Php 数据库查询不使用xeditable插件

我已经创建了一个表，其中包含使用xeditable插件进行弹出式未编辑，并尝试使用ajax调用将数据提交到数据库。以下是我的表格行代码：

echo "<td><a href=\"#\" id='sel' data-pk=\"$row[loan_id]\">" . $row['loan_status'] . "</a></td>";

我的th xeditable插件jquery代码如下：

$(document).ready(function() {
    //toggle `popup` / `inline` mode
    $.fn.editable.defaults.mode = 'popup';     

$('#loan a').editable({
    type: 'text',
    name: 'loan_amount',
    url: '../payments/post.php',
    title: 'Enter amount',
    ajaxOptions: {
     type: 'put'

   }   

});

});

但我无法理解这个问题，因为查询对我来说似乎很好，但数据库仍然没有得到更新

---

任何能指引我正确方向的人。感谢

xeditable在其github上提供的脚本的第一个问题是使用mysql，而现在使用mysql并不安全。所以我把它们都换成了mysqli。以下是供任何人参考的工作代码：

<?php
$server = "localhost";
$username = "user";
$password = "pass";
$database = "db";

$con = new mysqli($server,$username,$password,$database);
if (!$con){
die('Could not connect: ' . mysqli_connect_error($con));
echo "Cudnt cnnect";}
else{
echo "connected";
} 

    var_dump($_POST); 

    $pk = $_POST['pk'];
    $name = $_POST['name'];
    $value = $_POST['value'];
    /*
     Check submitted value
    */
    if(!empty($value)) {

if ($result =  mysqli_query($con,'update k_loans set '.mysqli_real_escape_string($con,$name).'="'.mysqli_real_escape_string($con,$value).'" where loan_id ="'.mysqli_real_escape_string($con,$pk).'"'));
{
        echo "Update successful";
    }

    mysqli_close($con);
       print_r($_POST);
    } else {

        echo "This field is required!";
    }
?>

---

我建议您在PHP中尝试三件事：（1）在查询数据库之前打印$\u POST数组-如果出现DB错误，您将看不到它（2）将查询格式化为PHP字符串变量并打印它-我认为这是您的问题，（3）始终关注mysqlerrors@bitfiddler谢谢你的来访和建议，但我已经解决了这个问题，并发布了解决方案。再次感谢
<?php
$server = "localhost";
$username = "user";
$password = "pass";
$database = "db";

$con = new mysqli($server,$username,$password,$database);
if (!$con){
die('Could not connect: ' . mysqli_connect_error($con));
echo "Cudnt cnnect";}
else{
echo "connected";
} 

    var_dump($_POST); 

    $pk = $_POST['pk'];
    $name = $_POST['name'];
    $value = $_POST['value'];
    /*
     Check submitted value
    */
    if(!empty($value)) {

if ($result =  mysqli_query($con,'update k_loans set '.mysqli_real_escape_string($con,$name).'="'.mysqli_real_escape_string($con,$value).'" where loan_id ="'.mysqli_real_escape_string($con,$pk).'"'));
{
        echo "Update successful";
    }

    mysqli_close($con);
       print_r($_POST);
    } else {

        echo "This field is required!";
    }
?>