echo image according to a condition

Question

I have an HTML tag here :

    

        

        
                      

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Answer 1

The output from your code was putting <img> tags inside the src attribute of the tag.
That by definition doesn't work in HTML. If everything else was right, this should work:

<?php

function get_random_elements( $array,$limit = 0 ) {

    shuffle($array);

    if ( $limit > 0 ) {
        $array = array_splice($array, 0, $limit);
    }
    return $array;
}

function render_images() {
    global $stmt3;
    $output = '';

    if ($count = sqlsrv_num_rows($stmt3) > 0) {
        while ($recentBadge = sqlsrv_fetch_array($stmt3)) {
            $result[] = $recentBadge;
        }

        if ($count > 3) {
            $result = get_random_elements(result, 3);
        }

        foreach ($result as $recentBadge) {
            $output .= $recentBadge['BadgeName'];
            $output .= '<img src="' . $recentBadge['BadgeImage'] . '" alt="">';
            $output .= '<br>';
        }
    } else {
        $output = 'no results';
    }

    return $output;
}
?>

<span class="fa-stack fa-5x has-badge" >

    <div class="badgesize">

        <?php echo render_images(); ?>

    </div>

</span>

As a tip: please try to keep your code separated, the logic separated from the view.