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up vote 0 down vote favorite

I'm using a PHP/MySQL to style a web app using dynamic css (style.php).

The MySQL values are determined from the URL:

$url = "http://".$_SERVER['HTTP_HOST'].$_SERVER['REQUEST_URI'];

if($url == "this") $style = "blue";
else( $style = "red"; )

The problem I seem to be having is that the style.php uses:

header('Content-type: text/css');

This causes $url to be equal to: "http://", also any other variables assigned outside of the style.php file are ignored.

Does anyone know how to get these $_SERVER (and other) variables to work?

Here is the full code

$url = "http://".$_SERVER['HTTP_HOST'].$_SERVER['REQUEST_URI']; // current URL

$key = true;

while($key){

mysql_select_db($database, $connection);
$query_rsTheme = "
SELECT      s.name, s.url, t.name as theme, t.colour, t.hover 
FROM        db_site as s
INNER JOIN  db_theme.theme as t ON t.name = s.theme
WHERE       s.url = '$url'";
$rsTheme = mysql_query($query_rsTheme, $connection) or die(mysql_error());
$row_rsTheme = mysql_fetch_assoc($rsTheme);
$totalRows_rsTheme = mysql_num_rows($rsTheme);

if($totalRows_rsTheme == 1){ // sucessful match
    $key = false;

    $site = $row_rsTheme['name'];
    $theme = $row_rsTheme['theme'];
    $default_state = $row_rsTheme['colour'];
    $hover_state = $row_rsTheme['hover'];
}

$tokens = explode('/', $url);
$remove = $tokens[sizeof($tokens)-2]."/";
$url = substr($url, 0, strpos($url, $remove));
}

header('Content-type: text/css');
header("Cache-Control: no-cache, must-revalidate");
header("Expires: Sat, 26 Jul 1997 05:00:00 GMT");

$stylesheet = 'style.css';

$content = preg_replace('/\$([\w]+)/e','$0',@file_get_contents($stylesheet));

echo $content;
share|improve this question
    
How is style.php invoked? –  hank Aug 12 '13 at 8:36
    
what is purpose of style.php? –  VIVEK-MDU Aug 12 '13 at 8:36
    
if($url == "this"){ $style = "blue";} else { $style = "red";} –  Vector Aug 12 '13 at 8:38
    
<link rel="stylesheet" href="/include/version-3/css/style.php" type="text/css" media="screen, projection"/> –  user715105 Aug 12 '13 at 8:38
    
I can't understand what your problem is but I can assure you that the header() function does not write anything in the $_SERVER array. –  Álvaro G. Vicario Aug 12 '13 at 8:40

3 Answers 3

up vote 1 down vote accepted

You mention several times that $_SERVER is empty but I suspect you don't really test it:

print_r($_SERVER);

Whatever, your style.php script assumes that certain global variables exist (namely $database and $connection). If you've really posted the complete script, you never define them.

You also mention:

any other variables assigned outside of the style.php file are ignored.

Of course. That's how PHP works: each script is independent. Thankfully, style.php will not pick variables from any other random script that runs on the same server.

My advice is:

  1. Enable full error reporting. It's clear that you aren't seeing notices and possibly warnings and errors.

  2. Test the script separately. Load http://example.com/include/version-3/css/style.php in your browser and see the generated code, rather than possibly relying on styles showing up in your HTML.

share|improve this answer
    
problem found: style.php is being accessed through http://, so the server variables were being set, i was calling the $url variable at a later stage where the $tokens were being removed. I added a landing page parameter to style.php?page=$currentPage; which solves my problem. sorry for the confusion.#] –  user715105 Aug 12 '13 at 9:35
up vote 0 down vote

You could check if the URI matches certain characters

if (strpos($_SERVER['REQUEST_URI'], 'this') !== FALSE ){
    $style = "blue";
} else {
    $style = "red";
}

This is particularly helpful if the file you are using is in fact an include into another file.

share|improve this answer
up vote 0 down vote

I believe the problem is not what you're describing. So long as style.php is accessed via http, then the $_SERVER variables will be set.

There is, however, some syntax error in the code you described:

if($url == "this") $style = "blue";
else( $style = "red"; )  // Incorrect syntax

The right way to write that would be:

if ($url == "this") { // $url will _never_ be "this"
    $style = "blue";
} else {
    $style = "red";
}

Edit: There's some funky code when evaluating the MySQL results:

$row_rsTheme = mysql_fetch_assoc($rsTheme);
$totalRows_rsTheme = mysql_num_rows($rsTheme);

if($totalRows_rsTheme == 1){ // sucessful match, but ONLY if there's only one row
    ...
}

You should replace it with:

if($row_rsTheme = mysql_fetch_assoc($rsTheme)){ // sucessful match
    ...
}

That way, there will be success even if there's more than one result.

share|improve this answer
    
yes, apologies, the syntax is wrong, this is corrected. –  user715105 Aug 12 '13 at 8:43
    
But still no cigar? Can you show us the whole php code? –  Nicklas Nygren Aug 12 '13 at 8:44
    
original post updated –  user715105 Aug 12 '13 at 8:53
    
sweet, thanks for the info –  user715105 Aug 12 '13 at 9:34

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