L6 Python Tkinter Load Entry 1 Open Button db_open() and filedialog.askopenfilename under Windows
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def db_open():
s1 = "script dir (db_location): "
s2 = app_executed_script_dir_path
l is the identifier
for the memory space
in which
there is an instance
of code of object type Label
l = Label(w, text=_s_)
l.pack(pady=5)
l.place(x=9, y=576)
l.config(text=s1+s2)
selected_filename = filedialog.askopenfilename(
filetypes=[("Text files","*.txt")],
initialdir=app_executed_script_dir_path,
title="e1 Open Keywords Fille")
if(selected_filename != None):
if (selected_filename.find('.txt') !=-1):
f = open(selected_filename, encoding ='utf-8')
l1.config(text=selected_filename)
set_text(selected_filename, e2)
s=f.read()
selected_basename_name_of_file = os.path.basename(selected_filename)
selected_dirname = os.path.dirname(selected_filename)
set_text(selected_basename_name_of_file, e1)
set_text(s, e2)
b_db_save_as["state"] = "active"
b_db_open = Button(w, text ="db Open", command = db_open)
b_db_open.pack(padx=10, pady=2)
b_db_open.place(x=482, y=573-26)
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