Currently when I write to a label, it will write only one line, and this sometimes disappears off the edge of the label, like so:
Here is an example of the code I am using to write to the label:
def updateLabel(self, event):
global string
global labelContents
global windowCommand
global currentEnvironment
if event != "no input":
windowCommand = self.getEntry(event)
labelDisplay = "> " + windowCommand
labelContents += labelDisplay
labelContents += "\n"
self.checkLabel()
string.set(labelContents)
self.textEntry.delete(0, END)
self.master.after(0, play)
else:
self.checkLabel()
string.set(labelContents)
labelContents += "You have died. Game over." + "\n"
labelContents += "You scored {0}.".format(score) + "\n"
app.updateLabel("no input")
I would like to know if there was any way to force it to a new line after a certain amount of pixels (the label width) without having to go through and add "\n" everywhere (as that last line is 1 of ~150 possibilities).
Label widget has a perfect option for that: wraplengt.
label = Label(parent, text="This is a really long text; " * 5, wraplengt=200)
From the Label's documentation on effbot.org:
Determines when a label’s text should be wrapped into multiple lines. This is given in screen units. Default is 0 (no wrapping).
Related
I have a List which I go through and generate a Button for each element in the List. Now when I press the Button I would like to know what Text is inside the Button. Sadly I was not able to figure out how to do this.
Here is the entire Loop however the Last 3 rows are the important ones.
Listezumsortieren = [*self.root.ids.Kommentar.text]
x = 0
stelle = 0
for i in Listezumsortieren:
if x == 17:
Listezumsortieren.insert(stelle," \n")
x = 0
x = x+1
stelle = stelle + 1
if stelle > 65:
return
Zitatselbst = ''.join(Listezumsortieren)
btn = Button(text = '\n"' + Zitatselbst + '"' + "\n ~" + self.root.ids.field.text + "\n", size_hint = (0.8 , None))
btn.bind(on_press=lambda x:self.kek())
self.root.ids.Zitate.add_widget(btn)
Thanks for helping
To specify because of the first answer:
so for example if I have a list of [1,2,3] and have three buttons with the text: (1,2,3) no matter which button I click the Button will return "3" and not the actual text of the Button, Thanks for trying to help
This is because "Self" is not referring to the Button but rather to the class the Function is in (MainApp) so to be more precise I need to know how to get to the Button
I know my question sounds very much like many previous questions but I can honestly not figure it out in the context of my program. I have an algorithm for the Collatz Conjecture that I would like to run through a Tkinter GUI (everything works just fine through the terminal).I have tried to bind the relevant function to the Return key and to a button but I get the same error message for both methods of entering data, which I will show below. I get the output to work perfectly on the GUI if I input through the terminal.
What I have tried is best explained through the code below. (The code above the #### line has mostly to do with making the GUI appear over my Spyder IDE and not hiding behind it.)
Code:
from tkinter import *
root = Tk()
import os
import subprocess
import platform
def raise_app(root: Tk):
root.attributes("-topmost", True)
if platform.system() == 'Darwin':
tmpl = 'tell application "System Events" to set frontmost of every process whose unix id is {} to true'
script = tmpl.format(os.getpid())
output = subprocess.check_call(['/usr/bin/osascript', '-e', script])
root.after(0, lambda: root.attributes("-topmost", False))
########################################################################
lst = []
def collatz(num):
while num != 1:
lst.append(num)
if num % 2 == 0:
num = int(num / 2)
else:
num = int(3 * num + 1)
def main(event):
collatz(num)
#Input Box
input = Entry(root, width = 10, bg = "light grey")
input.grid(row = 0, column = 0, sticky = W)
input.get()
input.bind("<Return>", main)
##Button
#button1 = Button(root, width = 10, text = "Run", command = main)
#button1.grid(row = 1, column = 0, sticky = W)
##Output box
output1 = Text(root, width = 100, height = 10, bg = "light grey")
output1.grid(row = 3, column = 0, sticky = W)
output2 = Text(root, width = 50, height = 1, bg = "white")
output2.grid(row = 2, column = 0, sticky = W)
output1.insert(END, lst)
output2.insert(END, "Number of iterations are: " + str(len(lst)))
########################################################################
raise_app(root)
root.mainloop()
When I run the code as is, the input box appears but when I click return, I get an error message:
Exception in Tkinter callback
Traceback (most recent call last):
File "/anaconda3/lib/python3.7/tkinter/__init__.py", line 1705, in __call__
return self.func(*args)
File "/Users/andrehuman/Desktop/Python/programs/Collatz Conjecture/Collatz_alt3.py", line 43, in main
collatz(num)
NameError: name 'num' is not defined
Exactly the same if I try and link a button to the "main" function.
When I comment the input and buttons out, and enter the number through the terminal, everything works as expected. The list of iteration numbers appear in the text box as it should. (And I can even get a Matplotlib graph to display the data visually in the terminal.) If I can get this problem sorted out I want to try and display (or embed) the Matplotlib graph in the GUI.
Anyway, that's it. Any help will be greatly appreciated.
Andre Human
name 'num' is not defined occurs because you're calling collatz(num), but the program does not understand what value you are referring to when you say num. You should assign a value to that name before using it. I assume you want the value to be the contents of your input box.
def main(event):
num = int(input.get())
collatz(num)
You will also need to copy your output1.insert and output2.insert lines to the inside of main. Right now, those lines execute before the window even appears to the user, so there's no way that they can enter a number fast enough to get collatz to trigger before the text gets written. And changing lst after the fact does nothing to the text, since it's not smart enough to notice that the list has changed.
def main(event):
num = int(input.get())
collatz(num)
#delete previous contents of text boxes
output1.delete(1.0, END)
output2.delete(1.0, END)
output1.insert(END, lst)
output2.insert(END, "Number of iterations are: " + str(len(lst)))
Another problem is that successive calls to collatz will cause lst to grow and grow, because the contents of the list from previous calls is still present. Try entering 4 into the text box, and press Enter a few times. The output will go from 2 to 4 to 6... That's not right.
This is something of a natural hazard when using mutable global state. One possible solution is to reset lst at the beginning of each collatz call.
def collatz(num):
lst.clear()
#rest of function goes here
... But I'm more inclined to make lst local to the function, and return it at the end.
def collatz(num):
lst = []
while num != 1:
lst.append(num)
if num % 2 == 0:
num = int(num / 2)
else:
num = int(3 * num + 1)
return lst
def main(event):
num = int(input.get())
lst = collatz(num)
#delete previous contents of text boxes
output1.delete(1.0, END)
output2.delete(1.0, END)
output1.insert(END, lst)
output2.insert(END, "Number of iterations are: " + str(len(lst)))
#later, just before mainloop is called...
#lst doesn't exist in this scope, so just set the text to a literal value
output1.insert(END, "[]")
output2.insert(END, "Number of iterations are: 0")
I participated in a local programming competition yesterday, where I was put on a team of four with members of varying skill levels. We were trying to make a text-based Zork style game, but since UI was one of the categories I was put on creating a GUI which I have no experience with in python. I used Tkinter and got everything mostly working inside the GUI itself, but can't figure out how to get the GUI to interact with the rest of the code.
I'm just including the relevant bits here, partially because the rest of the code is quite a mess, with no comments and lots of syntax errors, but I can post it if needed. Here's the code:
def pinput():
global e
player_input = e.get()
print(player_input)
def progress():
hallway_1()
def hallway_1():
global player_input
global prin
global e
prin.set("You are in a hallway. Yellow lockers line the sides of all walls and you see three doors. One leads into a Library, one to a History classroom, and one to a Math room. You can 1.) Search the hallway or 2.) Procede to one of the rooms.")
while player_input != "1" and player_input != "2": #Best guess at a way to get the code to wait for input, actually just causes code to freeze
pass
if player_input == "1": #Even if 1 is in the entry box beforehand, it never gets here
prin.set("You search through the lockers and find 1 health potion.\n Would you like to add it to your inventory? 1.) Yes 2.) No")
player_input = "999"
def player_attack():
global enemy_hp
enemy_hp -= player_strength
prin.set("You just attacked the " + enemy_name + "!")
def player_block():
global block
prin.set("You predict that your enemy is going to attack and decide to block")
block = 1
def player_use():
global player_item_use
global item_name
player_item_use = 100
prin.set("You decide to use one of your items. What do you use?")
while player_turn_input != 0 or 1 or 2 or 3 or 4 or 5 or 6 or 7 or 8 or 9 or 10 or 11 or 12 or 13 or 14 or 15 or 16 or 17 or 18 or 19 or 20 or 21 or 22 or 23 or 24 or 25 or 26:
player_item_use = e.get()
time.sleep(1)
item_name = Inventory[player_item_use]
item(item_name)
def run_away():
global player_hp
global lost_item_1
global lost_item_2
prin.set("You decide to run away from the " + enemy_name + "!\nThis causes you to lose 20 health points and two inventory items.")
player_hp -= 20
if backpack == True:
lost_item_1 = random.randint(1,27)
lost_item_2 = random.randint(1,27)
else:
lost_item_1 = random.randint(1,9)
lost_item_2 = random.randint(1,9)
RemoveInventory(lost_item_1)
RemoveInventory(lost_item_2)
library()
top = tk.Tk()
prin = tk.StringVar()
#path = "smalllibrary.png"
#tkimage = ImageTk.PhotoImage(Image.open(path))
#displayimage = tk.Label(top, image=tkimage).grid(row=0)
inv = tk.Text(top, width="24")
inv.grid(row=0, column=1)
for x in Inventory:
inv.insert("end", str(x) + str(invnum) + '\n')
invnum += 1
healthbar = tk.Label(top, text=("Health: " + str(player_hp))).grid(row=1, column=1)
text = tk.Label(top, bg="black", fg="white", textvariable=prin, justify="left", cursor="box_spiral").grid(row=1, sticky="w")
e = tk.Entry(top)
e.grid(row=2)
e.focus_set()
Attack = tk.Button(top, text ="Attack", activebackground="red", width="10", command = player_attack).grid(row=4, sticky="e")
Block = tk.Button(top, text ="Block", activebackground="red", width="10", command = player_block).grid(row=5, sticky="e")
Use = tk.Button(top, text ="Use Item", activebackground="red", width="10", command = player_use).grid(row=4, column=1, sticky="w")
Run = tk.Button(top, text ="Run", activebackground="red", width="10", command = run_away).grid(row=5, column=1, sticky="w")
Enter = tk.Button(top,text='Enter',command=pinput).grid(row=3)
Progress = tk.Button(top, text="Progress", activebackground="green",
prin.set("REDACTED")#This had my team's full names
top.mainloop()
So, there are a few problems. I added the "progress" button for testing purposes, ideally the game would start in Hallway 1. However, I couldn't figure out where to put the hallway_1() function for this to work. Before the top.mainloop() and the GUI wouldn't open. After it, and e.get() threw TCL errors.
So, I bound it to a button. However, now the code keeps getting stuck in the "while" loop even if 1 is inputted into the entry box before the code runs. I'm perplexed by this because Visual Studio reports player_input as being '1' in the Autos box, but it keeps running through the loop anyway.
You can't do a while loop like that in a GUI event handler. Until you return from the handler function, the GUI doesn't get to update its display, accept user input, or do anything else, which means player_input will never change once you get there. A handler function needs to do one thing, set up anything needed for future handlers, and return immediately.
It's hard to get your head around thinking in terms of the event loop and handler callbacks the first time you write a GUI, but it's absolutely essential; until you do, things seem bizarre and arbitrary.
So, the answer isn't to "wait for input", but to set up a handler that fires when that input is ready. (You can instead set up a handler on an after function so it fires every N milliseconds instead of firing on input, but that usually just makes things more complicated.)
One way to do this is to have the "Progress" button trigger "the next step", and keep track of some state that lets you determine what "the next step" is. An in fact, you already have that state: it's whether the player has already given correct input or not. So:
def pinput():
global player_input
player_input = e.get()
print(player_input)
def progress():
pinput()
if player_input != "1" and player_input != "2":
hallway_1_enter()
else:
hallway_1_doit()
def hallway_1_enter():
global player_input
global prin
global e
prin.set("You are in a hallway. Yellow lockers line the sides of all walls and you see three doors. One leads into a Library, one to a History classroom, and one to a Math room. You can 1.) Search the hallway or 2.) Procede to one of the rooms.")
def hallway_1_doit():
if player_input == "1": #Even if 1 is in the entry box beforehand, it never gets here
# etc.
This is obviously a bit of a clunky design, but it's the smallest change to your existing design that gets you past the first step. Once you get the hang of it, you should be able to improve it from there.
Also notice the global player_input I put in pinput. Your existing function didn't have that, so it was just creating a local variable with the same name, which then goes away immediately; the global never gets changed. Any function that wants to assign to a global needs the global statement.
So I have a currency which is increasing (that system is working fine). The first part updates the label every 100 ms. I have another button which triggers the second function which is supposed to clear the labels from the first. It sets home_status equal to 0 which should in theory run Money.place_forget() to clear the code. I have tested each part individually and it works but when I put the clears inside the elif statement it doesn't. It does not give me any errors, it just simply doesn't do anything (it does print END OF UPDATE HOME so the elif is triggered).
Any suggestions?
def updatehome(self):
print("UPDATE HOME")
global buy_button, home_status, currency
MoneyLabel = Label(self, text = "Money: ")
MoneyLabel.place(x = 5, y = 70)
Money = Label(self, text=currency)
Money.place(x = 50, y = 70)
if (home_status == 1):
self.after(100, self.updatehome)
elif (home_status == 0):
print("END OF UPDATE HOME")
Money.place_forget()
MoneyLabel.place_forget()
def clearhome(self):
print("CLEAR HOME")
global home_status
home_status = 0
you are creating ten labels every second, all stacked on top of each other, but you are only deleting the very last label you create.
I have made a coin flip button that shows the result in Label, every time the button is pressed, the result change according to the function.
def coin_flip(self):
sound_1 = SoundLoader.load('coin.wav')
res = random.randint(1,2)
if sound_1:
sound_1.play()
if res == 1:
self.coin_f_result.text = "HEAD"
else:
self.coin_f_result.text = "TAIL"
What I want to do is, showing the result in the Label, and then, after a second set the Label text as " ". Here's what I tried, but i only get the function calling delayed, and the label text is set directly to " ".
def coin_flip(self):
sound_1 = SoundLoader.load('dice.wav')
res = random.randint(1,2)
if sound_1:
sound_1.play()
if res == 1:
self.coin_f_result.text = "HEAD"
time.sleep(1)
self.coin_f_result.text = " "
else:
self.coin_f_result.text = "TAIL"
time.sleep(1)
self.coin_f_result.text = " "
Never use time.sleep() in an event driven framework such as kivy. It just blocks execution and as you saw, events are not handled. Use Clock.schedule_once() instead. For example, in the same class that has the coin_flip method, define
def reset_label(self, *args):
self.coin_f_result.text = ' '
And at the end of coin_flip() write
Clock.schedule_once(self.reset_label, 1)
For smooth transitions you can pair that with Animation, too.