如何使用php从mysql数据库中获取名称而不是id

我有一个用PHP和MYSQL和ajax编写的网站，其中包括3个依赖的dropdown列表，其中第一个下拉列表包含其他2个dropdown列表所需的几个值

在这个查询中，我根据ID进行搜索，以选择所需的值

代码：

---

在本次查询之前，您是否尝试打印
$owner\u name
和
$company\u name
？是的，在图像中，我显示的结果是ID而不是名称…因为在第一次查询中，我正在选择您所看到的ID，请确保将
$company\u name
和
$owner\u name
正确地传递到第二次查询好的，那么当用户从下拉列表中选择时，您需要获取所有三个名称，而不是仅获取siteNAME，对吗？您是否使用AJAX进行此事务？
       <?php
     $query_site_name =$wpdb->get_results("select DISTINCT 
        i.siteNAME, 
        i.ownerID, 
        i.companyID,
        o.ownerNAME,
        x.companyNAME
       from site_info i
       LEFT  
       JOIN owner_info o
       on i.ownerID = o.ownerID
       LEFT  
       JOIN company_info x
       on i.companyID = x.companyID
       ");

      foreach($query_site_name as $row)
      {   
         echo "<option id = '".$row ->ownerID."' name = '".$row ->companyID."' value = '".$row ->siteNAME."'>".$row->siteNAME."</option>";
;         
      } 
    ?>
    </select></td>

<?php
        // code for submit button action
        global $wpdb, $site_name, $data;
    //variables that handle the retrieved data from mysql database based on the ID of the variable in HTML (select)



    if(isset($_POST['query_submit']))
    {



        if(isset($_POST['site_name'])) 
       { 
         $site_name=$_POST['site_name']; 
       }
       else { $site_name=""; }


        if(isset($_POST['owner_name'])) 
         {
          $owner_name=$_POST['owner_name']; 

         } 
         else { $owner_name=""; }

         if(isset($_POST['Company_name'])) 
         {
          $company_name=$_POST['Company_name'];

         } 
         else { $company_name=""; }

        if(isset($_POST['Subcontractor_name'])) 
        { 
         $Subcontractor_name=$_POST['Subcontractor_name']; 

        }
        else { $Subcontractor_name="";}



    var_dump($site_name);
    var_dump($owner_name);
    var_dump($company_name);


$sql = $wpdb->prepare("select i.siteID
     , i.siteNAME
     , i.equipmentTYPE
     , c.latitude
     , c.longitude
     , c.height 
     , o.ownerNAME
     , o.ownerCONTACT
     , x.companyNAME
     , y.subcontractorCOMPANY
     , y.subcontractorNAME
     , y.subcontractorCONTACT
  from site_info i
  LEFT  
  JOIN owner_info o
    on i.ownerID = o.ownerID
  LEFT  
  JOIN company_info x
    on i.companyID = x.companyID
  LEFT 
  JOIN subcontractor_info y
    on i.subcontractorID = y.subcontractorID
    LEFT JOIN site_coordinates2 c
    on i.siteID=c.siteID 
    where 
    i.siteNAME = %s
    AND 
    o.ownerNAME = %s
    AND 
    x.companyNAME = %s
   ",$site_name,$owner_name,$company_name);

 $query_submit =$wpdb->get_results($sql, OBJECT);

    echo "<br>";
    echo "<br>";
echo $sql;

 <script type="text/javascript">

// make Dropdownlist depend on each other
$(document).ready(function(){


 // depend owner name on site name 

       $('#site_name').change(function(){
          var ownerID = $(this).children(":selected").attr("id");
          var companyID = $(this).children(":selected").attr("name");



     $.ajax({
      url:"<?php echo get_stylesheet_directory_uri(); ?>/dropdown_fetch_owner.php",
       method:"POST",
       data:{ownerID:ownerID,companyID:companyID},
       dataType:"text",
       success:function(data){
                 var Response = data.split("--");
                 $('#owner_name').html(Response[2]);
                 $('#Company_name').html(Response[4]);
        }
     });


 //  }


  });


});

</script>

<?php

 include_once($_SERVER['DOCUMENT_ROOT'].'/wordpress/wp-load.php');
 global $wpdb,$owner_name,$company_name;

   $sql =$wpdb->get_results("select ownerID, ownerNAME from owner_info where ownerID = '".$_POST['ownerID']."' ORDER BY ownerNAME");
   $owner_name = '--Owner--';
   var_dump($sql);
  foreach($sql as $row){


      $owner_name.= "<option value ='".$row ->ownerID."'>".$row->ownerNAME."</option>";
  }
  echo $owner_name;


   $sql =$wpdb->get_results("select companyID, companyNAME from company_info where companyID = '".$_POST['companyID']."' ORDER BY companyNAME");
   $company_name = '--Company--';
   var_dump($sql);


   foreach($sql as $row){

     $company_name.= "<option value ='".$row ->companyID."'>".$row->companyNAME."</option>";
   }
   echo $company_name;
   exit();




?>