因此,我正在尝试自动化我的网站,以显示与一项股票(车辆)相关的所有图像。我可以展示图片,但它们都出现在一个块中,因为有一个项目需要活动类。这有可能实现自动化吗?或者有没有在循环中使用一次“active”的方法?
下面是代码(更新!),并不是所有的代码,只是我正在做的部分。是的,我已经把jQuery和JavaScript拉进来了!:)谢谢!
<?php
include 'Login-System/db.php';
$query = 'SELECT * FROM forsale ORDER by StockID ASC';
$result = mysqli_query($conn, $query);
if($result):
if(mysqli_num_rows($result)>0):
while($forsale = mysqli_fetch_assoc($result)):
?>
<div class="col-md-6" >
<div class="forsale">
<?php
$match = $forsale['StockID'];
$queryimg = "SELECT * FROM forsaleimg WHERE StockID = ".$match." ORDER BY 'Order' ASC ";
$resultimg = mysqli_query($conn, $queryimg);
?>
<!--Image Carousel-->
<div id="myCarousel" class="carousel slide" data-ride="carousel">
<!-- Indicators -->
<ol class="carousel-indicators">
<?php
for($i=0; $i <4; $i++){
?>
<li data-target="#myCarousel" data-slide-to="<?php echo $i;?>" <?php if ($i == 0) {
echo 'class="active"';
} ?>></li>
<?php
}
?>
</ol>
<!-- Wrapper for slides -->
<div class="carousel-inner" role="listbox" style="width:auto; height:auto;" >
<?php
$active = 0;
if($resultimg):
if(mysqli_num_rows($resultimg)>0):
while($forsaleimg = mysqli_fetch_assoc($resultimg)):
?>
<div class="carosuel-item <?php if ($active == 0){echo 'active';}?>">
<img class="d-block w-100 img-responsive" src="<?php echo $forsaleimg['FileDestination']; ?>">
</div>
<?php
$active= $active + 1;
endwhile;
endif;
endif;
?>
</div>
<!-- Left and right controls -->
<a class="left carousel-control" href="#myCarousel" role="button" data-slide="prev">
<span class="glyphicon glyphicon-chevron-left" aria-hidden="true"></span>
<span class="sr-only">Previous</span>
</a>
<a class="right carousel-control" href="#myCarousel" role="button" data-slide="next">
<span class="glyphicon glyphicon-chevron-right" aria-hidden="true"></span>
<span class="sr-only">Next</span>
</a>
</div>发布于 2018-06-10 03:41:46
在你的班级旋转木马指示器里。只需从0循环到3,并在每次迭代中输出所需的li。如果你做了一个简单的for循环(i=0;i<4;i++)等,那么你可以检查I ==是否为0,如果是,将活动类添加到li中。
https://stackoverflow.com/questions/50777663
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