如何從數據庫中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數據庫檢索數據並將結果顯示在表中。

我有4個下拉列表，允許用戶選擇所需的輸入，然後根據所選輸入系統顯示用戶選擇的所有相關數據。

當我嘗試運行它顯示錯誤的代碼：

通知：Array對字符串的轉換

的碼第一部分：

<?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 reteive 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 } ?> 

第二部分代碼用於從數據庫檢索數據並將其包含在下拉列表中。

我會很感激任何幫助。

Answer 1

你可以擺脫「Array to string conversion」錯誤很容易。

在這些線路，要創建數組：

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

...您稍後嘗試呼應。你根本無法回顯數組。

只要改變以上是字符串，而不是：

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

注：正如其他人已經指出的那樣，你的代碼是敞開的，以SQL注入。在任何查詢中使用它之前，你都應該真的逃避你的數據。

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 }