显示项目后如何将其添加到数据库表中？-解网

问：

这是我用来显示项目的display.php

<?php
    session_start();
 
    if($_SESSION['status']!="Active")
    {
        header("location:index.php");
    }
        
?>

<?php
    
    include 'dbcon.php';
    $id = 1;
    $squery= mysqli_query($con,"select * from menuitems");
    while ($sq=mysqli_fetch_assoc($squery)) {
        $res[] = $sq;
    }
?>

<!DOCTYPE html>
<html lang="en">

<head>
    <meta charset="UTF-8">
    <meta name="viewport" content="width=device-width, initial-scale=1.0">
    <link href="style.css?v=<?php echo time(); ?>" rel="stylesheet" type="text/css" />
    <link rel="stylesheet" href="https://maxcdn.bootstrapcdn.com/bootstrap/4.5.0/css/bootstrap.min.css">
    <title>Menu</title>
    <h1>Menu</h1>
</head>

<body>
<div class="container">
    <div>
        <a href="restmainpage.php">Return to Main page</a>
    </div>
    <div class="log">
        <a href="logout.php">Logout</a> 
    </div>

<div class="restmenuitems">
    <?php foreach($res as $sq){ ?>
        
    <?php  
        echo '<form method="post" action="order.php">';
        echo '<div class="card" style="width: 18rem;">
        <div class="card-header">Featured</div>
        <ul class="list-group list-group-flush">';
        echo '<li class="list-group-item"><strong>Dish: </strong>' . $sq['name'] . '</li>';
        echo '<li class="list-group-item"><strong>Restaurent: </strong>' . $sq['restname'] . '</li>';
        echo '<li class="list-group-item"><strong>Quantity: </strong>' . $sq['quantity'] . '</li>';
        echo '<li class="list-group-item"><strong>Price: </strong>' . $sq['price'] . '</li>';
        echo '<li class="list-group-item"><strong>NonVeg/Veg: </strong>' . $sq['preference'] . '</li>';
        if ($_SESSION['iscust'] == "true") {
            echo 'Qty <input type="text" name="product_qty" value="1" size="1" />';
            echo '<button class="add_to_cart">Add To Cart</button>';
        }
        echo '</ul>';
        echo '</div>';
        echo '</form>';
    ?>
<br>    

    <?php 
        } 
    ?>
 </div>
 </div>
  
</div>
</body>
</html>

这是我用来检索和显示项目中的表格：menuitemsdisplay.php

而且，这是示例项的显示方式（我使用条件 [] 仅向客户显示按钮）：$_SESSION['iscust'] == "true"Add to cart

当我单击该按钮时，我想将与、、、的关联值添加到数据库中的表中。Add to cartDishRestaurantQuantityPriceNonVeg/Vegcart

我不知道该怎么做。我最大的疑问是，当我单击时，如何获取用于对表中的项目执行查询的单个 ID？由于 for 循环，即使我将其存储在变量中，ID 属性也会不断变化。insertcartAdd to cart

php mysql 数据库会话 mysqli

显示项目后如何将其添加到数据库表中？

How to add items into database table after displaying them?

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