Question

HTML

 <a href="samcheckdb.php"><img src="getdesserticecreamimage.php?itemId=oepd1007" alt="image" id="img1"></a></li>

samcheckdb.php

        <?php
        $hostname="localhost";
        $username="root";
        $password="tiger";
        $itemId=intval(\filter_input(\INPUT_GET,'itemId'));
        * @var $dbhandle type */
       $dbhandle = \mysqli_connect($hostname, $username, $password) 
        or die("Unable to connect to MySQL");

        /* @var $select type */
         $select= \mysqli_select_db($dbhandle,"sample")
     or mysqli_error($dbhandle);
     $sql="select subtitle,descript from dessert where itemId='oepd1007'"; 
      $result=mysqli_query($dbhandle,$sql);
    $row= mysqli_fetch_array($result);
    $subtitle=$row['subtitle'];
    $descript=$row['descript'];
     mysqli_close($dbhandle);
     ?>
  <html>
<head>
    <title>Customer menu card</title>
    <meta charset="UTF-8">
    <meta name="viewport" content="width=device-width">
    <link rel="stylesheet" type="text/css" href="dessert.css">
</head>
<body>
    <form id='custdisp' name='custdisp' method='post' action="samcheckdb.php"  enctype="multipart/form-data">
        <div id="d"><?php echo $descript ?></div>
        <div id="s"><?php echo $subtitle?></div>
    </form>
</body>
    </html>

我能够使用此代码从数据库中检索相应的数据。但是我有n个图像，不可能创建那么多页面，我不认为它是最优的。如何我为多个图像做这个???

Answer 1

你可以点击

<img src="getimage.php?itemId=oepsv1007" alt="image" id="img1" onclick="get_detail('oepsv1007');">

然后执行ajax调用以使用此id获取详细信息。

javascript功能

function get_detail(id) {
  // ajax call
}

如何从数据库中显示多个图像的数据

1 个答案: