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如何在PHP/MYSQL中获得不同的ID值?_Php_Mysql - Fatal编程技术网

如何在PHP/MYSQL中获得不同的ID值?

如何在PHP/MYSQL中获得不同的ID值?,php,mysql,Php,Mysql,我有PHP代码从数据库中获取两个表“tbl_products”和“tbl_productphotos”数据。我想获取不同的数据。 我的查询和代码是: <?php $sql = "select DISTINCT t1.*,t2.photo from tbl_products t1, tbl_productphotos t2 where t1.ProductID = t2.ProductID and t1.tilename = 'Glass

我有PHP代码从数据库中获取两个表“tbl_products”和“tbl_productphotos”数据。我想获取不同的数据。 我的查询和代码是:

<?php
$sql = "select DISTINCT t1.*,t2.photo 
        from tbl_products t1, tbl_productphotos t2 
        where t1.ProductID = t2.ProductID 
        and t1.tilename = 'Glass Tile' 
        and t1.ProductID = t2.ProductID 
        ORDER BY ProductID DESC";
$qex=mysql_query($sql); 
while($row=mysql_fetch_array($qex))
{
?> 

    <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/thumbs/<?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="glass-tile-details.php?ProductID=<?php echo base64_encode($row['ProductID']); ?> &quot;&amp;&lt;&nbsp; &Atilde;&Chi;!&upsih;&upsilon; &quot;&amp;&gt;&euro; &nbsp;&Aacute;&Acirc;&quot;&amp; &acute;&AElig;&copy;&quot;&amp;&amp; &quot;&Eacute;&piv;&Pi; 1&quot;&amp;&thetasym;&Theta; &quot;&amp;"><span class="thumb-info-action-icon thumb-info-action-icon-primary"><i class="fa fa-link"></i></span>
        </a> 

        <a href="images/products/thumbs/<?php echo $row['photo']; ?>" style="width: 500px; height: 500px;" 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
}
?>
第二表:“tbl_产品照片”

我们可以选择多个图像我选择3个图像和所有3个存储在所有三个文件夹中的图像。所以我只想用一个ProductID只获得一张带有详细信息的图片,但我在网页上显示了所有3张图片

请检查、回复并帮助我。 谢谢和问候 Ankit

尝试避免选择*。 (条件t1.ProductID=t2.ProductID有两次)


尽快
?我不记得你雇用过任何人。我们不是你的工作猴子。
t2。照片
在所有行中都是一样的。你不应该得到3份。它们应该是不同的照片吗?如果是,它应该返回哪一个?对不起,先生,我是这里的新用户,从现在起,我将永远不会尽快写这个。问候-ankit您不需要在
WHERE
子句中有
t1.ProductID=t2.ProductID
两次。使用
GROUP BY t1.ProductID
而不是
SELECT DISTINCT
。很抱歉,即使使用此查询,它也不起作用。我会得到一个空白屏幕。请更新您的问题,并添加您现在取消的确切查询代码..请。。anc检查错误和相关消息。。为了避免歧义,我还按列顺序更新了答案
ProductID      Title     Description     titlename
35             ABC       tile           Glass Tile
36             XYZ       tile 2          Glass Tile
id      ProductID          photo        Each Image store in 3 folders
5         35            image.jpeg   (in folder big,medium,thumbs)
6         35            image.jpeg   (in folder big,medium,thumbs)
7         35            image.jpeg   (in folder big,medium,thumbs)
8         36            image.jpeg   (in folder big,medium,thumbs)
9         36            image.jpeg   (in folder big,medium,thumbs)
10        36            image.jpeg   (in folder big,medium,thumbs)
$sql = "select DISTINCT 
            t1.ProductID, 
            t1.Title,
            t1.titlename,
            t1.Description,
            t2.photo 
        from tbl_products t1
        INNER JOIN  tbl_productphotos t2 on t1.ProductID = t2.ProductID 
        and t1.tilename = 'Glass Tile' 
        ORDER BY t1.ProductID DESC";