How to include derived images from database into css file

Question

I would like to derive my website background image from database.those background images are loading through banner.css file.

banner.css 

<?
header("Content-type: text/css");  
include("getfrontimage.php");
$theme = "images";
$size = "cover";
$height = "780px";

?>
<style>
.w3layouts-banner-top{
    background: url(<?php  echo $theme.'/'.$row1[image];?>) no-repeat 0px 0px;
    background-size: <?php echo $size; ?>;
    -webkit-background-size: <?php echo  $size; ?>;
    -moz-background-size: <?php echo $size; ?>;
    -o-background-size: <?php echo $size; ?>;     
    -moz-background-size: <?php echo $size; ?>;
    min-height: <?php echo $height; ?>;
}
.w3layouts-banner-top1{ 
    background: url(<?php echo  $theme.'/'.$row2[image]; ?>) no-repeat 0px 0px;
    background-size: <?php echo $size; ?>;
    -webkit-background-size: <?php echo $size; ?>;
    -moz-background-size: <?php echo $size; ?>;
    -o-background-size: <?php echo $size; ?>;     
    -moz-background-size: <?php echo $size; ?>;
    min-height: <?php echo   $height; ?>;
}
.w3layouts-banner-top2{
    background: url(<?php echo $theme.'/'.$row3[image]; ?>) no-repeat 0px 0px;
    background-size: <?php echo $size; ?>;
    -webkit-background-size: <?php echo $size; ?>;
    -moz-background-size: <?php echo $size; ?>;
    -o-background-size: <?php echo $size; ?>;     
    -moz-background-size: <?php echo $size; ?>;   
    min-height: <?php echo $height; ?>;
}
</style>

But background images are not loading in my website. I refer this links enter link description here

please give me any idea to resolve this problem.

I would assume you can only do it inside the markup, as php code run there, but not in the css files. — Rainbow, May 12 '18 at 12:16
Where's your code to initialise variables like $row1, $row2 etc? You may inspect the CSS file from your browser to see if all those variables are set properly. — art, May 12 '18 at 12:25
You might need to rename your file to `.php` unless otherwise you have set your server to parse the php code in `.css` file. — art, May 12 '18 at 12:29
@art06 i initialize into include("getfrontimage.php"); file.. and i tried that also — Arebhy Sridaran, May 12 '18 at 12:35

score 2 · Accepted Answer · answered May 12 '18 at 13:22

$row1[image] must be $row1['image'], because the array key can either be an integer or a string. If you enable display of warnings, you'll get a message in this case: Warning: Use of undefined constant image Correct the other variables.

Processing PHP code in files with the .css extension is not a not quite optimal solution, because you need to add special directives so that your web server processes css files as php files. It is easier to include CSS rules in a php file, and then add the css file to a web page using:

<link rel="stylesheet" href="css/styles.php">

By the way, short tags <?= $size ?> are more readable than <?php echo $size; ?>.

Did my answer help to solve the problem, @Arebhy Sri? – camelsWriteInCamelCase May 12 '18 at 20:27 — camelsWriteInCamelCase, May 12 '18 at 20:27

How to include derived images from database into css file

1 Answers1