如何使用已记录的用户会话ID从一个表中获取数据到另一个表

Question

我是MySQL数据库的新手。在我的管理面板中。当我添加子类别时，主类别将以登录用户类别的下拉菜单显示。现在在我的类别下拉列表中显示所有用户类别数据。我想显示数据库中唯一记录的用户类别，但我没有得到该做什么。请任何人都知道帮助我。

下面我分享一下我的代码

<div class="form-group">  
    <select class="form-control" name="input_cat" id="input_cat" onchange="category_select(this)"> 
          <?php 
            $query1 = "SELECT * FROM category"; 
            /* execute multi query */ 
            if (mysqli_multi_query($conn, $query1)) { 
             do { 
              echo '<option value=-1>select Category</option>'; 
              /* store first result set */ 
              if ($resulttest = mysqli_store_result($conn)) { 
               while ($test = mysqli_fetch_row($resulttest)) { 

                echo "<option value=" . $test[0] .">". $test[1]. 
                "</option>"; 
               } 
              } 

             } while (mysqli_next_result($conn)); 
            } 
            ?> 
         </select>  
       <input type="hidden" name="catid" id="catidval" value="">        
    </div> 
    below my query and database connection. 

<?php 
require 'dbconfig.php'; 
    if (session_status() == PHP_SESSION_NONE) 
    session_start();  

if (isset($_POST['submit'])) 
{ 
$venderID=$_SESSION['vid']; 
$subcategory = mysqli_real_escape_string($conn,$_POST['subcategory']); 
$catid = mysqli_real_escape_string($conn,$_POST['catid']); 
$item_type = mysqli_real_escape_string($conn,$_POST['item_type']); 
$price = mysqli_real_escape_string($conn,$_POST['price']); 

$sql = "INSERT INTO item 
(vendor_id,item_name,category_id,item_type,item_rate) VALUES 
('$venderID','$subcategory','$catid','$item_type','$price');"; 

$result = mysqli_query($conn, $sql); 


if ($result) { 
    echo '<script type="text/javascript">alert("Sub-category Added 
Successfully.");</script>'; } 
else 
    { 
    echo '<script type="text/javascript">alert("Oops! some error.");</script>'; 
    } 
} 
    ?> 

感谢

Answer 1

我修好了我的错误使用查询...

$query1 = "SELECT * FROM category WHERE vendor_id = ".$venderID;