因此,我正在尝试使轮播与数据库图像一起工作,但是它总是向我返回此错误。 obs:该连接位于include中,连接字符串为“ $ conexao”。 这是te代码:
<div class="container">
<div class="row">
<div id="mycarousel" class="carousel slide" dataride="carousel">
<ol class="carousel-indicators">
<?php
$a = 0;
$query = "select * from images";
$sql = mysqli_query($conexao,$query);
while($row = mysqli_fetch_array($sql))
{
?>
<li data-target="#mycarousel" data-slide-to="<?php echo
$a++; ?>"></li>
<?php } ?>
</ol>
<div class="carousel-inner" role="listbox">
<?php
$queryy = "select * from images";
$sqll = mysqli_query($conexao,$queryy);
while($img = mysqli_fetch_array($sqll))
{
?>
<div class="item">
<img src="img/<?php echo $img['img']; ?>" class="imgresponsive"
alt="<?php echo $img['img']; ?>"/>
</div>
<?php } ?>
</div><!--/carousel-inner-->
<a href="#mycarousel" class="left carousel-control" dataslide="prev"
role="button">
<i class="fa fa-angle-left prevSlide"></i>
</a>
<a href="#mycarousel" class="right carousel-control" dataslide="next"
role="button">
<i class="fa fa-angle-right nextSlide"></i>
</a>
</div><!--carousel slide-->
</div><!-- /row-->
</div>
几乎可以使用,但是轮播让我返回:
Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given in C:\xampp\htdocs\new\production\index.php on line 128
Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given in C:\xampp\htdocs\new\production\index.php on line 140
有任何线索吗?