单击链接时的Sql UPDATE - Sql UPDATE on link click

当用户单击链接时，我试图更新一个值。

<?php
$data = mysql_query("SELECT * FROM comments a LEFT JOIN pins b ON a.pin_id = b.id INNER     JOIN board c on b.board_id = c.id WHERE a.to_id = '$myid' AND a.status = 'unviewed'") 
or die(mysql_error()); while($info = mysql_fetch_array( $data )) 
{ 
Print "<li>";
Print "<a href='/board/pins/".$info['board_id']."/".$info['pin_id']."'>";
Print "<img src='".$info['pin_url']."' width='50' align='left'>";
Print "<font size='1'>comment received on ".$info['date']."</font></a>";
Print "Collection: ".$info['board_name']."</li>";
} 
?>

链接是上面的链接。

单击链接时，我如何编辑以上代码以包含更新查询"UPDATE comments SET status='viewed' WHERE to_id = '$myid' AND id='$postid'"？

编辑：

这是我的mark_viewed.php：

<?php
$con=mysqli_connect("XXX","XXX","XXX","XXX");
// Check connection
if (mysqli_connect_errno())
  {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
  }
// get values sent from address bar
$myid=$_GET['myid'];
$postid=$_GET['postid'];
mysqli_query($con,"UPDATE comments SET status='viewed' WHERE to_id ='$myid' AND id='$postid'");
mysqli_close($con);
?>

我的发送页面：

<?php  // display of the notifications dropdown menu
$query_select = "SELECT * FROM comments a LEFT JOIN pins b ON a.pin_id = b.id LEFT JOIN     board c on b.board_id = c.id WHERE a.to_id = '$myid' AND a.status = 'unviewed' ORDER BY     a.date DESC";
$result_select = mysql_query($query_select) or die(mysql_error());
$rows = array();
while($row = mysql_fetch_array($result_select))
    $rows[] = $row;
foreach($rows as $row){ 
    $myid = $row['user_id']; // my id
    $name = $row['board_name']; // collection name
    $boardid = $row['board_id']; // collection id
    $postid = $row['pin_id']; // post id
    $url = $row['pin_url']; // image url
    echo "<li><a href='/board/pins/$boardid/$postid' data-myid='$myid' data-    postid='$postid' class='markviewed'>";
    echo "<img src='$url' height='50' width='50'>";
    echo "New comment in $name.";
    echo "</a></li>";
}
?>

我在同一个页面的标题中有JS，但它不起作用。变量在此页面$myid和$postid上很好，但它们不会在mark_viewed.php 上发送/接收

试试这个代码（如注释中所建议的AJAX）。它假设它基本上做的是：

将相关id作为属性打印到html中
绑定侦听器以单击这些链接上的事件
ping之后http://domain.com/mark_viewed.php（执行查询的文件，在查看时更新注释）它将浏览器重定向到所需位置（/board/pin/…）

//html part
<?php
$myid = 1234;
$postid = 333;
//echo prepared link with ids in it's data-attributes
echo '<a href="/board/pins/"'.$info['board_id'].'/'.$info['pin_id'].'"  
   data-myid="' . $myid . '" 
   data-postid="' . $postid . '" 
   class="markviewed">mark ids as viewed and go to /board/pins/...</a>';
?>
<script src="//ajax.googleapis.com/ajax/libs/jquery/1.7.2/jquery.min.js"></script>
<script>
$(document).ready(function() {
   //bind a listener to "click" event on links with class "markviewed"
   $('a.markviewed').click(function(event) {
      //prevent default behavior just in case
      event.preventDefault();
      //get ids from clicked <a>
      var myid = $(this).attr('data-myid');
      var postid = $(this).attr('data-postid');
      //ping the address to mark clicked link as viewed
      $.ajax('http://domain.com/mark_viewed.php?myid=' + myid + '&postid=' + postid);
      //redirect to the link in the href attribute
      window.location.href = $(this).attr('href');
   });
});
</script>