显示来自特定位置的所有.jpg图像

Question

我有两个php文件，一个叫employeeDirectory.php，另一个叫api.php。 我正在尝试使用api.php获取所有图像，然后将其显示在employeeDirectry.php上。

employeeDirectory.php

<div class="container">
    <img src="<?php include 'api.php'; 
    echo getEmployees($images); ?>"/><br />
</div>

api.php

function getEmployees($images) {
    $directory = "/employeedirectory/";
    $images = glob($directory."*.jpg");
    foreach($images as $image){
        echo $image;
    }
return $images;

Answer 1

使用以下内容：

employeeDirectory.php

 <?php include_once ('./api.php'); ?>
 ...
 <div class="container">
    <?php 
       $images = getEmployees();
       foreach ($images as &$value) 
       {
          echo '<img src="'.$value.'"><br />';
          unset($value);
       }
    ?>
 </div>

api.php

function getEmployees() 
{
   $directory = "/employeedirectory/";
   $images = scandir($directory);
   foreach($images as $imagePath)
   {
      if (strpos($imagePath, '.jpg') !== false) 
      {
         $images[] = $imagePath;
      }
   }
   return $images;
}

Answer 2

我能做的最简单的方法

api.php

function showJpgs($dir, $bet = '') {
    $img = glob($dir.'*.jpg');
    foreach ($img as &$i)
        $i = '<img src="'.$i.'" alt=""/>';
    echo implode($bet, $img);
}

employeeDirectory.php

<div class="container">
    <?php showJpgs('images/', '<br/>'); //Insert what you want to be between the images where is '<br/>'
    ?>
</div>

Answer 3

更改

echo $image;

至

$returnimages .= "<img src='$image'>";

然后

return $returnimages;

Answer 4

这不是更好的代码吗？

<?php
include 'api.php'; ?>
<div class="container">
    <?php getEmployees(); ?>
</div>

api.php：

<?php
function getEmployees() {
    $directory = "/employeedirectory/";
    $images = glob($directory."*.jpg");
    foreach($images as $image){
        echo "<img src=\"".$image."\"/><br />";
    }
}
?>