Question

我使用此脚本从包含子文件夹的文件夹中显示一个随机图片：

<?php
$imagesDir = glob('folders/pics/*', GLOB_ONLYDIR);
$randomfolder = $imagesDir[array_rand($imagesDir)];
$images = glob($randomfolder . '/*.{jpg,jpeg,png,gif}', GLOB_BRACE);
$randomImage = $images[array_rand($images)];
echo '<img src="'.$randomImage.'" class="image">'; 
?>

一切正常！但是现在我想同时显示5张图片（对于一个caroussel-slider）。我使用了以下代码

$randomImage = $images[array_rand($images, 5)];

但它向我显示了这个警告：

警告：非法偏移类型[...]

我做错了什么？

Answer 1

您使用<div class="row" id="featuredcities"> <div class="col-sm-3"> <img id="image" src="images/singapore.jpg" alt="singapore"> <a href="#"><p id="text">Singapore</p></a> <p id="summarytext">Amazing culinary experience and <br>efficient business ecosystem in this <br>tiny cosmopolitan country.</p> </div> <div class="col-sm-3"> <img id="image" src="images/bangkok.jpg" alt="bangkok"> <a href="#"><p id="text">Bangkok</p></a> <p id="summarytext">A bustling neon-lit city that combines tradition and modernity.</p> </div> <div class="col-sm-3"> <img id="image" src="images/shanghai.jpg" alt="shanghai"> <a href="#"><p id="text">Shanghai</p></a> <p id="summarytext">Blend of East meets West <br>in this high energy metropolis.</p> </div> <div class="col-sm-3"> <img id="image" src="images/hcmc.jpg" alt="ho chi minh city"> <a href="#"><p id="text">HCMC</p></a> <p id="summarytext">The "Paris of Asia", Ho Chi Minh City is <br>as much historical as it is modern.</p> </div> </div>   <div class="row" style="padding-top: 50px">  <div class="col-sm-3"> <img id="image" src="images/seoul.jpg" alt="seoul"> <a href="#"><p id="text">Seoul</p></a> <p id="summarytext">Famous for pop culture, <br>vibrant shopping and <br>historical palaces.</p> </div> <div class="col-sm-3"> <img id="image" src="images/bangkok.jpg" alt="bangkok"> <a href="#"><p id="text">Bangkok</p></a> <p id="summarytext">A bustling neon-lit city that combines tradition and modernity.</p> </div> <div class="col-sm-3"> <img id="image" src="images/shanghai.jpg" alt="shanghai"> <a href="#"><p id="text">Shanghai</p></a> <p id="summarytext">Blend of East meets West <br>in this high energy metropolis.</p> </div> <div class="col-sm-3"> <img id="image" src="images/hcmc.jpg" alt="ho chi minh city"> <a href="#"><p id="text">HCMC</p></a> <p id="summarytext">The "Paris of Asia", Ho Chi Minh City is <br>as much historical as it is modern.</p> </div> </div> 功能错误。它将返回array_rand()个键，而不是单个数字。你能做的是：

array

但是，如果您的$randomImageKeys = array_rand($images, 5); for ($randomImageKeys as $key) { echo '<img src="'.$images[$key].' class="image">'; }数组包含的内容较少，那么冒着 E_WARNING 的风险超过5张图片 - 为避免这种情况，您可以使用以下内容：

$images

Answer 2

$randomImageIndexes = array_rand($images, 5);

foreach ($randomImageIndexes as $imageIndex){

   echo $images[$imageIndex];

}

Answer 3

由于array_rand($array, $n)返回一个数组$n > 1（c.f。php.net documentation : array_rand），因此您必须遍历结果才能使脚本正常工作。

它应该这样正常工作：

<?php
$imagesDir = glob('folders/pics/*', GLOB_ONLYDIR);
$randomfolder = $imagesDir[array_rand($imagesDir)];
$images = glob($randomfolder . '/*.{jpg,jpeg,png,gif}', GLOB_BRACE);
foreach(array_rand($images) as $imageIndex)
    echo '<img src="'.$images[$imageIndex].'" class="image">';

PHP：如何用array_rand显示几个随机图像？

3 个答案: