Wie erhalten Sie verschiedene Tabellendaten?

Question

Ich habe einen PHP-Code, um die Daten aus der Datenbank mit 2 verschiedenen Tabelle 'tbl_products' & 'tbl_productphotos' mit der gleichen Spalte "ProductID", aber ich kann nicht Bilder von der 'tbl_productphotos' bekommen.

tables data is: 'tbl_products' 
ProductID Title  Description tilename 
    36  T-shirt  T-Shirt Red Glass Tile 
    37   Pant   Pant Black Glass Tile 

'tbl_productphotos' 
id  ProductID   photo 
1  36   image1.jpeg 
2  36   image2.jpeg 
3  37   imagepant.jpeg 

meine Frage ist:

<?php 
$sql="select tbl_products.*,tbl_productphotos.* from tbl_products inner join tbl_productphotos on tbl_products.ProductID=tbl_productphotos.ProductID where tbl_products.tilename='Glass Tile' ";  
    $qex=mysql_query($sql); 
     while($row=mysql_fetch_array($qex)) 
      { 
?> 

& Druck hier:

<input type="hidden" value="<?php echo $row['ProductID'];?>"> 


         <li class="col-md-3 col-sm-6 col-xs-12 isotope-item websites" style="float: left"> 
          <div class="portfolio-item"> 
           <span class="thumb-info thumb-info-lighten thumb-info-bottom-info thumb-info-centered-icons"> 
            <span class="thumb-info-wrapper"> 

             <img src="images/products/big/<?php echo $row['photo'];?>" class="img-responsive" alt="" height="200px" width="200px"> 
             <span class="thumb-info-title"> 
              <span class="thumb-info-inner"><?php echo $row['Title'];?></span> 
              <span class="thumb-info-type"><?php echo substr($row['Description'],0 ,37);?></span> 
             </span> 
             <span class="thumb-info-action"> 
              <a href="portfolio-single-project.html"> 
               <span class="thumb-info-action-icon thumb-info-action-icon-primary"><i class="fa fa-link"></i></span> 
              </a> 
              <a href="img/projects/project.jpg" class="lightbox-portfolio"> 
               <span class="thumb-info-action-icon thumb-info-action-icon-light"><i class="fa fa-search-plus"></i></span> 
              </a> 
             </span> 
            </span> 
           </span> 
          </div> 
        </li> 
        <?php 
         } 
        ?> 

Answer 1

Sie alias für den gleichen Spaltennamen definieren könnte: products_ProductID und productphotos_ProductID.

$sql = "select tbl_products.*,tbl_productphotos.*, tbl_products.ProductID as products_ProductID, tbl_productphotos.ProductID as productphotos_ProductID from tbl_products inner join tbl_productphotos on tbl_products.ProductID=tbl_productphotos.ProductID where tbl_products.tilename='Glass Tile' ";  

Answer 2

Diese Abfrage

$sql = "select t1.*,t2.* 
from tbl_products t1, tbl_productphotos t2 
where t1.ProductID = t2.ProductID 
and t1.tilename = 'Glass Tile' 
and t1.ProductID = 36"; 

Answer 3

titlename Titel sein sollte. Überprüfung unterhalb Abfrage

$sql="select tbl_products.*,tbl_productphotos.* from tbl_products inner join tbl_productphotos on tbl_products.ProductID=tbl_productphotos.ProductID where tbl_products.Title='Glass Tile' "; 

Answer 4

Versuchen Sie Folgendes:

1. $ sql = "SELECT * FROM tbl_products INNER JOIN tbl_productphotos ON tbl_productphotos.ProductID = tbl_productphotos.ProductID";

2. $ sql = "SELECT * VON tbl_products, tbl_productphotos WHERE tbl_products.ProductID = tbl_productphotos.ProductID";