将 PHP 添加到背景图像 url

Question

This is what I have so far but it's not working. This is my css:

background-image: url("upload/blog/'.$row["urlImg"].'");

Here is the full code:

<?php 
  $stmt = $db->query
  ( 'SELECT blog.postID, blog.postTitle, blog.postSlug, blog.postDesc, blog.urlImg, blog.postTags, blog.postDate
     FROM blog, blog_post_cats
     WHERE blog.postID = blog_post_cats.postID AND blog_post_cats.catID = 4 ORDER BY postID DESC LIMIT 1'
  ); 
  while($row = $stmt->fetch())
  {
    echo '<article style="background-image: url(upload/blog/'.$row['urlImg'].'" );"> '; 
  }
?>

This PHP echo does not work.

clicke to view output screenshot ,the blank space is the error ,others are url images

Answer 1

Since you try to use double quotes in inline CSS, you actually are closing the HTML attribute style="background-image: url(" . Use single quotes in this case of nested quotes. Rather than escaping single quotes in single-qouted echo strings, just close <?php ?> tags and write plain HTML.

<?php
  while($row = $stmt->fetch())
  {
?>
    <article class="latestPost excerpt big-1" style="background-image: url('upload/blog/<?php echo $row["urlImg"];?>');">

    </article>
<?php
  }

Answer 2

The problem are the double-quotes around the URL of the background image. The double quote after url( closes the style atrribute it's part of. So you should use escaped quotes inside the url(...) parenthesis:

echo '<article class="latestPost excerpt big-1" style="background-image: url(\"upload/blog/'.$row["urlImg"].'\");"> ';

Answer 3

Use double quote here...

     echo '<article style="background-image: url(upload/blog/'.$row["urlImg"].'");"> ';

enter image description here

Answer 4

thank y'all for your contributions

i got this myself

i just added my website address to it @ https://mywebsite.com/upload/blog/...

 echo '<article style="background-image: url(https://mywebsite.com/upload/blog/'.$row["urlImg"].'");"> ';

Answer 5

为什么不为我工作这个代码