php - 如何修复错误数组到字符串的转换并显示获取的数据

我正在WordPress上使用MYSQL和PHP以便从数据库中检索数据，但是使用4下拉列表允许用户进行选择，并且基于用户的选择，SQL查询将检索所有相关数据。

问题是系统显示以下警告:



并且不要从数据库中检索任何内容。

注意:SQL查询正确，并且检索了所需的数据(在phpmyadmin上进行了测试)。

码:

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

    get_header();
    ?>

    <?php
    // code for submit button action
    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="";}

   $site_id = ['siteID'];
//$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;


//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);

// create a table  inside php code

echo "<table width='30%' ";
echo     " <tr>";
echo         " <td>$query_submit ['siteNAME']</td>";
echo         " <td>$query_submit  ['ownerNAME']</td>";
echo         " <td>$query_submit  ['companyNAME']</td>";
echo         " <td>$query_submit ['subcontractorNAME']</td>";
echo         " <td>$query_submit ['siteID'] </td>";
echo         " <td>$query_submit ['equipmentTYPE'] </td>";
echo         " <td>$query_submit ['latitude'] </td>";
echo         " <td>$query_submit ['longitude'] </td>";
echo         " <td>$query_submit ['height']</td>";
echo         " <td>$query_submit ['ownerCONTACT']</td>";
echo         " <td>$query_submit ['subcontractorCONTACT']</td>";
echo         " <td>$query_submit ['subcontractorCOMPANY']</td>";
echo     "</tr>";
echo "</table>";
?>

最佳答案

错误来自以下几行:

$site_id = ['siteID'];
//$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'];

我不确定是否要从帖子中获取这些变量。如果要从帖子值中获取这些内容，请使用:

$site_id = $_POST['siteID'];
           ^^^use global variable

但正如我看到的那样，您正在$query_submit数组变量中使用它。这意味着它来自数据库。

在这种情况下，您必须删除或注释这些行。