I’m using this code to display an image, I need to use the Full image size and use display its width and height in my markup.
If i set $size
to thumbnail/medium/large the code returns the size values, but does not if $size
is set to full.
Can anyone explain/advise?
<?php
$image = get_field('image_object');
$url = $image['url'];
$size = 'full';
$width = $image['sizes'][ $size . '-width' ];
$height = $image['sizes'][ $size . '-height' ];
?>
<img src="<?php echo $url; ?>" width="<?php echo $width; ?>" height="<?php echo $height; ?>"/>
Hi @muckleby
The problem is that fullsize “width” and “height” does not exists in the array.
<?php
// You will get the "full" size image with:
$url = $image['url'];
// Full-size Image Width:
$image['width'];
// Full-size Image Height:
$image['height'];
echo '<img src="'.$image['url'].'" width="'.$image['width'].'" height="'.$image['height'].'" />';
// You can check all available values with:
echo "<pre>";
var_dump($image);
echo "</pre>";
?>
Hope i could help you.
awesome i hoped it would be as simple as this.
big thanks rjantis!
seems like the $image['width'];
property should be mentioned in the documentation. perhaps i missed it, or maybe it seems elementary to a more experienced brain.
@muckleby you’re welcome!:)