使用jQuery＆amp;更新PHP表单显示中的文本值MySQL数据库

Question

我有一个表格显示，我希望用户能够使用jQuery以实时方式编辑表格的值。该表很简单，只有1列，列有名称。我把它全部建成了，但目前表格没有更新价值。所有代码都在下面，我仍在学习所有内容，因此它可能写得不是很好/正确！提前谢谢。

表格显示：

<?php
    echo "<div class=\"table-responsive\"><table class=\"table table-striped\">";
    echo "<thead><tr><th>Employee Name</th><th></th></tr></thead><tbody>"; 

    require_once 'connectionsettings.php'; // Gets connection settings

    $sql = "SELECT id, employees FROM Employees ORDER BY employees ASC";
    $result = $mysqli->query($sql);
    if ($result->num_rows > 0) {
        // what it's saying is that if there's rows do something
        while($row = $result->fetch_assoc()) {
        // now it's saying get the data and put it in rows
            echo "<tr><td data-id='{$row['id']}' contenteditable=\"true\">" . $row["employees"] . "</td><td><span data-id='{$row['id']}' name='remove_{$row['id']}' class='employee glyphicon glyphicon-remove' aria-hidden='true'></span></td></tr>";
        }
    }else{
        echo "No Employees! Let's add some.";
    }
    $mysqli->close();
    echo "</tbody></table></div>";
?>

Jquery信息：

$(function(s){
    $("td[contenteditable=true]").blur(function(){
        var id = $(this).attr("id") ;
        var name = $(this).text() ;
        var formURL = "updateemployeename.php"; 
        $.ajax({
            url : formURL,
            type: "POST",
            data : {name: name, id: id},
            success:function(data, textStatus, jqXHR) 
            {
                $('#successmessage2').slideDown('fast').delay(1500).slideUp('fast');
                $('div#employeedisplay').hide(); 
                $('div#updatedemployeedisplay').load('employeedisplay.php').fadeIn(3000); 
            },
            error: function(jqXHR, textStatus, errorThrown) 
            {
                //if fails      
            }
        });   
    });
    s.preventDefault(); //STOP default action
});

SQL信息：

<?php 
    require_once 'connectionsettings.php'; // Gets connection settings

    $name = htmlspecialchars(trim($_POST['name']));  
    $id = htmlspecialchars(trim($_POST['id']));  

    $sql = "UPDATE Employees SET employees='$name' WHERE id='$id'";

    if($mysqli->query($sql) === TRUE) {
        echo "status updated successfully";
    }else{
        echo "Error updating status" . $mysqli->error;
    }
    $mysqli->close();
?>