如何在尝试使其动态化的同时显示我的模态信息？

Question

在单击特定图像后，尝试打开一个模式，该模式应该从我的数据库中获取信息并将其呈现到模式上。模态弹出，但它显示在 html 输入上并且没有任何 php 输入。我怎样才能解决这个问题？ Rental_id 是数据库表中的主表。

<?php
 require_once '../core/init.php';
$id = $_POST['rental_id'];
$id = (int)$id;
$sql = "SELECT * FROM rental WHERE rental_id = '$id'";
$result = $db->query($sql);
$rental = mysqli_fetch_assoc($result);

 ?>

 <!-- the div that represents the modal for the form -->
 <?php ob_start();?>
            <div class="modal fade" id="quote" tabindex="-1" role="dialog" aria-labelledby="quote" aria-hidden="true">
                <div class="modal-dialog modal-lg">
                    <div class="modal-content">
                        <div class="modal-header">
                            <button class="close" type="button" data-dismiss='modal' aria-label="Close">
                                <span aria-hidden="true">&times;</span>
                            </button>
                            <h4 class="modal-title text-center">Quote</h4>
                        </div>
                        <div class="modal-body">
                            <div class="container-fluid">
                                <div class="row">
                                <div class="col-sm-6">
                                <div class="center-block">
                                    <img src="<?= $rental['img']; ?>" alt="<?= $rental['make']; ?>" class="w3-image img-responsive"/>
                                </div>...

还有……

<script>
    function detailsmodal(rental_id){
        var data = {"rental_id" : rental_id};
        jQuery.ajax({
            url : <?php echo BASEURL;?> + 'includes/detailsmodal.php',
            method : "post",
            data : data,
            success: function(data){
                jQuery('body').append(data);
                jQuery('#quote').modal('show');
            },
            error: function(){
                alert("Something went wrong!");
            }
        });
    }
</script>

Answer 1

你检查过 $id = (int)$id; ?它设置好了吗？ 任何返回错误信息？ (SO question)

<?php     
print_r($rental); /* seems ok when tested with ie: $id=3 */

$sql = " SELECT img, make FROM rental WHERE rental_id=? ";
$stmt = $mysqli->prepare($sql);
$stmt->bind_param("s", $id);

$results = $stmt->execute();
$stmt->bind_result($img, $make);
$stmt->store_result();

while($stmt->fetch()){
echo"[ $img / $make ]";
}
?>

有区别吗？