CODEWITHSUNDEEP
phpmysqlarrayswordpress
Jay-oh
Jay-oh

Reputation: 456

PHP Array values are correct but not displayed correct

I've been trying to combine two database queries in to one. Now I've finally done this (yeah). But One problem keeps me from completing this task. And I bet its really simple but I can't seem to get it to work.

I've a database query set-up and it works.

    $sql = "SELECT `guid` FROM `pf_posts` WHERE `id` IN (SELECT  `meta_value`
    FROM  `pf_postmeta`
    WHERE  `meta_key` LIKE  '_thumbnail_id'
    AND  `post_id` = $post_id)";

    $result = mysqli_query($wp_database, $sql);
    $images = mysqli_fetch_all($result, MYSQLI_ASSOC);
    mysqli_free_result($result);
    foreach ($images as $image): ?>
        <div>
             <?php print_r($image); ?>
        </div>    
    <?php endforeach; ?>

This get's me the correct value BUT in a print_r which get's me this:

Array ( [guid] => http://mydomein.nl/wp-content/images-001.jpg )

But I would like to remove the Array ( [guid] => and the last ) part from the print_r.

I've looked into string replace but I have no idea how to set correct. I would like to be able to echo it. I've gotten the code for the database from here: How can I retrieve posts with featured images from a WordPress database if WordPress is no longer installed? [closed]

Upvotes: 1

Views: 54

Answers (2)

Muhammed Mehdi
Muhammed Mehdi

Reputation: 1

Can you try this:

<?php print_r($image['guid']); ?>

Upvotes: 0

ishegg
ishegg

Reputation: 9947

print_r will print a human-readable representation of the argument. In your case, $image is an array with one element, key "guid" and value "http://mydomein.nl/wp-content/images-001.jpg".

So, what you need to do is just echo the element you want:

<?php
foreach ($images as $image): ?>
    <div>
            <?php echo $image["guid"] ?>
    </div>    
<?php endforeach; ?>

Upvotes: 4

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