如何從數據庫檢索數據並將其顯示在表中

Question

我正在使用WordPress和全局類$ wpdb以便從MySQL數據庫檢索數據並在表中顯示結果。

我有4個下拉列表，允許用戶選擇所需的輸入，然后基於所選輸入，系統顯示所有相關數據供用戶選擇。

當我嘗試運行代碼時，顯示錯誤：

注意：數組到字符串的轉換

代碼的第一部分：

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

    get_header();
    ?>

    <?php
    // code for submit button ation
    global $wpdb,$_POST;
//variables that handle the retrieved data from mysql database
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="";}


$site_id = ['siteID'];
$equipment_type = ['equipmentTYPE'];
$lat=['latitude'];
$long=['longitude'];
$height = ['height'];
$owner_contact = ['ownerCONTACT'];
$sub_contact = ['subcontractorCONTACT'];
$sub_company = ['subcontractorCOMPANY'];


   if(isset($_POST['query_submit']))
   {
//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`
 ");
 ?>
    <table width="30%" >
        <tr>
           <td>Site Name</td>
           <td>Owner Name</td>
           <td>Company Name</td>
           <td>Subcontractor Name</td>
         </tr>
         <tr>
            <td><?php echo $site_name ; ?></td>
            <td><?php echo $owner_name ; ?></td>
            <td><?php echo $company_name ; ?></td>
            <td><?php echo $Subcontractor_name ; ?></td>
            <td><?php echo $site_id ; ?></td>
            <td><?php echo $equipment_type ; ?></td>
            <td><?php echo $lat ; ?></td>
            <td><?php echo $long ; ?></td>
            <td><?php echo $height ; ?></td>
            <td><?php echo $owner_contact ; ?></td>
            <td><?php echo $sub_contact ; ?></td>
            <td><?php echo $sub_company ; ?></td>


         </tr>
    </table>
    <?php }  ?>

代碼的第二部分用於從數據庫檢索數據，並將其包括在下拉列表中。

我將不勝感激。

Answer 1

您可以輕松擺脫“數組到字符串轉換”錯誤。

在這些行中，您正在創建數組：

$site_id = ['siteID'];
$equipment_type = ['equipmentTYPE'];
$lat=['latitude'];
...
$sub_company = ['subcontractorCOMPANY'];

...您稍后會嘗試回應。 您根本無法回顯數組。

只需將上面的內容更改為字符串即可：

$site_id = 'siteID';
$equipment_type = 'equipmentTYPE';
$lat = 'latitude';
...
$sub_company = 'subcontractorCOMPANY';

注意：正如其他人已經指出的那樣，您的代碼對SQL注入很開放。 在任何查詢中使用數據之前，您實際上應該轉義數據。

Answer 2

<table border="1">
<tr>
 <th>Firstname</th>
 <th>Lastname</th>
 <th>Points</th>
</tr>
  <?php
    global $wpdb;
    $result = $wpdb->get_results ( "SELECT * FROM myTable" );
    foreach ( $result as $print )   {
    ?>
    <tr>
    <td><?php echo $print->firstname;?></td>
    </tr>
        <?php }