PHP - Path To Root Inside Iframes
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); ?> 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 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 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? 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? <td><label for='images'> <b>File to upload:</b> </label></td> <td><input type='file' name = 'drama_image' '<?php echo $row['drama_image']; ?>'/></ </tr> <?php $target_path = "images/"; $target_path = $target_path . basename( $_FILES['images']['name']); if(move_uploaded_file($_FILES['images']['tmp_name'], $target_path)) { echo "The file ". basename( $_FILES['images']['name']). " has been uploaded"; } else{ echo $row['drama_image']; } ?> ['drama_image'] is the name of the file I wanna echo it out in the box of file upload so when I save , the default picture will still be there instead of being overwritten as the box does not have any value in it. If I an deep inside of nested folders, how do I get back to the Web Root (as far as directory notation goes)?? Debbie Hi guys, I've got a simple script that I'm running from a folder above the root (as a cron job). I've tried the script in the httpdocs folder and it works fine. Code: [Select] mail('email@example.com', 'Before include', '', 'From:<email@example.com>'); // include the myriad class include($_SERVER['DOCUMENT_ROOT'].'/path/to/class'); mail('email@example.com', 'After include', '', 'From:<email@example.com>'); // initiate and instance of our class $obj = new class_name; $email = $obj->email; $file = $obj->file_name; mail('email@example.com', 'End of file', $email.' & '.$file, 'From:<email@example.com>'); If I run this file in httpdocs, it's fine and I get all emails. If I run this from 1 folder above the root (from a cron job), I get the first 2 emails but not the last one, suggesting that the include hasn't worked. Could this be a permissions problem? I have tried running the cron job as both root and apache and changed the owner/group of the script to match. Is there a reason why includes won't work above the root? Having problems pointing to where files are. Is this the root?: Code: [Select] include "../connectdb.php"; That file is locate din the root. If I am already in the root when I include that statment, the file isn't found. If I use the above code while I am in the subdirectory hello, (root/hello), it works. I use this in every page, so I think I need a way to point to the absolute root, not relative. Hi All, I googled this and there is endless results. I went through a lot of them but couldn't get this working properly. How do I link from within my web site root to files outside the root? It works for me using relative links i.e. ../../phpfiles/includes but that is going to get messy and I can't get a way of doing absolute links to work. If someone could lay that out so a newbie can get it clearly I would really appreciate it! Also - I understand why I should put all of my php files outside the web root but is this a guaranteed way to secure these files other than someone hacking my ftp access? I've looked at a few site hierarchy examples - Am I right that the only pages within the web site root should be template pages with calls to required files (outside the root), session checks, and content includes and all other includes that have php executable code should be outside the root? I really appreciate the advice and insight. Thank you! is there a way in php to link from the root dir ? like in html you just use the '/' at the start for the link " <a herf="/link.php" ></a> " but i noticed this does not work when using php like include or require. so is there anywey to tell a link to start from root dir? without using the ../../link.php I have been getting a lot more client requests to protect files. What is the easiest way to do this. So, basically I have tried doing it outside the public directory. There are too many things that cause issues with this. I haven't been able to get a successfull implementation of this since I started working with this. So I was thinking instead about password protecting a directory that is inside public view, but still get files via PHP. Is there a way to setup a password protected directory, then retreive stuff from that directory using PHP. Or, a good way to put them outside the public folder. Everything I have tried to do to get a file to save outside of public view, has not worked. It always says uploaded but the file is never there. Also, I have verified correct permission for this as well. The Script:
$desired_width = 110; if (isset($_POST['submit'])) { $j = 0; //Variable for indexing uploaded image for ($i = 0; $i < count($_FILES['file']['name']); $i++) {//loop to get individual element from the array $target_path = $_SERVER['DOCUMENT_ROOT'] . "/gallerysite/multiple_image_upload/uploads/"; //Declaring Path for uploaded images $validextensions = array("jpeg", "jpg", "png"); //Extensions which are allowed $ext = explode('.', basename($_FILES['file']['name'][$i]));//explode file name from dot(.) $file_extension = end($ext); //store extensions in the variable $new_image_name = md5(uniqid()) . "." . $ext[count($ext) - 1]; $target_path = $target_path . $new_image_name;//set the target path with a new name of image $j = $j + 1;//increment the number of uploaded images according to the files in array if (($_FILES["file"]["size"][$i] < 100000) //Approx. 100kb files can be uploaded. && in_array($file_extension, $validextensions)) { if (move_uploaded_file($_FILES['file']['tmp_name'][$i], $target_path)) {//if file moved to uploads folder echo $j. ').<span id="noerror">Image uploaded successfully!.</span><br/><br/>'; $tqs = "INSERT INTO images (`original_image_name`, `image_file`, `date_created`) VALUES ('" . $_FILES['file']['name'][$i] . "', '" . $new_image_name . "', now())"; $tqr = mysqli_query($dbc, $tqs); // Select the ID numbers of the last inserted images and store them inside an array. // Use the implode() function on the array to have a string of the ID numbers separated by commas. // Store the ID numbers in the "image_file_id" column of the "thread" table. $tqs = "SELECT `id` FROM `images` WHERE `image_file` IN ('$new_image_name')"; $tqr = mysqli_query($dbc, $tqs) or die(mysqli_error($dbc)); $fetch_array = array(); $row = mysqli_fetch_array($tqr); $fetch_array[] = $row['id']; /* * This prints e.g.: Array ( [0] => 542 ) Array ( [0] => 543 ) Array ( [0] => 544 ) */ print_r($fetch_array); // Goes over to create the thumbnail images. $src = $target_path; $dest = $_SERVER['DOCUMENT_ROOT'] . "/gallerysite/multiple_image_upload/thumbs/" . $new_image_name; make_thumb($src, $dest, $desired_width); } else {//if file was not moved. echo $j. ').<span id="error">please try again!.</span><br/><br/>'; } } else {//if file size and file type was incorrect. echo $j. ').<span id="error">***Invalid file Size or Type***</span><br/><br/>'; } } }Hey, sorry that I am posting this darn image upload script again, I have this almost finished and I am not looking to ask more questions when it comes to this script specifically. With the script above I have that part where the script should store the ID numbers (the auto_increment column of the table) of the image files inside of one array and then the "implode()" function would get used on the array and then the ID numbers would get inserted into the "image_file_id" column of the "thread" table. As you can see at the above part the script prints the following: Array ( [0] => 542 ) Array ( [0] => 543 ) Array ( [0] => 544 )And I am looking to insert into the column of the table the following: 542, 543, 544I thought of re-writing the whole image upload script since this happens inside the for loop, though I thought maybe I could be having this done with the script as it is right now. Any suggestions on how to do this? This topic has been moved to JavaScript Help. http://www.phpfreaks.com/forums/index.php?topic=316454.0 Hi Guys I am struggling with a shell_exec command in the long run, but I am trying to work down the root cause. The error I am getting with shell_exec is a file not found error, so I thought I would start with getting to the right folder. I have this Code: [Select] $sym_dir = "/root/"; $cwd = getcwd(); echo getcwd()."\n"; chdir ($sym_dir); echo getcwd()."\n"; Which in theory should 1st display '/var/www' as it does and then display the contents of root, at least that is my understanding! However my return is this /var/www /var/www Why cannot I move outside the web root? Cheers Dave Hi guys, I am trying to retrieve some information from sessions. I have this in my header.php <?php if(isset($_SESSION['username'])){ $username = $_SESSION['username']; echo $username; } ?> Hello! I was working on a simple url generator, but than i though to myself, why not put everything in the root. I've heard thats not a good idea, but i'm ot sure why. I can immagine it gets really messy when you have a lot of files, but its also easy, if you ask me. I was just wondering what your view on this is. What are the pros and what are the cons? Thanks Hey guys. Basically i have a file containing Database connection information stored at the root of my site www.worldwidelighthouses.com/Lighthouses-Database-Connection-Information.php I want to be able to connect to it using pages such as www.worldwidelighthouses.com/Lighthouses/English-Lighthouses/Beachy-Head-Lighthouse.php and www.worldwidelighthouses.com/Lightships/Lightship1.php Obviously 1 is 1 directory down, one is 2 directories down. My server doesnt allow for include 'www.worldwidelighthouses.com/Lighthouses-Database-Connection-Information.php' So whats the best way to always find this file, no matter what directory i am in? Finding out how far away i am from the server root and adding ../ to the include for each folder down was my first idea. Is this best? Thanks, Danny. Because I have include files in a subdirectory called "incl", I cannot rely on using relative paths because that include file is in a header on and the pages are in multiple subdirectories (or in the root). I need some way to get the location of the root (the one where my index file is located). I want to use something like this: $path= $_SERVER['DOCUMENT_ROOT']; include $path."/subdirectory/file.php"; However, when I use $_SERVER['DOCUMENT_ROOT'] I get : /home/content/j/p/f/name55/html/ which is not where the index file sits. Any ideas on how to get the directory where my index file sits? (i.e. the index that gets read when you got to www.domain.com) |