Empresas
Empregos
  • Sobre nós
  • Soluções
    • Publicação de vagas
      Publique sua vaga e receba candidatos qualificados em 48h.
    • Avaliações de candidatos
      Mais de 500 testes técnicos e psicológicos, mais anti-fraude.
    • Headhunting
      Busca executiva personalizada do início ao fim.
    • Folha de Pagamento + EOR
      Dispersão de folha e EOR em mais de 15 países da LATAM.
  • Preços
  • Empregos

0

170
Visualizações
Console error saying variable not defined, even though defined in same function

I have a variable called grid which contains variables called elements.

On click of a button I'd like to remove the grid and replace it with a fresh one. This is done via an event listener, a function and an if statement.

However I'm getting a console error telling me "grid is not defined".

Can anyone help?

//Create preset variables
let body = document.querySelector('body');
let content = document.querySelector('.content');
let numberOfSquares = 16;
let randomColorValue = '';
let colorShade = 10;
let buttonClicks = 0;
/* let grid = document.createElement('div'); */

//Create variable called container that contains a div
let container = document.createElement('div');
container.classList.add('container');
body.appendChild(container);

// Add a button &  position to top of screen
let button = document.createElement('button');
button.textContent = "Refresh!";
button.classList.add('refresh-button');
content.appendChild(button);
button.addEventListener('click', refreshGrid);

squaresInGrid();

function squaresInGrid() {
    if (buttonClicks > 0) {
        console.log(buttonClicks);
        container.removeChild(grid);
    }

    for (i = 0; i < numberOfSquares; ++i) {
        let grid = document.createElement('div');
        grid.classList.add('grid');
        container.appendChild(grid);

         let element = document.createElement('div');
         element.classList.add('grid-item');
         element.addEventListener('mouseenter', function mouseEnterFunctions() {
             randomColor();
             hoverStyle();
            --colorShade;
            element.removeEventListener('mouseenter', mouseEnterFunctions);
         });

        function hoverStyle() {
            element.style.backgroundColor = '#'+randomColorValue;
            element.style.opacity = (colorShade / 10); 
        }

        function randomColor() {
            randomColorValue = Math.floor(Math.random()*16777215).toString(16);
        }

        grid.appendChild(element);                
    }
} 

//Function to be called on button click removing existing grid and replacing with a new one
function refreshGrid() {
    ++buttonClicks;
    numberOfSquares = prompt("How many squares would you like? (Maximum 100)");
    colorShade = 0;
    squaresInGrid();
}
about 4 years ago · Juan Pablo Isaza
1 Respostas
Responde à pergunta

0

You defined your grid variable in the for loop, so it's not accessible anywhere but inside the loop.

You should instead uncomment your line at the top of your code to make it a global variable. Then in the for loop, remove the let keyword in front of your variable for grid so that it always references the outer variable.

about 4 years ago · Juan Pablo Isaza Relatório
Responde à pergunta
Encontrar trabalhos remotos

Descubra a nova forma de encontrar um emprego!

melhores empregos
Principais categorias de trabalho
Empresas
Postar vaga Preços Comercial
Jurídico
Termos e Condições Política de privacidade
© 2026 PeakU Inc. All Rights Reserved.
Andres GPT
Recomende algumas ofertas para mim
Preciso de ajuda