小编use*_*866的帖子

我有一个生成并显示表格的php页面.对于表格中的最后一行,我想显示附加了"onclick"功能的图像.这会将所选行的用户名发送到将使用AJAX更新数据库的脚本.该表显示正常,但AJAX无法正常工作.我的php显示图像是:

echo "<td> <img id='tblimg' 
    onclick='like('" . $row['Username'] . "')' 
    src='like.jpg' alt='like/dislike image' 
    width='80px' height='30px'></td>";

javascript函数是:

<script type="text/javascript" >

        function like(user) 
        {

            $.ajax(
                url: "update.php",
                type: "POST",
                data: { 'username': user, 'liked': '1' },                   
                success: function()
                            {
                                alert("ok");                                    
                            }
            );
        }

</script>       

这是update.php:

<?php

$con=mysqli_connect("","sam74","********","sam74");

// Check connection
if (mysqli_connect_errno())
{
    echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

$Username = $_POST['username'];
$Liked = $_POST['liked'];   

$sql = "UPDATE 'followers' SET 'Liked' = '$Liked' WHERE 'Username' = '$Username'";

if …