獲取價值

Question

我只是需要一些想法如何，我從下拉

<select id="ins" name="instructor" required> 
    <?php 
     //dropdown instructor name 
     echo "<option value=\"\">"."Select"."</option>"; 
     $qry = "select * from instructor"; 
     $result = mysqli_query($con,$qry); 
     while ($row = mysqli_fetch_array($result)){ 
     echo "<option value=".$row['insA_I'].">".$row['insname']."</option>"; 
     } 
    ?> 
</select> 

獲得的價值，並把價值這個PHP文件中$_POST['instructor']

<?php 
     session_start(); 
     require 'connection.php'; //connection 

     $qry = "select course from instructor where insA_I = '".$_POST['instructor']."'"; 
     $result = mysqli_query($con,$qry); 
     $_SESSION['row'] = mysqli_fetch_array($result); 
     $data = $_SESSION['row']['course']; 
     echo $data; 
    ?> 

上午還在新的AJAX

$(document).ready(function() { 
    $('#ins').change(function(){ 
     $.ajax({ 

      type: "POST", 
      url: "json_php.php", 
      data: '', 
      datatype: 'json', 
      success: function(result){ 
       alert(result); 

      } 
     }); 
    }); 
}); 

我已經在網上搜索過這個，但在修改後沒有得到我的答案 只需要想法，技術等。由於您使用jQuery的

Answer 1

試試這個Ajax代碼

$(document).ready(function() { 
     $('#ins').change(function(){ 

      $.ajax({ 

       type: "POST", 
       url: "json_php.php", 
       data: {instructor:$(this).val()}, 
       success: function(result){ 
        alert(result); 

       } 
      }); 
     }); 
    }); 

查詢

$qry = "select course from instructor where insA_I = '".$_POST['instructor']."'"; 
     $result = mysqli_query($con,$qry); 
     $fetch = mysqli_fetch_array($result); 
     $_SESSION['row']['course']=$fetch['course']; 
     $data = $_SESSION['row']['course']; 
     echo $data; 

Answer 2

PHP代碼，使用jQuery：

data: { instructor: $('#ins').val() } 

另外，請注意你的SQL查詢是開放給SQL注入。 This is a good place to start reading about that。