当我进行贝宝付款时,它不会移动到数据库中。这可以纠正吗?
When I make a paypal payment, it does not move to the database. Can this be corrected?
当我尝试使用 PayPal 付款时出现错误并且按钮不起作用,这是 JavaScript 代码:
$('.paypal-button-row').on('click', function(event) {
var account_id = $('#account_id').text();
var email = $('input[name="email"]').val();
var sesid = $('input[name="sesid"]').val();
$.ajax({
url: 'ajax.php?func=create_order',
data: {
'account_id': account_id,
'email': email,
'sesid': sesid
},
dataType: 'text',
type: 'post',
success: function(response) {
if (response == "success") {
console.log(success);
$(this).off("submit").trigger("submit");
}
}
});
});
这是按钮:
<div id="smart-button-container">
<div style="text-align: center;">
<div id="paypal-button-container"></div>
</div>
</div>
这是 php 按钮代码:
public function create_order() {
$sesid = $_POST["sesid"];
$account_id=(int) $_POST["account_id"];
print_r($account_id);
$user_id =(string) $_COOKIE['user_id'];
echo $user_id;
$accountsell = $this->query_executor("SELECT * FROM accountsell WHERE status=? AND account_id=?","ii",array(1,$account_id))->fetch_assoc();
$accountsell_id = (int) $accountsell["id"];
echo '-'.$accountsell_id;
$email = $_POST["email"];
echo "<br>".$email;
$this->query_executor("INSERT INTO ordered_accounts VALUES(NULL,?,?,?,?,?,0,NOW())","siiss",array($sesid,$accountsell_id,$account_id,$user_id,$email));
var_dump(array($sesid,$accountsell_id,$account_id,$user_id,$email));
$_SESSION["sesid"] = $sesid;
echo "success";
}
This is photo of PayPal pay
这是游戏网站,可以更正吗?
$('.paypal-button-row').on('click'...
这没有任何意义(将您自己的点击处理程序添加到该行是完全错误的),因此请在此处查看正确的示例:https://developer.paypal.com/demo/checkout/#/pattern/server
如果您需要在创建调用中传递额外的参数,请将正文对象添加到提取中。
您的服务器上需要两条相应的路由 return 它们自己的 JSON,一条用于第一个 'Create an order',一条用于 'Capture order',documented here.
捕获命令成功后,在 returning JSON 之前立即将您需要的任何内容写入数据库。
当我尝试使用 PayPal 付款时出现错误并且按钮不起作用,这是 JavaScript 代码:
$('.paypal-button-row').on('click', function(event) {
var account_id = $('#account_id').text();
var email = $('input[name="email"]').val();
var sesid = $('input[name="sesid"]').val();
$.ajax({
url: 'ajax.php?func=create_order',
data: {
'account_id': account_id,
'email': email,
'sesid': sesid
},
dataType: 'text',
type: 'post',
success: function(response) {
if (response == "success") {
console.log(success);
$(this).off("submit").trigger("submit");
}
}
});
});
这是按钮:
<div id="smart-button-container">
<div style="text-align: center;">
<div id="paypal-button-container"></div>
</div>
</div>
这是 php 按钮代码:
public function create_order() {
$sesid = $_POST["sesid"];
$account_id=(int) $_POST["account_id"];
print_r($account_id);
$user_id =(string) $_COOKIE['user_id'];
echo $user_id;
$accountsell = $this->query_executor("SELECT * FROM accountsell WHERE status=? AND account_id=?","ii",array(1,$account_id))->fetch_assoc();
$accountsell_id = (int) $accountsell["id"];
echo '-'.$accountsell_id;
$email = $_POST["email"];
echo "<br>".$email;
$this->query_executor("INSERT INTO ordered_accounts VALUES(NULL,?,?,?,?,?,0,NOW())","siiss",array($sesid,$accountsell_id,$account_id,$user_id,$email));
var_dump(array($sesid,$accountsell_id,$account_id,$user_id,$email));
$_SESSION["sesid"] = $sesid;
echo "success";
}
This is photo of PayPal pay
这是游戏网站,可以更正吗?
$('.paypal-button-row').on('click'...
这没有任何意义(将您自己的点击处理程序添加到该行是完全错误的),因此请在此处查看正确的示例:https://developer.paypal.com/demo/checkout/#/pattern/server
如果您需要在创建调用中传递额外的参数,请将正文对象添加到提取中。
您的服务器上需要两条相应的路由 return 它们自己的 JSON,一条用于第一个 'Create an order',一条用于 'Capture order',documented here.
捕获命令成功后,在 returning JSON 之前立即将您需要的任何内容写入数据库。