实际上我想在水平视图中显示数据库的前4个图像,当单击“下一步”和“上一个”按钮时,我们从数据库中获取下一个或上一个结果,我的代码显示结果,但它是垂直显示的,在页面加载时没有4张图像。请更正代码
<div class="row">
<div class="col-md-12">
<div class="well">
<div id="myCarousel" class="carousel slide">
<!-- Carousel items -->
<div class="carousel-inner">
<?php
$query2="select * from hotel INNER JOIN sponsored_pay on hotel.postid=sponsored_pay.hotelid where sponsored_pay.pagename='Side'";
$results2 = mysql_query($query2);
while($row_img2=mysql_fetch_array($results2))
{
?>
<div class="item active">
<div class="row">
<div class="col-sm-3"><a href="#x"><img src="<?php echo $row_img2['hotel_photo'];?>" alt="Image" class="img-responsive"></a>
</div>
</div>
<!--/row-->
</div>
<?php
}
?>
</div>
<!--/carousel-inner--> <a class="left carousel-control" href="#myCarousel" data-slide="prev">‹</a>
<a class="right carousel-control" href="#myCarousel" data-slide="next">›</a>
</div>
<!--/myCarousel-->
</div>