while($info = mysql_fetch_array( $result ))
{
//Outputs the data
Echo "<b>".$info['name'] . " </b> :-<br>";
include('db.php');
$postid = 1;
$data = mysql_fetch_object(mysql_query('SELECT `like` , `unlike` FROM posts WHERE id=" '.$postid.' " '));
Echo "<html><body><a href='javascript:;' onclick='doAction($postid,like);'>Like (<span id='<?php echo $postid;?>_likes'>$data->like</span>)</a></html></body>";
Echo "<html><body><a href='javascript:;' onclick='doAction($postid,unlike);'>Unlike (<span id='<?php echo $postid;?>_unlikes'>$data->unlike</span>)</a></html></body>";
}
在上面的脚本中,我在每个获取的结果中都包含了选项like
。unlike
我得到以下结果:like (0)unlike (0)
每件事都完美无缺。唯一的问题是单击时它不会增加。
请帮我。
它在下面显示的简单代码中完美运行:
<html>
<head>
<script src="http://code.jquery.com/jquery-1.3.2.min.js"></script>
<script>
function doAction(postid,type){
$.post('doAjax.php' , {postid:postid, type:type}, function(data){
$('#'+postid+'_'+type+'s').text(data);
});
}
</script>
</head>
<body>
<?php
include('db.php');
$postid = 1;
$data = mysql_fetch_object( mysql_query( 'SELECT `like` , `unlike` FROM posts WHERE id= " ' . $postid . ' " ' ) );
?>
<p>hello</p>
<a href="javascript:;" onclick="doAction('<?php echo $postid; ?>','like');">Like (<span id="<?php echo $postid; ?>_likes"><?php echo $data->like; ?></span>)</a>
<a href="javascript:;" onclick="doAction('<?php echo $postid; ?>','unlike');">Unlike (<span id="<?php echo $postid; ?>_unlikes"><?php echo $data->unlike; ?></span>)</a>
</body>
</html>
我也想使用PDO
而不是mysql_fetch
上面显示的脚本。
提前致谢。