Javascript 从文件夹中提取所有图像并创建引导旋转木马/幻灯片放映
我被这段代码困住了,我试图把所有的图片放到一个文件夹中,并把它们放在列表或div中,以创建引导旋转木马Javascript 从文件夹中提取所有图像并创建引导旋转木马/幻灯片放映,javascript,php,jquery,twitter-bootstrap-3,bootstrap-carousel,Javascript,Php,Jquery,Twitter Bootstrap 3,Bootstrap Carousel,我被这段代码困住了,我试图把所有的图片放到一个文件夹中,并把它们放在列表或div中,以创建引导旋转木马 <div id="spa_gallery" class="carousel slide" data-ride="carousel"> <!-- Carousel indicators --> <!-- Wrapper for carousel items --> <div class="carousel-inner"&g
<div id="spa_gallery" class="carousel slide" data-ride="carousel">
<!-- Carousel indicators -->
<!-- Wrapper for carousel items -->
<div class="carousel-inner">
<?php
$dirname = "spa_gallery/";
$images = glob($dirname."*.jpg");
foreach($images as $image) {
$imagelist.= '<li class="item"><a href=""><img src="'.$image.'" /></a> </li>';
}
?>
<!-- Carousel controls -->
<a class="carousel-control left" href="#spa_gallery" data-slide="prev"> <span class="glyphicon glyphicon-chevron-left"></span> </a>
<a class="carousel-control right" href="#spa_gallery" data-slide="next"> <span class="glyphicon glyphicon-chevron-right"></span> </a>
</div>
</div>
以前的代码是:
<div class="carousel-inner">
<div class="item active"> <img src="spa_gallery/01.jpg" alt="First Slide">
</div>
<div class="item"> <img src="spa_gallery/02.jpg" alt="Second Slide"> </div>
<div class="item"> <img src="spa_gallery/03.jpg" alt="Third Slide"> </div>
<div class="item"> <img src="spa_gallery/04.jpg" alt="Fourth Slide"> </div>
<div class="item"> <img src="spa_gallery/05.jpg" alt="Fifth Slide"> </div>
</div>
尝试使用
<?php
$dirname = "spa_gallery/";
$images = glob($dirname."*.jpg");
foreach($images as $image) {
?>
<div class="item"><img src="<?php echo $image; ?>" /></div>
<?php
}
因此,您的代码将是
<?php
$class_active = true;
$dirname = "spa_gallery/";
$images = glob($dirname."*.jpg");
foreach($images as $image) {
?>
<div class="item <?php if($class_active == true){ echo 'active' ; $class_active = false} ?>"><img src="<?php echo $dirname.$image; ?>" /></div>
<?php
}
运行代码时发生了什么?您希望发生什么?是否在某个位置回显$imagelist?如何为任何图像添加class=“item active”以开始幻灯片放映。幻灯片是从这个图像开始的吗?意思是代码是这样的:……引导程序要求第一个元素的class=“active”开始slideshow@luna.romania我明白你的意思了我更新了我的答案。。但您可以从我的回答中告诉我您使用的每个脚本,我会将其修改为您不工作:
$dir_imgs = "spa_gallery";
$files_imgs = scandir($dir_imgs);
foreach ($files_imgs as $key_imgs => $value_imgs){
$php_imgs = explode(".", $value_imgs);
$php_imgs = end($php_imgs);
if ($php_imgs == "jpg"){
?>
<div class="item"><img src="<?php echo $dir_imgs.'/'.$value_imgs; ?>" /></div>
<?php
}
}
$class_active = 0; // before foreach
<div class="item <?php echo (($class_active == 0)? 'active' : '') ?>">
$class_active = 1;
<?php
$class_active = true;
$dirname = "spa_gallery/";
$images = glob($dirname."*.jpg");
foreach($images as $image) {
?>
<div class="item <?php if($class_active == true){ echo 'active' ; $class_active = false} ?>"><img src="<?php echo $dirname.$image; ?>" /></div>
<?php
}