如何从数据库中检索数据并将其显示在表中

Question

我正在使用WordPress和全局类$ wpdb以便从MySQL数据库检索数据并将结果显示在表中。

我有4个下拉列表，允许用户选择所需的输入，然后根据所选输入系统显示用户选择的所有相关数据。

当我尝试运行它显示错误的代码：

通知：Array对字符串的转换

的码第一部分：

<?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 } ?> 

第二部分代码用于从数据库检索数据并将其包含在下拉列表中。

我会很感激任何帮助。

Answer 1

你可以摆脱“Array to string conversion”错误很容易。

在这些线路，要创建数组：

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

...您稍后尝试呼应。你根本无法回显数组。

只要改变以上是字符串，而不是：

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

注：正如其他人已经指出的那样，你的代码是敞开的，以SQL注入。在任何查询中使用它之前，你都应该真的逃避你的数据。

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 }