Question

大家好我的PHP有问题，我想要实现的是：

<li><a href="main.html" class="title">Product name from DB</a>
        <strong>&pound;499<a href="main.html"><img src="images/thumb.png" alt="Product name from DB"/></a></strong>
  </li>

这是我的代码用php：

<ul id="items">
    <?php while($product_data = mysql_fetch_array($query_product_result))
  {
    $num_rows_products = $num_rows_products - 1;
    Print "<li><a href = 'main.php?prodid=" . $product_data["product_id"] . " <strong> Name:  " . $product_data["title"] . "</strong></a>";
    Print "<strong>Price: &pound;" . $product_data["price"] .  "'><img src='images/" . $product_data["mainImageThumbnail"] . "' alt='Product image' /></a></strong></li>";
    if($num_rows_products > 0)
            Print '<p> nu products ? wtf</p>';
  }
  ?>

由于某些原因，我得到的是一个带有超链接，没有标题和价格的图像，你有没有机会发现错误？

Answer 1

您未正确关闭<a>标记。请参阅下面的代码。注意缺少的“＆gt;”你的代码？

Print "<li><a href = 'main.php?prodid=" . $product_data["product_id"] . "'> <strong> Name:  " . $product_data["title"] . "</strong></a>";
    Print "<strong>Price: &pound;" . $product_data["price"] .  "<img src='images/" . $product_data["mainImageThumbnail"] . "' alt='Product image' /></a></strong></li>";

Answer 2

请改用Echo。 Print不会显示连接字符串。

 <ul id="items">
        <?php while($product_data = mysql_fetch_array($query_product_result))
      {
        $num_rows_products = $num_rows_products - 1;
        echo "<li><a href = 'main.php?prodid=" . $product_data["product_id"] . " <strong> Name:  " . $product_data["title"] . "</strong></a>";
        echo "<strong>Price: &pound;" . $product_data["price"] .  "'><img src='images/" . $product_data["mainImageThumbnail"] . "' alt='Product image' /></a></strong></li>";
        if($num_rows_products > 0)
                echo '<p> nu products ? wtf</p>';
      }
      ?>

用于显示db数据的PHP代码

2 个答案: