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Keyron's questions

Martin Hope
Keyron
Asked: 2022-10-25 07:58:53 +0800 CST
  • 1

Hello! I have a website as a portfolio which I have uploaded to github pages, the website works impeccably, it has no problems except for one detail, everything works in my copy hosted on my computer, including what makes me most complex, which is that I have a function in js that measures distances between the menu and the top of the page to maintain its position and lower it when scrolling.

All good so far, but when uploading it to github pages, even though the page is clearly visible, the js functions don't work.

I leave the code of the function and the link of my github pages

Link code of my portfolio:

https://github.com/K3yr0nym0us/K3yr0nym0us.github.io

I also leave the link of my page visible in github pages

https://k3yr0nym0us.github.io/Index.html

$(document).ready(function(){
    /*creamos una variable llamada altura que contendra
    el valor en pixeles de la diferencia que hay entre
    el objeto de clase menu y el top de la pantalla*/
    var altura = $('.menu').offset().top;

    $(window).on('scroll', function(){
        /*si el scroll del sitio es mayor a altura*/
        if ( $(window).scrollTop() > altura ){
            /*entonces se añade la clase menu-fixed*/
            $('.menu').addClass('menu-fixed');
        } else {
            /*sino se remueve la clase*/
            $('.menu').removeClass('menu-fixed');
        }
    })
});

It should be noted that I uploaded the code exactly as it works on my computer and it works on it but on github pages it does not. I leave the code of my head where I have the call of my js.

<head>
    <meta charset="UTF-8">
    <title>Portafolio</title>
    <link rel="stylesheet" href="css/reset.css">
    <link rel="stylesheet" href="css/header.css">
    <link rel="stylesheet" href="css/index/index.css">
    <link rel="stylesheet" href="css/pie.css">
    <script src="http://code.jquery.com/jquery-latest.js"></script>
    <script src="js/main.js"></script>
</head>

any advice or criticism is welcome

javascript
Martin Hope
Keyron
Asked: 2022-08-09 15:43:55 +0800 CST
  • 0

hello! edit most of it to give executable and understandable code. I have the following code:

    from tkinter import * 
    
    class Tienda:
    
        def __init__(self,ventana):
            self.ventana=ventana
            self.ventana.title("Mi Tiendita") 
    
            #Marco de ventana
            marco=LabelFrame(self.ventana,text="Ingresar Producto")    
            marco.grid(columns=2,pady=10,padx=10,columnspan=4) 
    
            #Nombre
            Label(marco,text="Nombre").grid(row=0,column=0) 
            self.Nombre=Entry(marco)
            self.Nombre.grid(row=0,column=1) 
            self.Nombre.focus() 
    
            #marco2
            marco2=LabelFrame(self.ventana)    
            marco2.grid(column=0,columnspan=4)
    
            #BotonGuardar
            self.guardar=Button(marco2,text="Guardar Datos",command=self.agregarRegistro) 
            self.guardar.grid(row=2,column=0)
    
        def agregarRegistro(self):
            contenido = self.Nombre.get()
            print(contenido)
    
    if __name__=="__main__":
        ventana=Tk() 
        aplicacion=Tienda(ventana)
        ventana.mainloop()

Basically what I want to do is get the addRecord function from inside the class. I've tried several things but I can't get it to work since taking it out of the class I lose the use of self. Can someone guide me on how to replace the self to be able to execute the function from outside the class?

python-3.x