我为我的网站提供了一个页面结帐,因此我决定将它分离为5个部分,并使用javascript一次显示一个或两个显示。当页面重新载入时Javascript更改重置
看到下面的代码:
<fieldset id="vm-fieldset-pricelist" class="vm-fieldset-pricelist"> ... </fieldset>
<div id="coupon-section" class="coupon-section display-none"> ... </div>
<fieldset id="shipment-section" class="shipment-section display-none"> ... </fieldset>
<div id="edit-account-section" class="billto-shipto display-none"> ... </div>
<div id="part-five" class="display-none"> ... </div>
<?php
$user = JFactory::getUser();
if ($user->guest) {
$usercheckin = '_notloggedin';
} else {
$usercheckin = '_loggedin';
}
?>
<button type="button" id="one-to-two<?php echo $usercheckin ?>" class="sabz-button pull-right" onclick="cartButtonHandler(this.id); return false;"><?php echo vmText::_('COM_VIRTUEMART_CHECKOUT_TITLE')?> </button>
<script>
window.cartButtonHandler = function(button_id) {
switch(button_id) {
case 'one-to-two_notloggedin':
document.getElementById("vm-fieldset-pricelist").className = "display-none";
document.getElementById("com-form-login").className = "";
document.getElementById(button_id).innerHTML = "ثبت اطلاعات و ادامه خرید";
document.getElementById(button_id).className = "abi-button pull-right";
document.getElementById(button_id).id = 'two-to-three';
break;
case 'one-to-two_loggedin':
document.getElementById("vm-fieldset-pricelist").className = "display-none";
document.getElementById("edit-account-section").className = "billto-shipto display-none";
document.getElementById("shipment-section").className = "shipment-section";
document.getElementById(button_id).innerHTML = "ثبت اطلاعات و ادامه خرید";
document.getElementById(button_id).className = "abi-button pull-right";
document.getElementById(button_id).id = 'two-to-three';
break;
case 'two-to-three':
document.getElementById("edit-account-section").className = "display-none";
document.getElementById("shipment-section").className = "display-none";
document.getElementById("coupon-section").className = "coupon-section";
document.getElementById(button_id).innerHTML = "بازبینی و تایید سفارش";
document.getElementById(button_id).id = 'three-to-four';
break;
case 'three-to-four':
document.getElementById("part-chahar").className = "";
document.getElementById("vm-fieldset-pricelist").className = "vm-fieldset-pricelist";
document.getElementById("edit-account-section").className = "billto-shipto display-none";
document.getElementById("shipment-section").className = "shipment-section";
document.getElementById("coupon-section").className = "coupon-section";
document.getElementById("handlerbutton-section").className = "display-none";
document.getElementById(button_id).id = 'four-to-five';
break;
case 'four-to-five':
alert(button_id);
break;
}
}
</script>
首先,我不知道的JavaScript,有没有更好的办法在JavaScript或jQuery的还是要做到这一点...?
我的问题:当用户在“优惠券区”输入优惠券代码并点击提交按钮时,介意优惠券提交按钮与NEXT STEP按钮不同,页面重新加载并弹出第一步。 或当用户选中复选框或任何其他交互时,页面重新加载并弹出第一步。
如何保存javascript更改并将它们重新载入页面重新加载,以便用户可以在与表单交互后保留在同一步骤中? 我想这可能是使用cookie或其他方法解决的,但我不知道该怎么做。
本地存储或会话存储可以帮助你 –
如果你不想重新加载页面停止提交按钮默认动作 –
你不怎么编码的JavaScript? – Beginner