【发布时间】:2019-11-29 17:15:42
【问题描述】:
我正在尝试实施贝宝快速结帐。 所以我实现了按钮并将结果写入我的数据库。没什么特别的,就是他们给的剧本。
<script>
paypal.Buttons({
createOrder: function(data, actions) {
return actions.order.create({
purchase_units: [{
amount: {
value: '0.01'
}
}]
});
},
onApprove: function(data, actions) {
return actions.order.capture().then(function(details) {
// Call your server to save the transaction
return fetch('recordDatabase.php', {
method: 'post',
mode: "same-origin",
credentials: "same-origin",
headers: {"Content-Type": "application/json"},
body: JSON.stringify({
orderID: data.orderID,
time: details.create_time,
status: details.status,
nom: details.payer.name.given_name,
prenom: details.payer.name.surname,
pays: details.payer.address.country_code,
valeur:details.purchase_units[0].amount.value
})
})
});
}
}).render('#paypal-button-container');
</script>
要记录在数据库中的php:
<?php
$link = connect();
$date = date('Y-m-d H:i:s');
//Receive the RAW post data.
$contentType = isset($_SERVER["CONTENT_TYPE"]) ?trim($_SERVER["CONTENT_TYPE"]) : '';
if ($contentType === "application/json") {
//Receive the RAW post data.
$content = trim(file_get_contents("php://input"));
$decoded = json_decode($content, true);
//If json_decode failed, the JSON is invalid.
if(! is_array($decoded)) {
//echo "error";
} else {
$name = $decoded['nom'];
$time = $decoded['time'];
$id = $decoded['orderID'];
$stat = $decoded['status'];
$pays = $decoded['pays'];
$val = $decoded['valeur'];
$secQuery = "INSERT INTO myDatabase(PSEUDO,PASSWORD,CONNECTION,INSCRIPTION,ANNIVERSAIRE,MAIL,IDPAYPAL,STATPAYPAL,NOMPAYER,PAYS,VALEUR) VALUES ('essai','123456',0,'$date','$time','email@mail','$id','$stat','$name','$pays','$val') ";
if (mysqli_query($link,$secQuery)) {
//echo "ok";
} else {
//echo "error";
}
}
} else {
//echo "error";
}
所以,我数据库中的记录工作正常,但我的问题是:
如何在 javascript 中检索 echo 错误或 ok 以确认用户一切正常,或者是否发生错误。
我尝试了另一种解决方案,将用户从 php 重定向并添加到 php:
header("Location: confirmation web page"); 或echo "<script>window.location = 'confirmation web page'</script>";
但两种解决方案都不起作用。没有重定向发生
【问题讨论】:
标签: javascript php