如果数据库有变化，刷新页面

Question

如果使用 Ajax 和 PHP 的数据库中的 orderStatus 发生变化，我正在尝试刷新我的页面。我将当前 orderStatus 设置为预定义数据，然后使用 Ajax 从数据库中获取当前 orderStatus，最后比较它们是否不相同。如果它们不一样，我想刷新页面。

PHP (autorefresh.php)

<?php

$orderId = $_POST["orderId"];

$query = "SELECT * FROM orderinhomeonlinecall WHERE orderId='$orderId'";
$result = mysqli_query($db, $query);

  while($row = mysqli_fetch_array($result))
  {
      $orderStatus = $row['orderStatus'];

      $data = array(
        'orderStatus'   => $orderStatus
       );
       echo json_encode($data);
}
?>

Javascript

<script type="text/javascript" >
var predefined_val = '<?php echo $orderStatus; ?>';// your predefined value.
$.document(ready(function(){
    setInterval(function(){
        $.ajax({
            type:"POST",
            url:"autorefresh.php", //put relative url here, script which will return php
            data:{orderId: <?php echo $orderId; ?>}, // if any you would like to post any data
            success:function(response){
                var data = response; // response data from your php script
                if(predefined_val !== data){
                    window.location.href=window.location.href;
                }
            }
        });                     
    },5000);// function will run every 5 seconds
}));

Answer 1

下面的代码应该可以工作，需要提及dataType:"json" 否则使用JSON.stringify(data) 来解析响应

<script type="text/javascript">
var predefined_val = '<?php echo $orderStatus; ?>';// your predefined value.
$(document).ready(function () {
    setInterval(function () {
        $.ajax({
            type: "POST",
            url: "autorefresh.php", //put relative url here, script which will return php
            data: {orderId: <?php echo $orderId; ?>}, // if any you would like to post any data
            dataType: "json",
            success: function (response) {
                var data = response; // response data from your php script
                if (predefined_val !== data.orderStatus) {
                    window.location.href = window.location.href;
                }
            }
        });
    }, 5000);// function will run every 5 seconds
});
</script>

Answer 2

我已经通过创建两个文件（autorefresh.php，index.php）并使用表测试数据库来测试它，它对我有用。我认为下面的代码会有所帮助，如果没有请分享您的代码，我会检查并修复它。

autorefresh.php

// Create connection
$db = new mysqli("localhost", "root", "","test");

$orderId = $_POST["orderId"];

$query = "SELECT * FROM orderinhomeonlinecall WHERE orderId='$orderId'";

$result = mysqli_query($db, $query);

  while($row = mysqli_fetch_array($result))
  {
      $orderStatus = $row['orderStatus'];

      $data = array(
        'orderStatus'   => $orderStatus
       );
       echo json_encode($data);
}
?>

index.php

<?php
$orderStatus ='pending';
$orderId =1;
?>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.4.1/jquery.min.js"></script>
<script type="text/javascript">
var predefined_val = '<?php echo $orderStatus; ?>';// your predefined value.
$(document).ready(function () {
    setInterval(function () {
        $.ajax({
            type: "POST",
            url: "autorefresh.php", //put relative url here, script which will return php
            data: {orderId: <?php echo $orderId; ?>}, // if any you would like to post any data
            dataType: "json",
            success: function (response) {
                var data = response; // response data from your php script
                if (predefined_val !== data.orderStatus) {
                    window.location.href = window.location.href;
                }
            }
        });
    }, 5000);// function will run every 5 seconds
});
</script>