php - 代码和 SQL 查询中有什么错误？为什么网页上没有任何显示？

Question

我正在创建一个网页，向用户显示前 4 个下拉列表，其中包含从 MYSQL 数据库检索的数据。其中每个下拉列表代表不同表中的 1 列。

我正在使用 wordpress 和 $wpdb 类

我需要的是能够根据用户的选择向用户显示结果。

示例:

site id - site name - owner name - owner contact -  lat - long -  company name.... 

以及所有相关列。

问题是系统只是显示在下拉列表中选择的数据...看起来系统只是从下拉列表返回数据而不是从数据库返回数据。

我该如何解决这个问题？

我将显示部分代码和网页图像。

代码:

<?php
    /*
    Template Name: search info
    */

    get_header();
    ?>

    <?php
    // code for submit button ation
    global $wpdb;
//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="";}

//query to retrieve all related info of the selected data from the dropdown list  
$query_submit =$wpdb->get_results ("select 

site_info.siteID,site_info.siteNAME ,site_info.equipmentTYPE,site_coordinates.latitude,site_coordinates.longitude,site_coordinates.height ,owner_info.ownerNAME,owner_info.ownerCONTACT,company_info.companyNAME,subcontractor_info.subcontractorCOMPANY,subcontractor_info.subcontractorNAME,subcontractor_info.subcontractorCONTACT from `site_info`
LEFT JOIN `owner_info`
on site_info.ownerID = owner_info.ownerID
LEFT JOIN `company_info` 
on site_info.companyID = company_info.companyID
LEFT JOIN `subcontractor_info` 
on site_info.subcontractorID = subcontractor_info.subcontractorID
LEFT JOIN `site_coordinates` 
on site_info.siteID=site_coordinates.siteID 
where 
site_info.siteNAME = `$site_name` 
AND
owner_info.ownerNAME = `$owner_name`
AND
company_info.companyNAME = `$company_name`
AND
subcontractor_info.subcontractorNAME = `$Subcontractor_name`
 " , ARRAY_A);

$site_id = 'siteID';
$site_id = (array)$site_id;
$equipment_type = 'equipmentTYPE';
$equipment_type = (array)$equipment_type;
$lat='latitude';
$lat = (array)$lat;
$long='longitude';
$long = (array)$long;
$height = 'height';
$height = (array)$height;
$owner_contact = 'ownerCONTACT';
$owner_contact = (array)$owner_contact;
$sub_contact = 'subcontractorCONTACT';
$sub_contact = (array)$sub_contact;
$sub_company = 'subcontractorCOMPANY';
$sub_company = (array)$sub_company;

?>

    <table width="30%" >
        <tr>
           <td>Site Name</td>
           <td>Owner Name</td>
           <td>Company Name</td>
           <td>Subcontractor Name</td>
           <td>Site ID</td>
           <td>Equipment Type</td>
           <td> Lattitude</td>
           <td>Longitude </td>
           <td> Height</td>
           <td> Owner Contact</td>
           <td> Sub Contact</td>
           <td> Sub company Name</td>



         </tr>
         <tr>
  <?php   
         foreach ($query_submit as $query)
     {
     echo "<table>";
     echo   "<tr>";
     echo       "<td>" ,$query[siteNAME]. "</td>";
     echo       "<td>", $query[ownerNAME] ."</td>";
     echo       "<td>", $query[companyNAME] ."</td>";
     echo       "<td>", $query[subcontractorNAME]. "</td>";
     echo       "<td>" ,$query[siteID ]."</td>";
     echo       "<td>" ,$query[equipmentTYPE]. "</td>";
     echo       "<td>" ,$query[latitude]. "</td>";
     echo       "<td>" ,$query[longitude]. "</td>";
     echo       "<td>" ,$query[height]. "</td>";
     echo       "<td>" ,$query[ownerCONTACT]. "</td>";
     echo       "<td>" ,$query[subcontractorCONTACT ]."</td>";
     echo       "<td>" ,$query[subcontractorCOMPANY]. "</td>";
     echo  "</tr>";
     echo"</table>";   
     }
   ?>  
         </tr>
    </table>
 <?php    
}  
?>

点击提交按钮后没有任何显示...错误在哪里？

Answer 1

我修复了使用 echo 显示错误的问题。我应该使用此格式 echo"".$obj->siteNAME."";。

感谢您的回答和评论