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[Tkinter] Tkinter window pop up again when i click button
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[Tkinter] Tkinter window pop up again when i click button
#1
Hi guys, im new at python and new at programming code.
i got confused when run a little program that i created on.
so it pop up again when i click "run service", did i miss something ? thx.
[Image: ss.png]
source code : https://kuronekosan.web.id/source.txt

from tkinter import *
# from Outdated_Versions import server as outdated_Server



class HttpServer:

    def __init__(self,root):
        self.LabelProject = Label(root, text="HTTP Server Project Using Python")
        self.LabelProject2 = Label(root, text="Kelompok 2 - HTTP")
        self.label_1 = Label(root, text="IP Address / HOST")
        self.label_2 = Label(root, text="PORT")

        # self.entryIP = StringVar()
        # self.entryIP = IntVar()

        self.entryIP = Entry(root)
        self.entryPORT = Entry(root)

        # self.EntryIP_Str =
        # self.EntryPORT_Int =
        #
        # Normally, the characters that the
        # user types appear in the entry.To
        # make a.password.entry that echoes
        # each character as an asterisk, set
        # show = "*".
        #
        self.chkFrame = BooleanVar()
        self.FrameSettingsModule = Checkbutton(root, var=self.chkFrame )
        self.labelFrameSettingsModule = Label(root, text="Use FLASK Framework")
        self.chkDefault = BooleanVar()
        self.defaultSettingsModule = Checkbutton(root, var=self.chkDefault, command=self.cacheDefault )
        self.labeldefaultSettingsModule = Label(root, text="Use Default Settings")
        self.chkSave = BooleanVar()
        self.saveSettingsModule = Checkbutton(root, var=self.chkSave)
        self.labelsaveSettingsModule = Label(root, text="Save Settings")
        self.button_Process = Button(root, text="Run Service", fg="green", command=self.ButtonPress )
        self.button_Exit = Button(root, text="Stop Service", fg="red")


        self.LabelProject.grid(row=0, columnspan=2)
        self.LabelProject2.grid(row=1, columnspan=2)
        self.label_1.grid(row=2, sticky=W)
        self.label_2.grid(row=3, sticky=W)
        self.entryIP.grid(row=2, column=1)
        self.entryPORT.grid(row=3, column=1)
        self.FrameSettingsModule.grid(row=4)
        self.labelFrameSettingsModule.grid(row=4, column=1)
        self.defaultSettingsModule.grid(row=5)
        self.labeldefaultSettingsModule.grid(row=5, column=1)
        self.saveSettingsModule.grid(row=6)
        self.labelsaveSettingsModule.grid(row=6, column=1)
        self.button_Exit.grid(row=7, column=1)
        self.button_Process.grid(row=7)

    def cacheDefault(self):
        self.varCheckDefault = self.chkDefault.get()
        self.varCheckFrame = self.chkFrame.get()

        if self.varCheckDefault == True:
            self.entryIP.delete(0, END)
            self.entryIP.insert(0, "localhost")
            self.entryIP.config(state='disabled')
            self.entryPORT.delete(0, END)
            self.entryPORT.insert(0, 80)
            return
        else:
            self.entryIP.config(state='normal')
            self.entryIP.delete(0, END)
            self.entryPORT.delete(0, END)

    def ButtonPress(self):
        self.varCheckDefault = self.chkDefault.get()
        self.varCheckFrame = self.chkFrame.get()
        IP = self.entryIP.get()
        PORT = self.entryPORT.get()
        if self.varCheckFrame == True:
            if IP == 'localhost':
                IP = '127.0.0.1'
            self.Framework(IP,PORT)
        else:
            if self.varCheckDefault == True:
                # self.callServer(PORT)
                pass
            else:
                print("You need Checklist the Default Settings with the Old Version!")




    def callServer(self,PORT):
        from http.server import HTTPServer, BaseHTTPRequestHandler
        class Server(BaseHTTPRequestHandler):
            def do_GET(self):
                if self.path == '/':
                    self.path = '/index.html'
                try:
                    file_to_open = open(self.path[1:]).read()
                    self.send_response(200)
                except:
                    file_to_open = "File not found"
                    self.send_response(404)
                self.end_headers()
                self.wfile.write(bytes(file_to_open, 'utf-8'))
        httpd = HTTPServer(('',PORT), Server)
        httpd.serve_forever()

    def Framework(self,IP,PORT):
        IP = self.entryIP.get()
        PORT = self.entryPORT.get()
        try:
            from flask import Flask
            app = Flask(__name__)
            @app.route("/")
            def main():
                return "Welcome to my Flask page!"
            if __name__ == "__main__":
                app.run(debug=True, host=IP, port=PORT)
        except ModuleNotFoundError:
            print('Your Flask Framework is not installed!, please install it with "pip3 install flask"')
        except:
            print('Error! please check the code and connection!')

master = Tk()
frame = Frame(master, width=125, height=35)
master.resizable(False, False)
master.title("HTTP Server")

frame.grid()


callInit = HttpServer(master)
master.mainloop()
Reply
#2
worked for me with the callServer function commented.
I'm using linux, 3.5 idle
Did you run from console? or pycharm only?
I did notice a problem, when you get the port entry widget it's returning a string and port needs to be converted to integer. (int(PORT)) then the web server threw an error asking for permission.
Reply


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