Question

我有这段代码，但我不知道如何解决这个问题......

当我在我的网站上运行时，点击此链接

<a href="#add"> <img src="image/addtocart.png" width="50%" style="margin-left: 170px;"></a>

来自$row[product_name]

的数据

将不会存储在此课程中

<!----    POPUP #CHOOSE SIZE    ----->  

    <a href="#x" class="overlay" id="add"></a>

    <div class="box">
        <h2><span style="color:blue;font-weight:bold">Choose Size</span></h2>

        <?php echo "<h5>$row[product_name]</h5></br>"; ?>

    <?php
         $sql1="SELECT product_size, product_price FROM products where product_name='$row[product_name]'";

    foreach ($conn->query($sql1) as $row1){
    echo " $row1[product_price]";
    echo "<p><button name='submit' type='submit' class='a_demo_five' value='$row1[product_size]'>$row1[product_size] P$row1[product_price].00</button></p>"; 

    }
    ?>

这是我的原始代码：

<?php
         include 'core/admin/admin_db.php';

         $sql="SELECT DISTINCT product_name, product_description, product_image FROM products WHERE product_category = 'Chicken' GROUP BY product_name";
         echo"<div class='scroll2'>";
         foreach($conn->query($sql) as $row){
         echo "<li>";
         echo "<div class='grow2 pic'><img src='$row[product_image]'></div>";
     echo "<h5>$row[product_name]</h5></br>";



  ?>


<a href="#add"> <img src="image/addtocart.png" width="50%" style="margin-left: 170px;"></a>




<!----    POPUP #CHOOSE SIZE    ----->  

<a href="#x" class="overlay" id="add"></a>

<div class="box">
    <h2><span style="color:blue;font-weight:bold">Choose Size</span></h2>

    <?php echo "<h5>$row[product_name]</h5></br>"; ?>

<?php
     $sql1="SELECT product_size, product_price FROM products where product_name='$row[product_name]'";

foreach ($conn->query($sql1) as $row1){
echo " $row1[product_price]";
echo "<p><button name='submit' type='submit' class='a_demo_five' value='$row1[product_size]'>$row1[product_size] P$row1[product_price].00</button></p>"; 

}
?>

提前致谢！

Answer 1

您不应将变量括在双引号中，因为它们是用于字符串的。你也错过了一些结束括号，这使得整个代码无法运作。

请注意，您必须为不同的项使用不同的ID，否则页面只会在您单击“添加”时加载第一个类。

<?php
include 'core/admin/admin_db.php';
$sql = "SELECT DISTINCT product_name, product_description, product_image FROM products WHERE product_category = 'Chicken' GROUP BY product_name";
echo "<div class='scroll2'>";
foreach($conn->query($sql) as $row){
    $product_name = $row['product_name'];
    echo "<li>";
    echo "<div class='grow2 pic'><img src='".$row['product_image']."'></div>";
    echo "<h5>".$product_name."</h5>";
    echo "<a href='#add_".$row['product_id']."'><img src='image/addtocart.png' style='width:50%;margin-left:170px;'></a>";
    // POPUP #CHOOSE SIZE
    echo "<a href='#x' class='overlay' id='add_".$row["product_id"]."'></a>";
    echo "<div class='box'>";
    echo "<h2><span style='color:blue;font-weight:bold'>Choose Size</span></h2>";
    echo "<h5>".$product_name."</h5>"; 
    $sql1="SELECT product_size, product_price FROM products WHERE product_name='$product_name'";
    foreach ($conn->query($sql1) as $row1){
        echo $row1['product_price'];
        echo "<p><button name='submit' type='submit' class='a_demo_five' value='".$row1['product_size']."'>".$row1['product_size']." P".$row1['product_price'].".00</button></p>"; 
    }
    echo "</div>";
    echo "</li>";
}
echo "</div>";
?>

如何使用PHP将数据放到jquery并放入php？

1 个答案: