RazorPay 表单在页面加载时自动加载,无需单击“立即付款”按钮
RazorPay form auto load on page load without clicking Pay Now button
我正在尝试将 Razorpay 集成到我的应用程序中。这是一个代码,用于在单击 "Pay Now" 按钮时控制支付模型的加载。我想在页面加载时加载模态而不是单击按钮。我尝试通过 javascript 提交表单以查看它是否加载了模态,但它不起作用。
代码如下:
<form action="verify.php" method="POST" id="gateway">
<script
src="https://checkout.razorpay.com/v1/checkout.js"
data-key="<?php echo $data['key']?>"
data-amount="<?php echo $data['amount']?>"
data-currency="INR"
data-name="<?php echo $data['name']?>"
data-image="<?php echo $data['image']?>"
data-description="<?php echo $data['description']?>"
data-prefill.name="<?php echo $data['prefill']['name']?>"
data-prefill.email="<?php echo $data['prefill']['email']?>"
data-prefill.contact="<?php echo $data['prefill']['contact']?>"
data-notes.shopping_order_id="3456"
data-order_id="<?php echo $data['order_id']?>"
<?php if ($displayCurrency !== 'INR') { ?> data-display_amount="<?php echo $data['display_amount']?>" <?php } ?>
<?php if ($displayCurrency !== 'INR') { ?> data-display_currency="<?php echo $data['display_currency']?>" <?php } ?>
>
document.getElementById("gateway").submit(); // Not working
</script>
<!-- Any extra fields to be submitted with the form but not sent to Razorpay -->
<input type="hidden" name="shopping_order_id" value="3456">
</form>
但是,我还注意到页面 verify.php
的表单操作,因此如果提交表单,它将转到我只想加载支付模式的页面。因此,我尝试的方法不能成为解决方案。
我找到了解决方案。发布可能对以后的人有帮助:
在 razorpay/checkout/automatic.php (or manual.php)
中,将其紧跟在 <form>
元素之后:
<script>
$(window).on('load', function() {
$('.razorpay-payment-button').click();
});
</script>
<script src="{{url('/')}}/frontendtheme/js/plugins/jquery-3.3.1.min.js"></script>
<script>
$(window).on('load', function() {
jQuery('#gateway').submit();
});
</script>
注意:使用脚本源上的 min.js 文件夹
将 onload 事件附加到脚本并在处理程序中模拟点击
<script>
function loadPaymentModal() {
const elem = document.getElementsByClassName('razorpay-payment-button')[0];
elem.click();
}
</script>
<form action="verify.php" method="POST" id="gateway">
<script onLoad="loadPaymentModal()" src="https://checkout.razorpay.com/v1/checkout.js"
data-key="<?php echo $data['key']?>"
data-amount="<?php echo $data['amount']?>"
data-currency="INR"
data-name="<?php echo $data['name']?>"
data-image="<?php echo $data['image']?>"
data-description="<?php echo $data['description']?>"
data-prefill.name="<?php echo $data['prefill']['name']?>"
data-prefill.email="<?php echo $data['prefill']['email']?>"
data-prefill.contact="<?php echo $data['prefill']['contact']?>"
data-notes.shopping_order_id="3456"
data-order_id="<?php echo $data['order_id']?>"
<?php if ($displayCurrency !=='INR' ) { ?> data - display_amount="<?php echo $data['display_amount']?>" <? php } ?>
<? php if ($displayCurrency !== 'INR') { ?> data - display_currency="<?php echo $data['display_currency']?>" <? php } ?>
>
</script>
<!-- Any extra fields to be submitted with the form but not sent to Razorpay -->
<input type="hidden" name="shopping_order_id" value="3456">
</form>
我正在尝试将 Razorpay 集成到我的应用程序中。这是一个代码,用于在单击 "Pay Now" 按钮时控制支付模型的加载。我想在页面加载时加载模态而不是单击按钮。我尝试通过 javascript 提交表单以查看它是否加载了模态,但它不起作用。
代码如下:
<form action="verify.php" method="POST" id="gateway">
<script
src="https://checkout.razorpay.com/v1/checkout.js"
data-key="<?php echo $data['key']?>"
data-amount="<?php echo $data['amount']?>"
data-currency="INR"
data-name="<?php echo $data['name']?>"
data-image="<?php echo $data['image']?>"
data-description="<?php echo $data['description']?>"
data-prefill.name="<?php echo $data['prefill']['name']?>"
data-prefill.email="<?php echo $data['prefill']['email']?>"
data-prefill.contact="<?php echo $data['prefill']['contact']?>"
data-notes.shopping_order_id="3456"
data-order_id="<?php echo $data['order_id']?>"
<?php if ($displayCurrency !== 'INR') { ?> data-display_amount="<?php echo $data['display_amount']?>" <?php } ?>
<?php if ($displayCurrency !== 'INR') { ?> data-display_currency="<?php echo $data['display_currency']?>" <?php } ?>
>
document.getElementById("gateway").submit(); // Not working
</script>
<!-- Any extra fields to be submitted with the form but not sent to Razorpay -->
<input type="hidden" name="shopping_order_id" value="3456">
</form>
但是,我还注意到页面 verify.php
的表单操作,因此如果提交表单,它将转到我只想加载支付模式的页面。因此,我尝试的方法不能成为解决方案。
我找到了解决方案。发布可能对以后的人有帮助:
在 razorpay/checkout/automatic.php (or manual.php)
中,将其紧跟在 <form>
元素之后:
<script>
$(window).on('load', function() {
$('.razorpay-payment-button').click();
});
</script>
<script src="{{url('/')}}/frontendtheme/js/plugins/jquery-3.3.1.min.js"></script>
<script>
$(window).on('load', function() {
jQuery('#gateway').submit();
});
</script>
注意:使用脚本源上的 min.js 文件夹
将 onload 事件附加到脚本并在处理程序中模拟点击
<script>
function loadPaymentModal() {
const elem = document.getElementsByClassName('razorpay-payment-button')[0];
elem.click();
}
</script>
<form action="verify.php" method="POST" id="gateway">
<script onLoad="loadPaymentModal()" src="https://checkout.razorpay.com/v1/checkout.js"
data-key="<?php echo $data['key']?>"
data-amount="<?php echo $data['amount']?>"
data-currency="INR"
data-name="<?php echo $data['name']?>"
data-image="<?php echo $data['image']?>"
data-description="<?php echo $data['description']?>"
data-prefill.name="<?php echo $data['prefill']['name']?>"
data-prefill.email="<?php echo $data['prefill']['email']?>"
data-prefill.contact="<?php echo $data['prefill']['contact']?>"
data-notes.shopping_order_id="3456"
data-order_id="<?php echo $data['order_id']?>"
<?php if ($displayCurrency !=='INR' ) { ?> data - display_amount="<?php echo $data['display_amount']?>" <? php } ?>
<? php if ($displayCurrency !== 'INR') { ?> data - display_currency="<?php echo $data['display_currency']?>" <? php } ?>
>
</script>
<!-- Any extra fields to be submitted with the form but not sent to Razorpay -->
<input type="hidden" name="shopping_order_id" value="3456">
</form>