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PHP - Resolve Relative Path Outside Document Root
I need to find a way to resolve a relative path outside the document root, in a cross-platform friendly manner.
My users have a settings page where they are able to set the path to a folder where files should be included. This path may not exist at the time of saving the setting. The given path is then retrieved from the database when files are being saved, the path is checked to see if a folder needs to be created, and the file is saved to the path. Two possible paths they may use a * files (This is the webpath: http://site.com/files or absolute path /home/user/public_html/files) * ../files (This is the absolute path: /home/user/files where the webroot is /home/user/public_html/) The first path is easy to deal with. However, I'm having a rough time resolving the second path into a usable system path (i.e. /home/user/files). This needs to be cross platform compatible (windows/'Nix). I've played around with realpath(), but I'm just not finding something that works for me. Any suggestions? Similar Tutorialshey 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, 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 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" , $_SERVER['HTTP_HOST']); //and used it like this <a href='<?php echo SERVER_NAME; ?>/filename.php'>go to that page</a> it works when we're on root directory. but when we go to sub directory it added sub directory name with it. which i don't want. is there any way? Thanks im using this script to upload files i want to use it inside an iframe but get an error it looks like a path to root issue it works fine with no errors NOt in an iframe Code: [Select] if (!empty($_FILES)) { $tempFile = $_FILES['Filedata']['tmp_name']; $targetPath = $_SERVER['DOCUMENT_ROOT'] . $_REQUEST['folder'] . '/'; // $fileTypes = str_replace('*.','',$_REQUEST['fileext']); // $fileTypes = str_replace(';','|',$fileTypes); // $typesArray = split('\|',$fileTypes); // $fileParts = pathinfo($_FILES['Filedata']['name']); $tempName = $_FILES['Filedata']['name']; //$tempNameEnd = explode('.',$tempName); //$tempNameEnd = $tempNameEnd[1]; $tempName = basename($tempName); if (isset($_POST['apartmentID']) && is_numeric($_POST['apartmentID'])) { $tempName = $_POST['apartmentID'].'-'.rand(0,999999).'-'.$tempName; //.$tempNameEnd; } else { $tempName = rand(0,999999).'-'.$tempName; //.$tempNameEnd; } $targetFile = str_replace('//','/',$targetPath) . $tempName; // if (in_array($fileParts['extension'],$typesArray)) { // Uncomment the following line if you want to make the directory if it doesn't exist // mkdir(str_replace('//','/',$targetPath), 0755, true); $dbTargetFile = $targetFile; $targetFile = '/srv/disk1/744444/www/example.com'.$targetFile; $success = move_uploaded_file($tempFile,$targetFile); echo str_replace($_SERVER['DOCUMENT_ROOT'],'',$targetFile); if (isset($_POST['apartmentID']) && is_numeric($_POST['apartmentID'])) { $query = 'INSERT INTO images SET ID = \''.mysql_real_escape_string($_POST['apartmentID']).'\', ImageURL = \''.$dbTargetFile.'\', InternalSupplierID = \'100\''; fwrite($fp,$query.PHP_EOL); mysql_query($query); fwrite($fp,var_export(mysql_error(),true).PHP_EOL); //security risk!! :O //can only fix if we change image uploading script $query = 'UPDATE apartments SET mainImage = \''.$dbTargetFile.'\' WHERE ID = \''.mysql_real_escape_string($_POST['apartmentID']).'\''; mysql_query($query); } // } else { // echo 'Invalid file type.'; // } } fclose($fp); ?> Directory structu /home/~test/public_html/soap/ .../xml/ .../bin/ .../products/browser/ In file /home/~test/public_html/soap/bin/products/browser/test.php, I include another file and call it's function: Code: [Select] require("../../xml/processXML.php"); $xmlData = generateXML(....); In the /home/~test/public_html/soap/xml/processXML.php, I have the following code: Code: [Select] function generateXML() { exec("../bin/code.cur"); } The problem is the exec("../bin/code.cur") fails of the relative path issue. I did a getcwd() before the exec() and it returned /home/~test/public_html/soap/bin/products/browser. I have tried using dirname() and no help. I hope this is clear enought to understand where I am coming from. Thank you. AM 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. Hey y'all.
I'll gladly admit that I'm not a server guy - I have the utmost respect for people comfortable enough with *nix to handle Apache setup and configuration in a production environment. I can noodle about in *nix typically without killing anything vitally important, but trust me when I say no-one wants me to be a server administrator. That having been said, I do have a development server behind my home network firewall running Ubuntu that I use to test and develop the sites that I create on my separate Windows box. So I try to keep up with the way a shared server would be set up in a real-world situation, to a degree. For instance, I haven't bothered setting up the mail server or FTP, but I do create virtual host files for each of my projects.
So, I've got a virtual host set up on the directory /var/www/myAwesomeSite.com/ with the .conf file containing the following configuration:
<VirtualHost *:80>
ServerAdmin webmaster@localhost
ServerName myAwesomeSite.com
DocumentRoot /var/www/myAwesomeSite.com/www
<Directory /var/www/myAwesomeSite.com/www>
AllowOverride All
</Directory>
ErrorLog ${APACHE_LOG_DIR}/error.log
CustomLog ${APACHE_LOG_DIR}/access.log combined
</VirtualHost>
Now, in my php scripts echoing out $_SERVER['DOCUMENT_ROOT'] gives me 'myAwesomeSite/www' as I would expect. And using $_SERVER['DOCUMENT_ROOT'] in any require() statements actually does include the requested file, so that part seems to be working fine. My question is this: when I use a leading slash ('/'), shouldn't that equate to the document root? I use it at work all the time and it works flawlessly, but on the dev server I have set up here, it's a no-go. Is there another config command that I need to issue to make it work, or am I crazy?
Basically, am I missing a setting somewhere that will make this:
require_once('/path/to/my/include.php');
work like this?
require_once($_SERVER['DOCUMENT_ROOT'].'/to/my/include.php');*edit* I just realized that I used the wrong slash in the title of this post, but I don't see a way to change the topic title. crap. Edited by maxxd, 23 November 2014 - 11:20 AM. Any thoughts on this function? Basically what it does is take the current users location ($location) and a relative path ($link) and convert it to a valid URL. I do know that this does work on my example, but does anyone have any suggestions on this? Code: [Select] <?php $location = "http://mysite.com/pages/images/docs/myfile.html"; $link = "../../games/game.html"; function resolve_url($current, $link){ $ui = (object)parse_url($current); $domain = $ui->scheme."://".$ui->host; $levelsA = preg_split("/\.\.\//", $link); $levels = 0; foreach($levelsA as $l){ if(empty($l)) $levels++; } $currentA = preg_split("/\//", $current); $length = count($currentA); array_splice($currentA, -$levels - 1); $currentA[] = preg_replace("/\.\.\//", "", $link); return implode("/", $currentA); } echo resolve_url($location, $link); ?> Thanks! Howdy,
So I am creating a website with SQLite and this time I decided to follow the rule of putting the SQLite DB outside of the document root. However, the admin would need to admin, basically, the DB, so I have two choices:
a) Put PhpLiteAdmin (The SQLite DB manager like PHPmyAdmin is for MySQL) in the document root while giving it a path to the actually DB that is outside of the document root;
b) Set a subdomain to a directory outside of the document root of course and put both PhpLiteAdmin and the SQLite DB there and the admin would access it like this - db.Awesome-Website.com;
Which one of those two options are safer?
Thank you very much!
I have been trying to learn how to build a php/mysql website using my local computer for testing. I'm at the point now where I need to move my files over to our server so I can give an update and show how much I have done. On my computer I have been setting up my include lines like this. include $_SERVER['DOCUMENT_ROOT'] . '/includes/head.inc.php'; the problem is that when I move my files over to the server I want to put them in a .htpassword protected directory called dbsite until the new site is ready to go live at which point I will move the files out of that directory and put them in the root dir. (there is currently I live site which I am replacing with this new one in the root right now, but until I'm ready I can't delete/overwrite the current one) My question is, is there a way to make $_SERVER['DOCUMENT_ROOT'] = the dbsite dir instead of the actual root dir? Just temporarily while the site is in the testing directory on the server. Otherwise I have to change all of my lines on my local computer from include $_SERVER['DOCUMENT_ROOT'] . '/includes/head.inc.php'; to include $_SERVER['DOCUMENT_ROOT'] . '/dbsite/includes/head.inc.php'; which will then cause the files on my local computer to not work correctly. Am I making sense as to my dilemma? Is there an easier/better solution? Thanks. I found one post that related to this topic but I want to make sure I understand. If I have a site with the structu /root/ ____pageone.php ____common/ ___________base.php ___________header.php ____inc/ ______constants.inc.php ____admin/ _________login.php and base.php includes: <?php // Include site constants include_once "../inc/constants.inc.php"; ?> and login.php includes: <?php include_once "../common/base.php"; $pageTitle = "Log in"; include_once "../common/header.php"; ?> and pageone.php includes: Code: [Select] <?php include_once "common/base.php"; $pageTitle = "Page One"; include_once "common/header.php"; ?>the login page loads normally, but pageone.php is broken; I get error messages like Quote Warning: include_once(../inc/constants.inc.php) [function.include-once]: failed to open stream: No such file or directory in /Applications/XAMPP/xamppfiles/htdocs/rootdirectory/common/base.php on line 10 . From what I understand, PHP doesn't want to include the same files from different relative paths; you either need to use absolute paths or change the directory structure so that the included files are accessed through the exact same relative path. Is this correct? I find that if I move pageone.php to the admin folder and make the relative path the same, it does work. I have to say, I'm leery of using absolute paths; and if I'm understanding this correctly, you're really constrained on how you set up your directory structure. Comments? This might be better for the Apache forum, but I'll explain anyways. I'm switching up my local dev environment so I can use SVN. I have a directory for all my SVN stuff on my local machine now in /var/svn. When I call $_SERVER['DOCUMENT_ROOT'] its listing the document root as /var/www. For this alias I've setup in apache I want the document root to be /var/svn/myproject. Alot of my includes are failing because of this now. Given a directory - folder1/subfolder1, I would like to loop through all subdirectories under the given directory and return the file path relative to the given directory. Example: Code: [Select] Given Directory FolderA FolderB SubFolderA SubFolderB SubSubFolderA File1 FolderC the file path for File1 should be FolderB/SubFolderB/SubSubFolderA/ Any ideas? What is an equivalent super global to $_SERVER['PHP_SELF'] that gives me JUST the currently opened document name without the path? e.g. profile.php but NOT: directory1/directory2/profile.php What I mean by that is, if I echo out $_SERVER['PHP_SELF'], it will give me the directories as well, but I'm just interested in the document name. I've looked for the $_SERVER super global in the PHP manual, but no results come up. 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 + $unknowncount;
?>
Can anyone tell me the error here, can how to fix it? <?php session_start(); if (!$_SESSION["user_name"]) { // User not logged in, redirect to login page Header("Location: login.php"); } // Member only content // ... $con = mysql_connect('localhost','root',''); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("login", $con); $result = mysql_query("select * from reference,users where reference.username=users.user_name AND reference.refid = '".$_POST['refid']."'"); while($row = mysql_fetch_array($result)) { echo $row['refid'];?><br><? echo $row['origin']; echo $row['dest']; echo $row['date']; echo $row['exdate']; echo $row['user_name']; } // ... // ... // Display Member information // echo "<p>User ID: " . $_SESSION["valid_id"]; //echo "<p>Username: " . $_SESSION["valid_user"]; //echo "<p>Logged in: " . date("m/d/Y", $_SESSION["valid_time"]); // Display logout link echo "<p><a href=\"logout.php\">Click here to logout!</a></p>"; ?> for the above code i got error Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in solve my problem Haven't been here in a while but I have stumped myself at the moment or maybe I've looked at it too much LOL In the given code below I am getting this error Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in C:\xampp\htdocs\samples\getcustomers.php on line 20 WHY? Code: [Select] <?php $value = $_GET["q"]; $con = mysql_connect('localhost', 'root', ''); //db_connect(); if (!$con) { die('Could not connect: ' . mysql_error()); } $db_selected = mysql_select_db("ajaxDb",$con); $sql = "SELECT * FROM 'cutomers' WHERE Name = '".$value."'"; // echo $sql; $result = mysql_query($sql); echo "<table border='1'> <tr> <th>Name</th> <th>Address</th> <th>Country</th> </tr>"; while($row = mysql_fetch_array($result)) { echo "<tr>"; echo "<td>" . $row['Name'] . "</td>"; echo "<td>" . $row['Address'] . "</td>"; echo "<td>" . $row['Country'] . "</td>"; echo "</tr>"; } echo "</table>"; mysql_close($con); ?> Thanks for reading this. I have just begun using php, and I cannot, for my life, figure out what I am doing wrong here. I want to submit a form for a person to register with my site. It's simple, just name, email and password. I keep getting an error that the connection failed, so I will just show you what I have up until that point. My form is Post method, and it looks like: Code: [Select] <form method="post" action="signup.php" enctype="multipart/form-data"> <input name="name" type="text" id="name"> <input name="email" type="text" id="email"> <input name="pass" type="password"> <input type="submit" name="Submit" value="Submit"> </form> And the connection script is like so: Code: [Select] <?php $username="db_username"; $password="db_password"; $database="db_name"; $dbc=mysql_connect ("localhost", "$username", "$password") or die ('I cannot connect to the database.'); mysql_select_db ("$database", "$dbc"); It's the 'I cannot connect to the database.' error that I keep getting. In case it helps, the rest of the thing looks like this: Code: [Select] <?php $name= $_POST['name']; $email= $_POST['email']; $pass= $_POST['pass']; $query = "INSERT INTO table (name, email, password) VALUES ('$name', '$email', '$pass')"; mysql_query($query); mysql_close(); I have deliberately left out the php opening and closing tags (<?php...?>). Thanks so much for any help. Adam MOD EDIT: [code] . . . [/code] BBCode tags added. This is how looks bug looks : Code echo "<div align='center' class='thread'>Forum</div>"; $forum_select = mysql_query("SELECT * FROM forum") or die (mysql_error()); while ($forum_show = mysql_fetch_array($forum_select)) { $forum_name = mysql_query("SELECT * FROM forum_group"); while($forum_group_name = mysql_fetch_array($forum_name)) { if ($forum_group_name['hosting_forum_group_id'] == $forum_show['hosting_forum_group_id']) { echo "<table width='900' border='1' align='center' cellpadding='0' cellspacing='0'>"; echo "<tr><td colspan='6' style='padding: 7px;color:red'>".$forum_group_name['hosting_forum_group_name']."</td></tr>"; echo "<tr><td style='white-space: nowrap;' align='center' width='1%'>".$forum_show['hosting_forum_icon']."</td>"; echo "<td>"."<a href='/?section=forumview&id=".$forum_show['hosting_forum_id']." '>".$forum_show['hosting_forum_title']."</a><br />".$forum_show['hosting_forum_description']."</td>"; echo "<td width='230'> </td></tr>"; echo "</table><br>"; } } } I Hope to resolve the problem maybe somebody can explain where is my mistake to fix it and resolve it ! |
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