其中是sql查詢中的錯誤？

Question

我有一個PHP代碼，其中包括SQL查詢，用戶從下拉列表中選擇並根據其選擇SQL查詢檢索數據。

問題是，當我試圖比較選定的數據與數據庫中的現有值查詢不起作用。我試圖在查詢之前和之後顯示變量的值，它與用戶選擇相同。所以我相信這個問題是在SQL QUERY裏我比較i.siteNAME = '".$site_name."'

如何解決這個問題我一直堅持使用這段代碼5天，仍然在計數。

這是代碼的一部分：

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

      get_header(); 
      ?> 

      <?php 
      // code for submit button action 
      global $wpdb, $site_name; 
     //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="";} 

     var_dump($site_name); 

     $query_submit =$wpdb->get_results (" 
    select i.siteID 
     , i.siteNAME 
     , i.equipmentTYPE 
     , c.latitude 
     , c.longitude 
     , c.height 
     , o.ownerNAME 
     , o.ownerCONTACT 
     , x.companyNAME 
     , y.subcontractorCOMPANY 
     , y.subcontractorNAME 
     , y.subcontractorCONTACT 
     from site_info i 
     LEFT 
     JOIN owner_info o 
     on i.ownerID = o.ownerID 
     LEFT 
     JOIN company_info x 
     on i.companyID = x.companyID 
     LEFT 
     JOIN subcontractor_info y 
     on i.subcontractorID = y.subcontractorID 
     LEFT JOIN `site_coordinates` c 
     on i.siteID=c.siteID 
     where 
     i.siteNAME = '".$site_name."' 
     AND 
     o.ownerNAME = '".$owner_name." ' 
     AND 
     x.companyNAME = '".$company_name."' 
     "); 
     var_dump($_POST['site_name']); 
     echo "<br>"; 
     echo "<br>"; 
      echo $site_name; 
      echo $owner_name; 
      echo $company_name; 
      echo $Subcontractor_name; 


    foreach ($query_submit as $obj) { 
       echo "query is working"; 
     echo "<table width='30%' "; 
    echo  "<tr>"; 
    echo   "<td>".$obj->siteNAME."</td>"; 
    echo   "<td>".$obj->ownerNAME."</td>"; 
    echo   "<td>".$obj->companyNAME."</td>"; 
    echo   "<td>".$obj->subcontractorNAME."</td>"; 
    echo   "<td>".$obj->siteID."</td>"; 
    echo   "<td>".$obj->equipmentTYPE."</td>"; 
    echo   "<td>".$obj->latitude."</td>"; 
    echo   "<td>".$obj->longitude."</td>"; 
    echo   "<td>".$obj->height."</td>"; 
    echo   "<td>".$obj->ownerCONTACT."</td>"; 
    echo   "<td>".$obj->subcontractorCONTACT."</td>"; 
    echo   "<td>".$obj->subcontractorCOMPANY."</td>"; 
    echo  "</tr>"; 
    echo "</table>"; 
     } 
    ?> 
<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 } 
?> 
<!-- the below part of code work as it should --!> 
    <!--create dropdown list site names--> 

    <form method ="post" action ="" name="submit_form"> 
     <table width="30%"> 
      <tr> 
       <td>Site Name</td> 
       <td>Owner Name</td> 
       <td>Company Name</td> 
       <td>Subcontractor Name</td> 
      </tr> 
      <tr> 
       <td><select id="site_name" name = "site_name"> 

       <?php 


        $query_site_name =$wpdb->get_results ("select DISTINCT siteNAME from site_info"); 
         foreach($query_site_name as $site_name) 
         { 
         $site_name = (array)$site_name; 
         echo "<option value = '{".$site_name ['siteNAME']."}'>". $site_name['siteNAME']."</option>"; 
         } 
       ?> 

       <!--create dropdown list owner names--> 
       </select></td> 

       <td><select id="owner_name" name ="owner_name"> 
       <?php 
       global $owner_name; 
         $query_owner_name =$wpdb->get_results ("select DISTINCT ownerNAME from owner_info"); 
         foreach($query_owner_name as $owner_name) 
         { 
         $owner_name = (array)$owner_name; 
         echo "<option value = '{".$owner_name ['ownerNAME']."}'>". $owner_name['ownerNAME']."</option>"; 
         } 
        ?> 
       </select></td> 

       <!--create dropdown list Company names--> 
       </select></td> 

       <td><select id="Company_name" name ="Company_name"> 
       <?php 
       global $Company_name; 
        $query_Company_name =$wpdb->get_results ("select DISTINCT companyNAME from company_info"); 
        foreach($query_Company_name as $Company_name) 
        { 
         $Company_name = (array)$Company_name; 
         echo "<option value = '{".$Company_name ['companyNAME']."}'>". $Company_name['companyNAME']."</option>"; 
        } 
       ?> 
       </select></td> 

       <!--create dropdown list Subcontractor names--> 
       </select></td> 

       <td><select id="Subcontractor_name" name ="Subcontractor_name"> 
       <?php 
       global $Subcontractor_name; 
        $query_Subcontractor_name =$wpdb->get_results ("select DISTINCT subcontractorNAME from subcontractor_info"); 
        foreach($query_Subcontractor_name as $Subcontractor_name) 
        { 
         $Subcontractor_name = (array)$Subcontractor_name; 
         echo "<option value = '{".$Subcontractor_name ['subcontractorNAME']."}'>". $Subcontractor_name['subcontractorNAME']."</option>"; 
         } 
        ?> 
       </select></td> 
      <tr> 
     <td></td> 
     <td></td> 
     <td></td> 
     <td></td> 
     <td> 
      <input type ="submit" name="query_submit" value ="Search" /> 

     </td> 
     </tr> 

     </table> 
    </form> 


<?php 
get_footer(); 
?> 

Answer 1

我解決我的問題有沒有在查詢中的任何錯誤，問題是在的foreach循環內的迭代過程。

我被提取作爲查詢陣列這是它必須被取出作爲對象 行，我改變的錯誤是：

echo "<option value = '{".$owner_name ['ownerNAME']."}'>". $owner_name['ownerNAME']."</option>"; 

變得

echo "<option value = '".$owner_name->ownerNAME."'>". $owner_name->ownerNAME."</option>";