Changing CSS with PHP

Question

So i read article on css-tricks.com˙(http://css-tricks.com/css-variables-with-php/) about PHP in CSS and I tried it myself. I'm having problem structuring it. To be specific, i'm using Sql to take image from database and use result as a background image. Like this:

<?php

header("Content-type: text/css; charset: UTF-8");

$con = mysql_connect("localhost","root","");
if (!$con)
{
    die('Could not connect: ' . mysql_error());
}
mysql_select_db("login", $con);
$result = mysql_query("SELECT * FROM users");
while($row = mysql_fetch_array($result))
{
?>

Then goes my css, i will just write needed ID:

#user {
background: url(../img/<?php echo $row['cover']; ?>) fixed;
}

And at the end, I close connection:

<?php
};

mysql_close($con);

?>

Everything works except css is outputted double, that is listed twice.Can anyone point the problem? Thanks.

after $result = run echo mysql_num_rows($result) and tell us the output. — Andrew Brown
– Andrew Brown, Commented Mar 14, 2014 at 17:54
It's a 5 year old article. I suggest you use a more recent tutorial for learning. Also, use the MySQLi or PDO extensions with prepared statements and bind variables, rather than learning using the old, deprecated MySQL extension. — Amal
– Amal, Commented Mar 14, 2014 at 17:55
I haven't even looked at the date of article -.- i guess I need to start. — Emir Dupovac
– Emir Dupovac, Commented Mar 14, 2014 at 18:00

andrew · Accepted Answer · 2014-03-14 17:58:17Z

1

you're going to create:

   background: url(../img/<?php echo $row['cover']; ?>) fixed;

for each row returned by the database. you need to limit the results, either in the sql like:

 $result = mysql_query("SELECT * FROM users where userId = $userId");

or in the php like:

if ($row['userId']==$userId){ ?>
     background: url(../img/<?php echo $row['cover']; ?>) fixed;
<?php}

answered Mar 14, 2014 at 17:58

andrew

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mathius1 · Accepted Answer · 2014-03-14 18:06:10Z

0

It will return as many values as you have in your DB table "users"

If you only want 1 result you need to write

"SELECT * FROM users LIMIT 1"

You dont have a "WHERE" statement for the query?

"SELECT * FROM users WHERE SOME_COLUMN = $your_variable "

edited Mar 14, 2014 at 18:06

answered Mar 14, 2014 at 17:58

mathius1

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