css中mySQL的变量不起作用

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.

Answer 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'].