Variables from mySQL within css not working

Question

I am trying to use a form to set colors in a table in my database, then call those colors into my stylesheet (style.php)

Defining the color variables manually works:

<?php 
    header("Content-type: text/css"); 
    $color1  = '#cc0000';
?>

#thisdiv { background-color: <?php echo $color1; ?>; }

But this does not:

<?php 
    header("Content-type: text/css"); 
    $getSettings = mysqli_query($db, "SELECT * FROM settings WHERE setting_id = 1");
    $setting     = mysqli_fetch_assoc($getSettings);
    $color1      = $setting['setting_color1'];
?>

#thisdiv { background-color: <?php echo $color1; ?>; }

My 'settings' table looks like this:

setting_id | setting_color1 | setting_color2 | setting_color3
1          | #cc0000        | #000000        | #ffffff

How can I get this working properly? Maybe I am doing something wrong, or maybe I am just overlooking something stupid. Any help is appreciated.

Try inspecting the code and verify what is in there. You can do this by pressing CTRL+U — Phiter
– Phiter, Commented Mar 17, 2016 at 17:25
How this code is supposed to work without a database connection? — Your Common Sense
– Your Common Sense, Commented Mar 17, 2016 at 17:26
@PhiterFernandes there is nothing in background-color - it's just empty. @YourCommonSense $db connection is defined elsewhere. It was tested and is working properly. I figured that would be assumed, sorry. The query was checked (as per a comment below) and is working also. — pruoccojr
– pruoccojr, Commented Mar 17, 2016 at 17:42
Run a var_dump($setting) before you echo it and check the values. — Phiter
– Phiter, Commented Mar 17, 2016 at 17:51
A db connection doesn't work if you define it elsewhere and have not a single reference to it in your code — Your Common Sense
– Your Common Sense, Commented Mar 17, 2016 at 18:01

Xorifelse · Accepted Answer · 2016-03-17 17:50:55Z

1

<?php 
    header("Content-type: text/css"); 
    $getSettings = mysqli_query($db, "SELECT * FROM settings WHERE setting_id = 1");
    // --------------------------^ and where is that defined?
    $setting     = mysqli_fetch_assoc($getSettings);
    $color1      = $setting['setting_color1'];
?>

#thisdiv { background-color: <?php echo $color1; ?>; }

You should always verify if your query ran successfully, isn't that hard.

<?php
  if($result = mysqli_query($db, 'SELECT * FROM table') != false){
    var_dump(mysqli_fetch_assoc($result));
    // hurray, query ran successfully.
    // Now execute the code that sets the variable.
  } else {
    // On fail, perhaps load from cache?
  }
?>

Secondly, your DB structure is wrong in my opinion. Try something like the following.

settings:

id (primary key)| key (varchar)| value (varchar)| type (varchar)| decryption (text) | theme (int)
1                 mycolor        AABBCC           hex             Background of we..   1

$sql = 'SELECT key, value, type FROM settings WHERE theme = 1'; // etc..

This way, you have much more control over your 'theme'. Then you can while-fetch the results and set: $arr[$row['key']] = $row['value'].

edited Mar 17, 2016 at 17:50

answered Mar 17, 2016 at 17:30

Xorifelse

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2 Comments

pruoccojr Over a year ago

Hi, thanks for the reply. I have $db defined elsewhere, but the connection was tested and is working properly. I checked my query with your code above and it was successful as well.

Xorifelse Over a year ago

If that's the case, try the updated code it would show debug output that is actually quite important to know.

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