使用Variable设置后台URL - 我做错了什么？

Question

So in my code the $image is randomised like so:

$image = get_rand_img('images/featured_images/csgo/');

if I echo $image, (bearing in mind I'm already inside PHP brackets) i.e:

echo'div id="featured-image">'
echo '<img src="/images/featured_images/dota/' . $image . '" alt="" />'
echo '</div>'

Then it works.

What I actually want to do, is assign this image to the div it's in

echo '<div id="featured-image" style="height: 267px; width: 292px">#
echo '<style="background:url('. $image .') "></style>'
echo '</div>'

This doesn't work. It draws the div but doesn't apply the background. I've tried various different things like "

What is wrong?

EDIT

The following code suggestions work.

<?php
try {
    $conn = new PDO("mysql:host=$dbhost;dbname=$dbname", $dbuser, $dbpass);
}
catch (PDOException $e) {
    echo $e->getMessage();
    echo 'Could not establish a connection to the database.';
}

$query = $conn->prepare('SELECT `articleid`,`title`  FROM `news_articles` WHERE  featured = 1 ORDER BY RAND() LIMIT 1');
$array = array(
    'N'
);
$query->execute();

$results = $query->fetchAll(PDO::FETCH_COLUMN, 0);
foreach ($results as $row) {
}

$image = get_rand_img('images/featured_images/csgo/');
$title = $result['title'];
echo '<a href="index.php?viewarticle=1&articleid=' . $row . '">';?>
<div id="featured-image" style="height: 267px; width: 292px; background:url(/images/featured_images/csgo/<?php echo $image ?>)"></div>;
<?php
echo '</a>';
?>

Thanks guys!

Answer 1

There are a few things wrong here.

The </style> tag in <style="background:url('. $image .') "></style> will trigger a Stray end tag "style" warning in HTML source.

You also need to use PHP tagging and echoing the variable.

Sidenote: I doubt you want to use 2x <style> tags here, rather than using all styles set as inlined.

<div id="featured-image" style="height: 267px; width: 292px; background:url('<?php echo $image;?>')">

If this isn't what you're looking for, I'll simply delete my answer.

Edit: As per OP's comment that they're inside PHP, use and as an example:

$image = "image.jpg";

echo '
<div id="featured-image" style="height: 267px; width: 292px; background:url('.$image.')">
</div>
';

---

Edit #2: and some questions in here too, because I fear commenting will go on forever.

You're using what looks to be 2 paths here:

$image = get_rand_img('images/featured_images/csgo/');

and in (/images/featured_images/<?php echo $image ?>) which by the way, you're already in PHP here, so drop those <?php echo - ?>.

You need to use the get_rand_img() function's path here and just the $image variable, not both paths.

Sidenote: It's hard to say which path you want to use here.

Do you really have a path called

images/featured_images/csgo/images/featured_images/ - ??

• images/featured_images/csgo/ ?? (and no slash like the one below)?
• /images/featured_images/ ??

Only you know what that is. So I suggest you do/try either

$image = get_rand_img('images/featured_images/csgo/');

$image = get_rand_img('images/featured_images/');

Depending on which folder you want to extract the images from.

This

echo '<a href="index.php?viewarticle=1&articleid=' . $row . '">';?>
<div id="featured-image" style="height: 267px; width: 292px; background:url(/images/featured_images/<?php echo $image ?>)"></div>;

Should have thrown you an error. You have a ;?> after $row . '">'; and that is closing your statement and also the PHP.

So just do this:

echo '<a href="index.php?viewarticle=1&articleid=' . $row . '">
<div id="featured-image" style="height: 267px; width: 292px; background:url('.$image.');"></div>';

Add error reporting to the top of your file(s) which will help find errors.

<?php 
error_reporting(E_ALL);
ini_set('display_errors', 1);

// rest of your code

Sidenote: Displaying errors should only be done in staging, and never production.