如何在Jquery中发布数据到数据库

Question

目前我有这个jquery帮助我从我的数据库中获取数据，当选择这个部门时它只显示不同类型的选项。但是现在我想再次将这些数据发布到数据库。有没有解决方案？这里是我的代码：

<script type="text/javascript" src="js/jquery-1.7.min.js"></script> 
    <script type="text/javascript"> 
     $(document).ready(function() { 
      $("#cat").change(function() { 
       $.ajax({ 
        type: "GET", 
        url: "getPositionTitle.php", 
        data: "category_id=" + $(this).find(":selected").val(), 
        cache: false, 
        success: function(msg){ 
         $("#position_title").empty(); 
         my_position_title_array = $.parseJSON(msg); 
         for (i = 0; i < my_position_title_array.length; i ++) { 
          $("#position_title").append('<option value="' + my_position_title_array[i].id + '">' 
           + my_position_title_array[i].position_title + '</option>'); 
         } 
         $("#position_title").trigger('change'); 
        } 
       }); 
      }); 
      $("#cat").trigger('change'); 
     }); 
    </script> 


<form id="offeredjob" method="post" action="doOfferedJob.php"> 

<tr> 
    <td><label for="applied_department">Department:</label></td> 
    <td> 
     <select id="cat" name ="applied_department" applied_position_title="category"> 
      <?php 
       $query = "SELECT id, department FROM department"; 
       $result = mysqli_query($link, $query) or die(mysqli_error()); 
        while ($row = mysqli_fetch_assoc($result)) { 
         echo "<option value ='" . $row['id'] . "'>" . $row['department'] . "</option>"; 
          } 
          ?> 
         </select> 

        </td> 
       </tr> 
       <tr> 
        <td><label for = "applied_position_title">Position Title:</label></td> 
        <td> 
         <select id="position_title" applied_position_title="applied_position_title"> 
          <option value="1"></option> 
         </select> 
        </td> 
       </tr> 

这就是我如何张贴到我的数据库：

$query = "UPDATE job_application SET applied_department = '$applied_department', applied_position_title = '$applied_position_title' WHERE id = '$id'"; 

$result = mysqli_query($link, $query) or die(mysqli_error($link)); 

Answer 1

添加这样的事情你JQuery的。

// Post data to postPositionData.php when user changes form 
$("#offeredjob").change(function() { 
    // Serialize form data 
    var yourFormData = $(this).serialize(); 
    // POST 
    $.ajax({ 
     type: "POST", 
     url: "postPositionData.php", 
     data: yourFormData, 
     success: function(msg){ 
      // do something 
     } 
    }); 
}); 

然后文件postPositionData.php将执行数据库插入/更新。