I’m echoing the post thumbnail here:
echo '<div class="post-image right">' . the_post_thumbnail('featured-portrait-large') . '</div>';
And it works, but it’s outputting the image outside the div like this:
<div class="post-header-text"> <img width="500" height="700" src="[IMAGE URL DELETED]" class="attachment-featured-portrait-large size-featured-portrait-large wp-post-image" alt="" loading="lazy"> <div class="post-image right"></div> </div>
What I want is:
<div class="post-header-text">
<div class="post-image right">
<img width="500" height="700" src="[IMAGE URL DELETED]" class="attachment-featured-portrait-large size-featured-portrait-large wp-post-image" alt="" loading="lazy">
</div>
</div>
Could someone explain why and what I’ve done wrong?
Answers:
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Method 1
the_post_thumbnail() echo the output, hence that’s why the thumbnail is misplaced.
To manually echo the output, you should use get_the_post_thumbnail() instead:
echo '<div class="post-image right">' . // wrapped
get_the_post_thumbnail(null, 'featured-portrait-large') . '</div>';
Or you can instead do this:
echo '<div class="post-image right">';
the_post_thumbnail('featured-portrait-large');
echo '</div>';
Method 2
You can simply get image url then add this url to image tag, I hope this will help you.
<?php $thumb_id = get_post_thumbnail_id(); $thumb_url = wp_get_attachment_image_src($thumb_id,'medium', true); $imageURL = $thumb_url[0]; ?>
All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0