我正在尝试做类似网站的操作,每当用户单击某种产品下的购买时,他们都将被带到购物车,然后他们将在那里看到其产品的图像,我希望图像从衬衫上改变运动鞋,但事实并非如此。因此,我想到了使用localstorage的想法,并在我的addToCart脚本文件中做了以下操作:
$(document).ready(function()
{
var isShoes = false
var isShirt = false
$('#purchaseShoes').on('click', function()
{
alert('Sneakers')
var Shoes = localStorage.setItem('img_shoes', '/static/Images/Sneakers.png')
isShoes = localStorage.setItem('isShoes', true)
})
$('#purchaseShirt').on('click', function()
{
alert('T-shirt')
var Shirt = localStorage.setItem('img_shirt', '/static/Images/Images/T-Shirt.png')
isShirt = localStorage.setItem('isShoes', true)
})
})
但是很明显,脚本将两个按钮都注册为单击状态,并且每当我转到我的cartControl脚本并对其进行测试时,它都将我的两个值都恢复为true。当我仅单击其中一个按钮时,为什么同时显示两个?我检查了HTML和他们的ID是不同的。我应该在addToScript脚本中使用if吗??? 这是cartControl脚本:
$(document).ready(function()
{
var Shoes = localStorage.getItem('img_shoes')
var Shirt = localStorage.getItem('img_shirt')
var isShirt = localStorage.getItem('isShirt')
var isShoes = localStorage.getItem('isShoes')
var quantityNumber = 1;
var shirtPrice = 25.00;
var shirt = document.getElementById('shirtTotal')
var Quantity = document.getElementById('Quantity')
alert(isShoes)
alert(isShirt)
$('#btnQuantity').on('click', function()
{
quantityNumber+=1;
Quantity.innerText = quantityNumber;
shirt.innerText = "Total Price: " + "$" + (quantityNumber*shirtPrice);
})
})