":
App().mainloop()

Tkinter Button - How to get the values and display it using a single module for different Buttons

I am creating a GUI, where I have placed three buttons for selecting a file. When I click the button it browse for a file and display the selected file path and I am trying to do it with a single browse function since the operation is same. I am bit confused to how I should proceed . Can anyone please help me out.
Thanks in advance.
Here is my code:
Browse_Files.py
from tkinter import filedialog
def Browse_File():
global Bfilepath
Bfilepath = filedialog.askopenfilename(filetypes = (("Please select the required file", "*"), ("All files", "*")))
return Bfilepath
Main.py
from tkinter import *
import sys
import fileinput
import Browse_Files
root = Tk()
root.geometry('1400x800')
root.title('Dr Configuration')
Heading = Label(root, font=('Times New Roman',50,'bold'),text = "Arxml
Configuration Tool").place(x=300,y=25)
BasepathLabel = Label(root,font=('Times New Roman',20,'bold'),text = " Base
arxml").place(x=200,y=150)
NewpathLabel= Label(root,font=('Times New Roman',20,'bold'),text = "
New/Unedited arxml").place(x=200,y=250)
InterfaceLabel = Label(root,font=('Times New Roman',20,'bold'),text = "
Interface_File").place(x=200,y=350)
BpathtoDisp = StringVar(None)
BpathEntry = Entry(root,font=('Times New Roman',18),textvariable=
BpathtoDisp,justify='left',width=48).place(x=500,y=150)
NpathtoDisp = StringVar(None)
NpathEntry = Entry(root,font=('Times New Roman',18),textvariable=
NpathtoDisp,justify='left',width=48).place(x=500,y=250)
InterPathtoDisp = StringVar(None)
InterPathEntry = Entry(root,font=('Times New Roman',18),textvariable=
NpathtoDisp,justify='left',width=48).place(x=500,y=350)
button1 = Button(root,text="...",height=1,width=3,command=lambda:Browse_Files.Browse_File()).place(x=1100,y=150)
button2 = Button(root,text="...",height=1,width=3,command=lambda:Browse_Files.Browse_File()).place(x=1100,y=250)
button3 = Button(root,text="...",height=1,width=3,command=lambda:Browse_Files.Browse_File()).place(x=1100,y=350)
root.mainloop()

You could create a class that contains the functionality to have a button, open a file dialog, store it and make it available to other parts of the program.
I have a similar widget that I use in many different programs
from tkinter import *
from tkinter import filedialog
class FileSelect(Frame):
def __init__(self,master,label="",**kw):
Frame.__init__(self,master)
self.configure(**kw)
self.file = StringVar()
self.Label = Label(self, text=label)
self.Label.config(width=10,anchor=E)
self.filenamebox = Entry(self,text=self.file)
self.filenamebox.config(width=50)
self.btnBrowse = Button(self,text='Browse',command=self.browse_file)
self.btnBrowse.config(width=10)
self.Label.grid(row=0,column=0,pady=5,sticky=E)
self.filenamebox.grid(row=0,column=1,pady=5)
self.btnBrowse.grid(row=0,column=2,pady=5,padx=5)
def browse_file(self):
filename = filedialog.askopenfilename(filetypes=[('All File','*.*')])
self.file.set(filename)
def get_filename(self):
return self.file.get()
def main():
root = Tk()
root.title("Example Widget")
fileSelect1 = FileSelect(root,label="My File 1")
fileSelect1.grid()
fileSelect2 = FileSelect(root,label="My File 2")
fileSelect2.grid()
root.mainloop()
if __name__ == '__main__':
main()
In your code if you want the value of the file selected use
fileSelect1.get_filename()
EDIT: I've created a new version that is only a Button 'with memory' to remember which item was selected by the file dialog. This will allow you to place using the place geometry manager (which I don't recommend). Its not complete but you should get the idea.
from tkinter import *
from tkinter import filedialog
class FileSelectButton(Button):
def __init__(self,master,**kw):
Button.__init__(self,master,text='Browse',command=self.browse_file,width=10,**kw)
self.file = StringVar()
def browse_file(self):
filename = filedialog.askopenfilename(filetypes=[('All File','*.*')])
self.file.set(filename)
def get_filename(self):
return self.file.get()
def main():
root = Tk()
root.geometry('1400x800')
root.title("Example Widget")
Label(root, font=('Times New Roman',50,'bold'),text = "Label1").place(x=200,y=150)
fileSelect1 = FileSelectButton(root)
fileSelect1.place(x=500,y=150)
Label(root, font=('Times New Roman',50,'bold'),text = "Label2").place(x=200,y=250)
fileSelect2 = FileSelectButton(root)
fileSelect2.place(x=500,y=250)
root.mainloop()
if __name__ == '__main__':
main()

The code runs in the CMD, but not in the GUI?

Here is the code that i run in the python CMD:
import mailbox
import pprint
f = open("results.txt","w")
mbox = mailbox.mbox('c:\documents and settings\student\desktop\mail\mailall.mbox')
count = 0
for msg in mbox:
pprint.pprint(msg._headers, stream = f)
if msg['Delivered-To'] == 'example#example.co.uk':
count += 1
f.close()
print(count)
However when i run it in the GUI, it makes the GUI crash (Flashes black and doesn't open). Here is the GUI:
import datetime
from tkinter import filedialog
import tkinter
class App:
def __init__(self, master):
self.master = master
# call start to initialize to create the UI elemets
self.start()
def start(self):
self.master.title("Extract Email Headers")
self.now = datetime.datetime.now()
# CREATE A TEXT/LABEL
# create a variable with text
label01 = "Please select the .mbox file you would like to analyse"
# put "label01" in "self.master" which is the window/frame
# then, put in the first row (row=0) and in the 2nd column (column=1),
# align it to "West"/"W"
tkinter.Label(
self.master, text=label01).grid(row=0, column=0, sticky=tkinter.W)
# CREATE A TEXTBOX
self.filelocation = tkinter.Entry(self.master)
self.filelocation["width"] = 60
self.filelocation.focus_set()
self.filelocation.grid(row=1, column=0)
# CREATE A BUTTON WITH "ASK TO OPEN A FILE"
# see: def browse_file(self)
self.open_file = tkinter.Button(
self.master, text="Browse...", command=self.browse_file)
# put it beside the filelocation textbox
self.open_file.grid(row=1, column=1)
# now for a button
self.submit = tkinter.Button(
self.master, text="Execute!", command=self.start_processing,
fg="red")
self.submit.grid(row=3, column=0)
def start_processing(self):
import mailbox
import pprint
f = open("results.txt","w")
mbox = mailbox.mbox('c:\documents and settings\student\desktop\mail\mailall.mbox')
count = 0
for msg in mbox:
pprint.pprint(msg._headers, stream = f)
if msg['Delivered-To'] == 'example#example.co.uk':
count += 1
f.close()
print(count)
def browse_file(self):
# put the result in self.filename
self.filename = filedialog.askopenfilename(title="Open a file...")
# this will set the text of the self.filelocation
self.filelocation.insert(0, self.filename)
root = tkinter.Tk()
app = App(root)
root.mainloop()
If i take the process code out and put "pass", the GUI works so i know its the code that is throwing it off.? Thanks

Python / Tkinter status bar not updating correctly

I have developed a simple app in Python (2.7) with Tkinter. But my status bar is only sort of working. Here's the stripped down code:
from Tkinter import *
import os
import sys
def fixFiles():
inputFilePath= input_dir.get()
#Build a list of files in a directory
fileList = os.listdir(inputFilePath)
#Loop through those files, open the file, do something, close the file
for filename in fileList:
infile = open(inputfilepath + "/" + filename,'r')
#Update the status with the filename
status_string = 'Status: Working on file: ' + str(filename)
status.set(status_string)
for line in infile:
#Do some stuff here
infile.close()
class App:
def __init__(self, master):
i = 0
status.set("Status: Press 'Fix Files!'")
statuslabel = Label(master, textvariable=status, relief = RIDGE, width = 65, pady = 5, anchor=W)
bFixFiles = Button(root, text='Fix Files!', command = fixFiles)
bQuit = Button(root, text='Quit', command = root.destroy)
statuslabel.grid(row=i, column = 0, columnspan = 2)
bFixFiles.grid(row=i, column=2, sticky=E)
bQuit.grid(row=i, column=3, sticky=W)
root = Tk()
root.title("FIX Files")
input_dir = StringVar()
status = StringVar()
choice = IntVar()
app = App(root)
root.mainloop()
Currently what's happening is that the status bar reads "Status: Press 'Fix Files!'" until the program is finished looping through the files, at which point it reads "Status: Working on file: XXXXX.txt" (which is the name of the last file to be opened and closed by the program.
I would like the status bar to update with the file name each time the program opens a new file. Any help is appreciated!

The goofy way is to use root.update_idletasks():
#Update the status with the filename
status_string = 'Status: Working on file: ' + str(filename)
status.set(status_string)
root.update_idletasks()
To its credit, it is simple, but it does not really work -- although the statuslabel gets updated, the Quit button is frozen until fixFiles is completed. That's not very GUI-friendly. Here are some more reasons why update and update_idletasks are considered harmful.
So how should we run a long-running task without freezing the GUI?
The key is to make your callback functions end quickly. Instead of having a long-running for-loop, make a function that runs through the innards of the for-loop once. Hopefully that ends quickly enough for the user to not feel the GUI has been frozen.
Then, to replace the for-loop, you could use calls to root.after to call your quick-running function multiple times.
from Tkinter import *
import tkFileDialog
import os
import sys
import time
def startFixFiles():
inputFilePath = tkFileDialog.askdirectory()
# inputFilePath= input_dir.get()
# Build a list of files in a directory
fileList = os.listdir(inputFilePath)
def fixFiles():
try:
filename = fileList.pop()
except IndexError:
return
try:
with open(os.path.join(inputFilePath, filename), 'r') as infile:
# Update the status with the filename
status_string = 'Status: Working on file: ' + str(filename)
status.set(status_string)
for line in infile:
# Do some stuff here
pass
except IOError:
# You might get here if file is unreadable, you don't have read permission,
# or the file might be a directory...
pass
root.after(250, fixFiles)
root.after(10, fixFiles)
class App:
def __init__(self, master):
i = 0
status.set("Status: Press 'Fix Files!'")
statuslabel = Label(
master, textvariable=status, relief=RIDGE, width=65,
pady=5, anchor=W)
bFixFiles = Button(root, text='Fix Files!', command=startFixFiles)
bQuit = Button(root, text='Quit', command=root.destroy)
statuslabel.grid(row=i, column=0, columnspan=2)
bFixFiles.grid(row=i, column=2, sticky=E)
bQuit.grid(row=i, column=3, sticky=W)
root = Tk()
root.title("FIX Files")
input_dir = StringVar()
status = StringVar()
choice = IntVar()
app = App(root)
root.mainloop()
The above begs the question, What should we do if our long-running task has no loop? or if even one pass through the loop requires a long time?
Here is a way to run the long-running task in a separate process (or thread), and have it communicate information through a queue which the main process can periodically poll (using root.after) to update the GUI status bar. I think this design is more easily applicable to this problem in general since it does not require you to break apart the for-loop.
Note carefully that all Tkinter GUI-related function calls must occur from a single thread. That is why the long-running process simply sends strings through the queue instead of trying to call status.set directly.
import Tkinter as tk
import multiprocessing as mp
import tkFileDialog
import os
import Queue
sentinel = None
def long_running_worker(inputFilePath, outqueue):
# Build a list of files in a directory
fileList = os.listdir(inputFilePath)
for filename in fileList:
try:
with open(os.path.join(inputFilePath, filename), 'r') as infile:
# Update the status with the filename
status_string = 'Status: Working on file: ' + str(filename)
outqueue.put(status_string)
for line in infile:
# Do some stuff here
pass
except IOError:
# You might get here if file is unreadable, you don't have read permission,
# or the file might be a directory...
pass
# Put the sentinel in the queue to tell update_status to end
outqueue.put(sentinel)
class App(object):
def __init__(self, master):
self.status = tk.StringVar()
self.status.set("Status: Press 'Fix Files!'")
self.statuslabel = tk.Label(
master, textvariable=self.status, relief=tk.RIDGE, width=65,
pady=5, anchor='w')
bFixFiles = tk.Button(root, text='Fix Files!', command=self.startFixFiles)
bQuit = tk.Button(root, text='Quit', command=root.destroy)
self.statuslabel.grid(row=1, column=0, columnspan=2)
bFixFiles.grid(row=0, column=0, sticky='e')
bQuit.grid(row=0, column=1, sticky='e')
def update_status(self, outqueue):
try:
status_string = outqueue.get_nowait()
if status_string is not sentinel:
self.status.set(status_string)
root.after(250, self.update_status, outqueue)
else:
# By not calling root.after here, we allow update_status to truly end
pass
except Queue.Empty:
root.after(250, self.update_status, outqueue)
def startFixFiles(self):
inputFilePath = tkFileDialog.askdirectory()
# Start long running process
outqueue = mp.Queue()
proc = mp.Process(target=long_running_worker, args=(inputFilePath, outqueue))
proc.daemon = True
proc.start()
# Start a function to check a queue for GUI-related updates
root.after(250, self.update_status, outqueue)
root = tk.Tk()
root.title("FIX Files")
app = App(root)
root.mainloop()