How to loop a php code?

Question

I wanted to do the DRY approach in my code but I'm having a hard time figuring it out. And also, I want to hide the entire code if there's no image_1. Hope you could help me do the trick.

Here's the code

<div class="col-md-4">
    <?php
    $image = get_field('image_1');
    if(get_field('image_1'))
    {
        echo '<a href="' . get_field('image_link_1') . '">';?>
        <img src="<?php echo $image['url']; ?>" title="<?php echo $image['title']; ?>" alt="<?php echo $image['alt']; ?>" />
        <?php echo '</a>';
    } else {
        echo '<img src="http://localhost/image.png">';
    } ?>
</div>
<div class="col-md-4">
    <?php
    $image = get_field('image_2');
    if(get_field('image_2'))
    {
        echo '<a href="' . get_field('image_link_2') . '">';?>
        <img src="<?php echo $image['url']; ?>" title="<?php echo $image['title']; ?>" alt="<?php echo $image['alt']; ?>" />
        <?php echo '</a>';
    } else {
        echo '<img src="http://localhost/image.png">';
    } ?>
</div>
<div class="col-md-4">
    <?php
    $image = get_field('image_3');
    if(get_field('image_3'))
    {
        echo '<a href="' . get_field('image_link_3') . '">';?>
        <img src="<?php echo $image['url']; ?>" title="<?php echo $image['title']; ?>" alt="<?php echo $image['alt']; ?>" />
        <?php echo '</a>';
    } else {
        echo '<img src="http://localhost/image.png">';
    } ?>
</div>

As Jannis is saying, it's great that you already spent some time trying to figure it out! If you show us how far you've come, we can answer your question what you should do differently! — stuXnet
– stuXnet, Commented May 17, 2016 at 7:46

cdm · Accepted Answer · 2016-05-17 07:55:41Z

2

You should put differences to arrays and then wrap everything into for loop:

<?php
  $images = array('image_1', 'image_2', 'image_3');
  $links = array('image_link_1', 'image_link_2', 'image_link_3');

  for($i=0; $i<3; $i++){
?>
<div class="col-md-4">
  <?php
    $image = get_field($images[$i]);
    if(get_field($images[$i])){
      echo '<a href="' . get_field($links[$i]) . '">';
  ?>
  <img src="<?php echo $image['url']; ?>" title="<?php echo $image['title']; ?>" alt="<?php echo $image['alt']; ?>" />
  <?php echo '</a>';
    } else {
      echo '<img src="http://localhost/image.png">';
    }
  ?>
</div>
<?php
  }
?>

answered May 17, 2016 at 7:55

cdm

1,36011 silver badges18 bronze badges

Sign up to request clarification or add additional context in comments.

3 Comments

Ruby Ann Rosales Over a year ago

Btw, how can I hide it if no image_1 is uploaded?

cdm Over a year ago

@RubyAnnRosales i can think of multiple solutions but the easiest one is probably to change for condition to: for($i=0; $images[0] && $i<3; $i++)

Ruby Ann Rosales Over a year ago

it's not working. the images echoed in else are still showing.

MarcoS · Accepted Answer · 2016-05-17 07:55:11Z

0

Just a hint...:

<?php
  for ($i = 0; $i < 3; $i++) {
    echo "<div class='col-md-4'>" . "\n";
    $image = get_field("image_" . ($i + 1));
    ...
    echo "</div>" . "\n";
  }
?>

answered May 17, 2016 at 7:55

MarcoS

17.8k27 gold badges101 silver badges189 bronze badges

Comments

Nev Stokes · Accepted Answer · 2016-05-17 07:55:43Z

0

Something along these lines should get you started if I'm understanding you correctly:

<?php for ($q = 1; $q <= 3; $q++) {
    $image_loop = 'image_' . $q;

    echo '<div class="col-md-4">';

    if ($image = get_field($image_loop)) {
        echo '<a href="' . get_field('image_link_' . $q) . '">';
?>
        <img src="<?php echo $image['url']; ?>" title="<?php echo $image['title']; ?>" alt="<?php echo $image['alt']; ?>" />
        <?php echo '</a>';
    } else {
        echo '<img src="http://localhost/image.png">';
    } ?>
</div>

<?php } // end loop ?>

answered May 17, 2016 at 7:55

Nev Stokes

9,9475 gold badges45 silver badges46 bronze badges

2 Comments

Ruby Ann Rosales Over a year ago

But, how can I hide it if no image_1 is uploaded?

Nev Stokes Over a year ago

That's left as an exercise for you. Stackoverflow isn't a place to get your code written for you! Hint: maybe test for image_1 before the for loop.

apokryfos · Accepted Answer · 2016-05-17 08:06:44Z

The other suggestions here will work as well but here's what I would do:

First arrange the images in an associative array with keys being the image name and values being the image link and then iterate via a foreach loop.

I generally try to not echo HTML unless strictly necessary.

<?php
$array = [
    "image_1" => "image_link_1",
    "image_2" => "image_link_2",
    "image_3" => "image_link_3",
    "image_4" => "image_link_4"
];

foreach ($array as $name => $link):
    $image = get_field($name); 
    if ($image): ?>
    <div class="col-md-4">
        <a href="<?=get_field($link)?>">
            <img src="<?= $image['url']; ?>" title="<?= $image['title']; ?>" alt="<?= $image['alt']; ?>" />        
        </a> 
        <?php else: ?>
            <img src="http://localhost/image.png">
        <?php endif; ?>
    </div>
<?php endforeach; ?>

Collectives™ on Stack Overflow

How to loop a php code?

4 Answers 4

3 Comments

Comments

2 Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

4 Answers 4

3 Comments

Comments

2 Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related