使用 PHP 从列表中随机显示 HTML 片段

Question

As the title implies, I have a few different lines of HTML code (Example Below), and I would like the server to randomly select one and show it when the page loads.

I have these three HTML snippets, and I want one to randomly show when the page is loaded.

<img src=“image1.png” alt=“image1”>
<p class=“class1”>Image Caption<a href=“link1”> And Link</a></p>

<img src=“image2.png” alt=“image2”>
<p class=“class2”>Image Caption<a href=“link2”> And Link</a></p>

<img src=“image3.png” alt=“image3”>
<p class=“class3”>Image Caption<a href=“link3”> And Link</a></p>

I found this example using the PHP array function, but I don't know how to put the array in a separate PHP file, or if I can change the image links to full HTML code.

The example I found:

<?php
    $images = array('img1.jpg', 'img2.jpg', 'img3.jpg');
?>

Then, you just get a random element from the array

<?php
    $random_image = array_rand($images);
?>

Then you can go right away displaying the image:

<img src="<?php echo $images[$random_image]; ?>" />

How can I make this so the HTML snippets above are stored in the array, and the array is in a different PHP file (I can use the “require” function for that, right?).

Answer 1

You can echo any amount of html code just like you would echo the value of a property.

You're right that you can use require to access an array defined in another file.

<?php
    $images = require 'images_array.php';
    echo $images[array_rand($images)];
?>

images_array.php:

array(
    '<img src=“image1.png” alt=“image1”><p class=“class1”>Image Caption<a href=“link1”> And Link</a></p>',
    '<img src=“image2.png” alt=“image2”><p class=“class2”>Image Caption<a href=“link2”> And Link</a></p>',
    '<img src=“image3.png” alt=“image3”><p class=“class3”>Image Caption<a href=“link3”> And Link</a></p>',
);

Answer 2

I recommend that you avoid bloating your file with redundant html markup. Instead of repeating yourself, practice the DRY principle and separate all of the variable parts from the static parts.

Using a lookup array and a template string with improve maintainability and reduce the potential for typos.

Code: ( Demo )

$lookup = [
    1 => [
        'img' => 'image1.png',
        'alt' => 'image1',
        'class' => 'class1',
        'caption' => 'Image Caption',
        'href' => 'link1',
        'linkText' => 'And Link',
    ],
    2 => [
        'img' => 'image2.png',
        'alt' => 'image2',
        'class' => 'class2',
        'caption' => 'Image Caption',
        'href' => 'link2',
        'linkText' => 'And Link',
    ],
    3 => [
        'img' => 'image3.png',
        'alt' => 'image3',
        'class' => 'class3',
        'caption' => 'Image Caption',
        'href' => 'link3',
        'linkText' => 'And Link',
    ],
];

$markup = '<img src="%1$s" alt="%2$s">
<p class="%3$s">%4$s <a href="%5$s">%6$s</a></p>';

vprintf($markup, $lookup[array_rand($lookup)]);

Potential Output:

<img src="image3.png" alt="image3">
<p class="class3">Image Caption <a href="link3">And Link</a></p>

---

If you want to store the array in another file, okay. Just be sure to return the array in the file being require ed. Example

lookup.php

<?php
return [1 => [...], 2 => [...], 3 => [...]];

markup.php

<?php
$lookup = require 'lookup.php';
$markup = '<img src="%1$s" alt="%2$s"><p class="%3$s">%4$s <a href="%5$s">%6$s</a></p>';
vprintf($markup, $lookup[array_rand($lookup)]);

Answer 3

As you seems to have pure PHP and no template engine you can do this :

<?php
$random = rand(0,60);
if ($random < 20) {
?>
<img src=“image1.png” alt=“image1”>
<p class=“class1”>Image Caption<a href=“link1”> And Link</a></p>
<?php
} elseif ($random < 40) {
?>
<img src=“image2.png” alt=“image2”>
<p class=“class2”>Image Caption<a href=“link2”> And Link</a></p>
<?php
} else {
?>
<img src=“image3.png” alt=“image3”>
<p class=“class3”>Image Caption<a href=“link3”> And Link</a></p>
<?php
}
?>

It'll display a random part at each page load.