Tkinter Filedialog.askopenfilename iteration
I am developing program to load up files and perform some calculations with those loaded files.
For that I wrote a simple iteration code to load the tkinter variables. The window, label, entry and button positions are already done. So far the code I have is:
import tkinter as tk
from tkinter import ttk, filedialog
LARGE_FONT = ("Arial", 12)
MEDIUM_FONT = ("Arial", 11)
REGULAR_FONT = ("Arial", 10)
text_z = ["Select file 1", "Select the file 2", "Select file 3", "Select file 4"]
window=tk.Tk()
def click():
z = tk.filedialog.askopenfilename(initialdir = "/",title = "Select file", filetypes = ( ("Excel file", "*.xlsx"), ("All files", "*.*") ) )
a[i-2].insert(tk.END, z)
z[i] = a[i-2].get()
##Main program
#There is an image I will add at the end on row=0
ttk.Label(window, text="file load", font = LARGE_FONT, background = "white").grid(row=1, column=1, columnspan=3, padx=20, pady = 10, sticky="W")
a = [tk.StringVar(window) for i in range(len(text_z))]
for i in range(2,len(text_z)+2):
Label_z = ttk.Label(window, text=text_z[i-2], background="white").grid(row= 2*i, column=0,columnspan=3, padx=10, pady=2, sticky="W")
a[i-2] = ttk.Entry(window, width=60, background="gray")
a[i-2].grid(row= 2*i+1, column=0, columnspan=3, padx=10, sticky="WE")
ttk.Button(window, text="Search", width=10, command=click).grid(row= 2*i+1, column=3, padx=5, sticky="W")
window.mainloop()
My problem is on the click button. It was supposed to during a click run the askopenfilename, get the file path and present on the entrybox, but all the buttons direct that to the last created Entrybox.
Can someone help me with this issue?
Thanks alot!
python tkinter iteration openfiledialog
add a comment |
I am developing program to load up files and perform some calculations with those loaded files.
For that I wrote a simple iteration code to load the tkinter variables. The window, label, entry and button positions are already done. So far the code I have is:
import tkinter as tk
from tkinter import ttk, filedialog
LARGE_FONT = ("Arial", 12)
MEDIUM_FONT = ("Arial", 11)
REGULAR_FONT = ("Arial", 10)
text_z = ["Select file 1", "Select the file 2", "Select file 3", "Select file 4"]
window=tk.Tk()
def click():
z = tk.filedialog.askopenfilename(initialdir = "/",title = "Select file", filetypes = ( ("Excel file", "*.xlsx"), ("All files", "*.*") ) )
a[i-2].insert(tk.END, z)
z[i] = a[i-2].get()
##Main program
#There is an image I will add at the end on row=0
ttk.Label(window, text="file load", font = LARGE_FONT, background = "white").grid(row=1, column=1, columnspan=3, padx=20, pady = 10, sticky="W")
a = [tk.StringVar(window) for i in range(len(text_z))]
for i in range(2,len(text_z)+2):
Label_z = ttk.Label(window, text=text_z[i-2], background="white").grid(row= 2*i, column=0,columnspan=3, padx=10, pady=2, sticky="W")
a[i-2] = ttk.Entry(window, width=60, background="gray")
a[i-2].grid(row= 2*i+1, column=0, columnspan=3, padx=10, sticky="WE")
ttk.Button(window, text="Search", width=10, command=click).grid(row= 2*i+1, column=3, padx=5, sticky="W")
window.mainloop()
My problem is on the click button. It was supposed to during a click run the askopenfilename, get the file path and present on the entrybox, but all the buttons direct that to the last created Entrybox.
Can someone help me with this issue?
Thanks alot!
python tkinter iteration openfiledialog
Sorry @Miraj50, edited the question with those items
– Rafael Castelo Branco
Nov 14 '18 at 13:12
add a comment |
I am developing program to load up files and perform some calculations with those loaded files.
For that I wrote a simple iteration code to load the tkinter variables. The window, label, entry and button positions are already done. So far the code I have is:
import tkinter as tk
from tkinter import ttk, filedialog
LARGE_FONT = ("Arial", 12)
MEDIUM_FONT = ("Arial", 11)
REGULAR_FONT = ("Arial", 10)
text_z = ["Select file 1", "Select the file 2", "Select file 3", "Select file 4"]
window=tk.Tk()
def click():
z = tk.filedialog.askopenfilename(initialdir = "/",title = "Select file", filetypes = ( ("Excel file", "*.xlsx"), ("All files", "*.*") ) )
a[i-2].insert(tk.END, z)
z[i] = a[i-2].get()
##Main program
#There is an image I will add at the end on row=0
ttk.Label(window, text="file load", font = LARGE_FONT, background = "white").grid(row=1, column=1, columnspan=3, padx=20, pady = 10, sticky="W")
a = [tk.StringVar(window) for i in range(len(text_z))]
for i in range(2,len(text_z)+2):
Label_z = ttk.Label(window, text=text_z[i-2], background="white").grid(row= 2*i, column=0,columnspan=3, padx=10, pady=2, sticky="W")
a[i-2] = ttk.Entry(window, width=60, background="gray")
a[i-2].grid(row= 2*i+1, column=0, columnspan=3, padx=10, sticky="WE")
ttk.Button(window, text="Search", width=10, command=click).grid(row= 2*i+1, column=3, padx=5, sticky="W")
window.mainloop()
My problem is on the click button. It was supposed to during a click run the askopenfilename, get the file path and present on the entrybox, but all the buttons direct that to the last created Entrybox.
Can someone help me with this issue?
Thanks alot!
python tkinter iteration openfiledialog
I am developing program to load up files and perform some calculations with those loaded files.
For that I wrote a simple iteration code to load the tkinter variables. The window, label, entry and button positions are already done. So far the code I have is:
import tkinter as tk
from tkinter import ttk, filedialog
LARGE_FONT = ("Arial", 12)
MEDIUM_FONT = ("Arial", 11)
REGULAR_FONT = ("Arial", 10)
text_z = ["Select file 1", "Select the file 2", "Select file 3", "Select file 4"]
window=tk.Tk()
def click():
z = tk.filedialog.askopenfilename(initialdir = "/",title = "Select file", filetypes = ( ("Excel file", "*.xlsx"), ("All files", "*.*") ) )
a[i-2].insert(tk.END, z)
z[i] = a[i-2].get()
##Main program
#There is an image I will add at the end on row=0
ttk.Label(window, text="file load", font = LARGE_FONT, background = "white").grid(row=1, column=1, columnspan=3, padx=20, pady = 10, sticky="W")
a = [tk.StringVar(window) for i in range(len(text_z))]
for i in range(2,len(text_z)+2):
Label_z = ttk.Label(window, text=text_z[i-2], background="white").grid(row= 2*i, column=0,columnspan=3, padx=10, pady=2, sticky="W")
a[i-2] = ttk.Entry(window, width=60, background="gray")
a[i-2].grid(row= 2*i+1, column=0, columnspan=3, padx=10, sticky="WE")
ttk.Button(window, text="Search", width=10, command=click).grid(row= 2*i+1, column=3, padx=5, sticky="W")
window.mainloop()
My problem is on the click button. It was supposed to during a click run the askopenfilename, get the file path and present on the entrybox, but all the buttons direct that to the last created Entrybox.
Can someone help me with this issue?
Thanks alot!
python tkinter iteration openfiledialog
python tkinter iteration openfiledialog
edited Nov 14 '18 at 14:44
Miraj50
2,7701924
2,7701924
asked Nov 14 '18 at 12:54
Rafael Castelo BrancoRafael Castelo Branco
395
395
Sorry @Miraj50, edited the question with those items
– Rafael Castelo Branco
Nov 14 '18 at 13:12
add a comment |
Sorry @Miraj50, edited the question with those items
– Rafael Castelo Branco
Nov 14 '18 at 13:12
Sorry @Miraj50, edited the question with those items
– Rafael Castelo Branco
Nov 14 '18 at 13:12
Sorry @Miraj50, edited the question with those items
– Rafael Castelo Branco
Nov 14 '18 at 13:12
add a comment |
2 Answers
2
active
oldest
votes
Lambda
to the rescue. One needs to know the right Button-Entry pair to update. So pass the value of the corresponding index when a button is pressed.
import tkinter as tk
from tkinter import ttk, filedialog
LARGE_FONT = ("Arial", 12)
MEDIUM_FONT = ("Arial", 11)
REGULAR_FONT = ("Arial", 10)
text_z = ["Select file 1", "Select the file 2", "Select file 3", "Select file 4"]
window=tk.Tk()
def click(m):
z = tk.filedialog.askopenfilename(initialdir = "~",title = "Select file", filetypes = ( ("Text files", "*.txt"), ("All files", "*.*") ) )
a[m].insert(tk.END, z)
ttk.Label(window, text="file load", font = LARGE_FONT, background = "white").grid(row=1, column=1, columnspan=3, padx=20, pady = 10, sticky="W")
a = [None for i in range(len(text_z))]
for i in range(2,len(text_z)+2):
Label_z = ttk.Label(window, text=text_z[i-2], background="white").grid(row= 2*i, column=0,columnspan=3, padx=10, pady=2, sticky="W")
a[i-2] = ttk.Entry(window, width=60, background="gray")
a[i-2].grid(row= 2*i+1, column=0, columnspan=3, padx=10, sticky="WE")
ttk.Button(window, text="Search", width=10, command=lambda m=i-2:click(m)).grid(row= 2*i+1, column=3, padx=5, sticky="W")
window.mainloop()
add a comment |
I think you should simplify things a bit by use a list to store your entry fields.
To do this I think it would be best to add frames for each set of widgets and to use the index of range to get what we need.
I have changed up your code a little to make it easier to work with list index as well as added a button that will print out each selected path on each entry field to show these values are accessible.
import tkinter as tk
from tkinter import ttk, filedialog
LARGE_FONT = ("Arial", 12)
MEDIUM_FONT = ("Arial", 11)
REGULAR_FONT = ("Arial", 10)
text_z = ["Select file 1", "Select the file 2", "Select file 3", "Select file 4"]
window = tk.Tk()
def click(x):
z = tk.filedialog.askopenfilename(initialdir="/", title="Select file", filetypes=(("Excel file", "*.xlsx"), ("All files", "*.*")))
a[x].insert(tk.END, z)
ttk.Label(window, text="file load", font=LARGE_FONT, background="white").grid(row=1, column=0, padx=20, pady=10, sticky="w")
a=
for i in range(len(text_z)):
frame = tk.Frame(window)
frame.grid(row=i+2, column=0, sticky="nsew")
ttk.Label(frame, text=text_z[i], background="white").grid(row=0, column=0, columnspan=3, padx=10, pady=2, sticky="w")
a.append(ttk.Entry(frame, width=60, background="gray"))
a[i].grid(row=1, column=0, columnspan=3, padx=10, sticky="ew")
ttk.Button(frame, text="Search", width=10, command=lambda x=i: click(x)).grid(row=1, column=3, padx=5, sticky="w")
def pring_current_paths():
for ndex, entry in enumerate(a):
print("Entry {}: ".format(ndex, entry.get()))
tk.Button(window, text="Print gurrent paths!", command=pring_current_paths).grid()
window.mainloop()
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Lambda
to the rescue. One needs to know the right Button-Entry pair to update. So pass the value of the corresponding index when a button is pressed.
import tkinter as tk
from tkinter import ttk, filedialog
LARGE_FONT = ("Arial", 12)
MEDIUM_FONT = ("Arial", 11)
REGULAR_FONT = ("Arial", 10)
text_z = ["Select file 1", "Select the file 2", "Select file 3", "Select file 4"]
window=tk.Tk()
def click(m):
z = tk.filedialog.askopenfilename(initialdir = "~",title = "Select file", filetypes = ( ("Text files", "*.txt"), ("All files", "*.*") ) )
a[m].insert(tk.END, z)
ttk.Label(window, text="file load", font = LARGE_FONT, background = "white").grid(row=1, column=1, columnspan=3, padx=20, pady = 10, sticky="W")
a = [None for i in range(len(text_z))]
for i in range(2,len(text_z)+2):
Label_z = ttk.Label(window, text=text_z[i-2], background="white").grid(row= 2*i, column=0,columnspan=3, padx=10, pady=2, sticky="W")
a[i-2] = ttk.Entry(window, width=60, background="gray")
a[i-2].grid(row= 2*i+1, column=0, columnspan=3, padx=10, sticky="WE")
ttk.Button(window, text="Search", width=10, command=lambda m=i-2:click(m)).grid(row= 2*i+1, column=3, padx=5, sticky="W")
window.mainloop()
add a comment |
Lambda
to the rescue. One needs to know the right Button-Entry pair to update. So pass the value of the corresponding index when a button is pressed.
import tkinter as tk
from tkinter import ttk, filedialog
LARGE_FONT = ("Arial", 12)
MEDIUM_FONT = ("Arial", 11)
REGULAR_FONT = ("Arial", 10)
text_z = ["Select file 1", "Select the file 2", "Select file 3", "Select file 4"]
window=tk.Tk()
def click(m):
z = tk.filedialog.askopenfilename(initialdir = "~",title = "Select file", filetypes = ( ("Text files", "*.txt"), ("All files", "*.*") ) )
a[m].insert(tk.END, z)
ttk.Label(window, text="file load", font = LARGE_FONT, background = "white").grid(row=1, column=1, columnspan=3, padx=20, pady = 10, sticky="W")
a = [None for i in range(len(text_z))]
for i in range(2,len(text_z)+2):
Label_z = ttk.Label(window, text=text_z[i-2], background="white").grid(row= 2*i, column=0,columnspan=3, padx=10, pady=2, sticky="W")
a[i-2] = ttk.Entry(window, width=60, background="gray")
a[i-2].grid(row= 2*i+1, column=0, columnspan=3, padx=10, sticky="WE")
ttk.Button(window, text="Search", width=10, command=lambda m=i-2:click(m)).grid(row= 2*i+1, column=3, padx=5, sticky="W")
window.mainloop()
add a comment |
Lambda
to the rescue. One needs to know the right Button-Entry pair to update. So pass the value of the corresponding index when a button is pressed.
import tkinter as tk
from tkinter import ttk, filedialog
LARGE_FONT = ("Arial", 12)
MEDIUM_FONT = ("Arial", 11)
REGULAR_FONT = ("Arial", 10)
text_z = ["Select file 1", "Select the file 2", "Select file 3", "Select file 4"]
window=tk.Tk()
def click(m):
z = tk.filedialog.askopenfilename(initialdir = "~",title = "Select file", filetypes = ( ("Text files", "*.txt"), ("All files", "*.*") ) )
a[m].insert(tk.END, z)
ttk.Label(window, text="file load", font = LARGE_FONT, background = "white").grid(row=1, column=1, columnspan=3, padx=20, pady = 10, sticky="W")
a = [None for i in range(len(text_z))]
for i in range(2,len(text_z)+2):
Label_z = ttk.Label(window, text=text_z[i-2], background="white").grid(row= 2*i, column=0,columnspan=3, padx=10, pady=2, sticky="W")
a[i-2] = ttk.Entry(window, width=60, background="gray")
a[i-2].grid(row= 2*i+1, column=0, columnspan=3, padx=10, sticky="WE")
ttk.Button(window, text="Search", width=10, command=lambda m=i-2:click(m)).grid(row= 2*i+1, column=3, padx=5, sticky="W")
window.mainloop()
Lambda
to the rescue. One needs to know the right Button-Entry pair to update. So pass the value of the corresponding index when a button is pressed.
import tkinter as tk
from tkinter import ttk, filedialog
LARGE_FONT = ("Arial", 12)
MEDIUM_FONT = ("Arial", 11)
REGULAR_FONT = ("Arial", 10)
text_z = ["Select file 1", "Select the file 2", "Select file 3", "Select file 4"]
window=tk.Tk()
def click(m):
z = tk.filedialog.askopenfilename(initialdir = "~",title = "Select file", filetypes = ( ("Text files", "*.txt"), ("All files", "*.*") ) )
a[m].insert(tk.END, z)
ttk.Label(window, text="file load", font = LARGE_FONT, background = "white").grid(row=1, column=1, columnspan=3, padx=20, pady = 10, sticky="W")
a = [None for i in range(len(text_z))]
for i in range(2,len(text_z)+2):
Label_z = ttk.Label(window, text=text_z[i-2], background="white").grid(row= 2*i, column=0,columnspan=3, padx=10, pady=2, sticky="W")
a[i-2] = ttk.Entry(window, width=60, background="gray")
a[i-2].grid(row= 2*i+1, column=0, columnspan=3, padx=10, sticky="WE")
ttk.Button(window, text="Search", width=10, command=lambda m=i-2:click(m)).grid(row= 2*i+1, column=3, padx=5, sticky="W")
window.mainloop()
edited Nov 14 '18 at 14:03
answered Nov 14 '18 at 13:46
Miraj50Miraj50
2,7701924
2,7701924
add a comment |
add a comment |
I think you should simplify things a bit by use a list to store your entry fields.
To do this I think it would be best to add frames for each set of widgets and to use the index of range to get what we need.
I have changed up your code a little to make it easier to work with list index as well as added a button that will print out each selected path on each entry field to show these values are accessible.
import tkinter as tk
from tkinter import ttk, filedialog
LARGE_FONT = ("Arial", 12)
MEDIUM_FONT = ("Arial", 11)
REGULAR_FONT = ("Arial", 10)
text_z = ["Select file 1", "Select the file 2", "Select file 3", "Select file 4"]
window = tk.Tk()
def click(x):
z = tk.filedialog.askopenfilename(initialdir="/", title="Select file", filetypes=(("Excel file", "*.xlsx"), ("All files", "*.*")))
a[x].insert(tk.END, z)
ttk.Label(window, text="file load", font=LARGE_FONT, background="white").grid(row=1, column=0, padx=20, pady=10, sticky="w")
a=
for i in range(len(text_z)):
frame = tk.Frame(window)
frame.grid(row=i+2, column=0, sticky="nsew")
ttk.Label(frame, text=text_z[i], background="white").grid(row=0, column=0, columnspan=3, padx=10, pady=2, sticky="w")
a.append(ttk.Entry(frame, width=60, background="gray"))
a[i].grid(row=1, column=0, columnspan=3, padx=10, sticky="ew")
ttk.Button(frame, text="Search", width=10, command=lambda x=i: click(x)).grid(row=1, column=3, padx=5, sticky="w")
def pring_current_paths():
for ndex, entry in enumerate(a):
print("Entry {}: ".format(ndex, entry.get()))
tk.Button(window, text="Print gurrent paths!", command=pring_current_paths).grid()
window.mainloop()
add a comment |
I think you should simplify things a bit by use a list to store your entry fields.
To do this I think it would be best to add frames for each set of widgets and to use the index of range to get what we need.
I have changed up your code a little to make it easier to work with list index as well as added a button that will print out each selected path on each entry field to show these values are accessible.
import tkinter as tk
from tkinter import ttk, filedialog
LARGE_FONT = ("Arial", 12)
MEDIUM_FONT = ("Arial", 11)
REGULAR_FONT = ("Arial", 10)
text_z = ["Select file 1", "Select the file 2", "Select file 3", "Select file 4"]
window = tk.Tk()
def click(x):
z = tk.filedialog.askopenfilename(initialdir="/", title="Select file", filetypes=(("Excel file", "*.xlsx"), ("All files", "*.*")))
a[x].insert(tk.END, z)
ttk.Label(window, text="file load", font=LARGE_FONT, background="white").grid(row=1, column=0, padx=20, pady=10, sticky="w")
a=
for i in range(len(text_z)):
frame = tk.Frame(window)
frame.grid(row=i+2, column=0, sticky="nsew")
ttk.Label(frame, text=text_z[i], background="white").grid(row=0, column=0, columnspan=3, padx=10, pady=2, sticky="w")
a.append(ttk.Entry(frame, width=60, background="gray"))
a[i].grid(row=1, column=0, columnspan=3, padx=10, sticky="ew")
ttk.Button(frame, text="Search", width=10, command=lambda x=i: click(x)).grid(row=1, column=3, padx=5, sticky="w")
def pring_current_paths():
for ndex, entry in enumerate(a):
print("Entry {}: ".format(ndex, entry.get()))
tk.Button(window, text="Print gurrent paths!", command=pring_current_paths).grid()
window.mainloop()
add a comment |
I think you should simplify things a bit by use a list to store your entry fields.
To do this I think it would be best to add frames for each set of widgets and to use the index of range to get what we need.
I have changed up your code a little to make it easier to work with list index as well as added a button that will print out each selected path on each entry field to show these values are accessible.
import tkinter as tk
from tkinter import ttk, filedialog
LARGE_FONT = ("Arial", 12)
MEDIUM_FONT = ("Arial", 11)
REGULAR_FONT = ("Arial", 10)
text_z = ["Select file 1", "Select the file 2", "Select file 3", "Select file 4"]
window = tk.Tk()
def click(x):
z = tk.filedialog.askopenfilename(initialdir="/", title="Select file", filetypes=(("Excel file", "*.xlsx"), ("All files", "*.*")))
a[x].insert(tk.END, z)
ttk.Label(window, text="file load", font=LARGE_FONT, background="white").grid(row=1, column=0, padx=20, pady=10, sticky="w")
a=
for i in range(len(text_z)):
frame = tk.Frame(window)
frame.grid(row=i+2, column=0, sticky="nsew")
ttk.Label(frame, text=text_z[i], background="white").grid(row=0, column=0, columnspan=3, padx=10, pady=2, sticky="w")
a.append(ttk.Entry(frame, width=60, background="gray"))
a[i].grid(row=1, column=0, columnspan=3, padx=10, sticky="ew")
ttk.Button(frame, text="Search", width=10, command=lambda x=i: click(x)).grid(row=1, column=3, padx=5, sticky="w")
def pring_current_paths():
for ndex, entry in enumerate(a):
print("Entry {}: ".format(ndex, entry.get()))
tk.Button(window, text="Print gurrent paths!", command=pring_current_paths).grid()
window.mainloop()
I think you should simplify things a bit by use a list to store your entry fields.
To do this I think it would be best to add frames for each set of widgets and to use the index of range to get what we need.
I have changed up your code a little to make it easier to work with list index as well as added a button that will print out each selected path on each entry field to show these values are accessible.
import tkinter as tk
from tkinter import ttk, filedialog
LARGE_FONT = ("Arial", 12)
MEDIUM_FONT = ("Arial", 11)
REGULAR_FONT = ("Arial", 10)
text_z = ["Select file 1", "Select the file 2", "Select file 3", "Select file 4"]
window = tk.Tk()
def click(x):
z = tk.filedialog.askopenfilename(initialdir="/", title="Select file", filetypes=(("Excel file", "*.xlsx"), ("All files", "*.*")))
a[x].insert(tk.END, z)
ttk.Label(window, text="file load", font=LARGE_FONT, background="white").grid(row=1, column=0, padx=20, pady=10, sticky="w")
a=
for i in range(len(text_z)):
frame = tk.Frame(window)
frame.grid(row=i+2, column=0, sticky="nsew")
ttk.Label(frame, text=text_z[i], background="white").grid(row=0, column=0, columnspan=3, padx=10, pady=2, sticky="w")
a.append(ttk.Entry(frame, width=60, background="gray"))
a[i].grid(row=1, column=0, columnspan=3, padx=10, sticky="ew")
ttk.Button(frame, text="Search", width=10, command=lambda x=i: click(x)).grid(row=1, column=3, padx=5, sticky="w")
def pring_current_paths():
for ndex, entry in enumerate(a):
print("Entry {}: ".format(ndex, entry.get()))
tk.Button(window, text="Print gurrent paths!", command=pring_current_paths).grid()
window.mainloop()
answered Nov 14 '18 at 13:50
Mike - SMTMike - SMT
9,55921434
9,55921434
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Sorry @Miraj50, edited the question with those items
– Rafael Castelo Branco
Nov 14 '18 at 13:12