Question

我有一个PHP代码来从数据库中获取数据，其中包含2个不同的表'tbl_products'＆amp; 'tbl_productphotos'具有相同的列“ProductID”，但我无法从'tbl_productphotos'获取图像。

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

我的查询是：

<?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))
          {
?>

＆安培;在这里打印：

<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

您可以为同一列名定义别名：productphotos_ProductID和$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' ";。

 var skill_f = $("#table_display_skills tr").find('td:nth-child(1)').text().trim();
 var importanceSkill_f = $("#table_display_skills tr").find('td:nth-child(2)').text().trim();

 var industry_f = $("#table_display_industry tr").find('td:nth-child(1)').text().trim();
 var importanceIndustry_f = $("#table_display_industry tr").find('td:nth-child(2)').text().trim();

 var education_f = $("#table_display_education tr").find('td:nth-child(1)').text().trim();
 var importanceEducation_f = $("#table_display_education tr").find('td:nth-child(2)').text().trim();

Answer 2

尝试此查询

$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应该是Title。检查以下查询

$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

尝试以下方法：

$ sql =＆＃34; SELECT * FROM tbl_products INNER JOIN tbl_productphotos ON tbl_productphotos.ProductID = tbl_productphotos.ProductID＆＃34;;
$ sql =＆＃34; SELECT * FROM tbl_products，tbl_productphotos WHERE tbl_products.ProductID = tbl_productphotos.ProductID＆＃34 ;;

如何获取不同的表数据？

4 个答案: