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PHP - How To Include Files From A Parent Directory
How do you include files from a higher up directory? I'm currently working on a file in public_html/Directory/otherdirectory and want to include a config file that's in public_html/Directory so how would I include public_html/Directory/config.php in the public_html_Directory/otherdirectory/index.php file? I've tried using ../ and ../Directory/ in the includes line but got errors both times Edited April 11, 2020 by Nematode128Similar Tutorialsi have made an delete files script which works for only one directory but not sub directory so i want to delete files of same extention from directory and subdirectory. My current code is Code: [Select] <? $dir = 'hmm/'; function scanr($dir){ $arr = glob($dir.'/*.jpg'); foreach($arr as $vv){ //check if $vv is a file if(is_file($vv)){ //if file, get the filename $vx=explode('/',$vv); $file=$vx[count($vx)-1]; // if no extension delete the file unlink($vv); // print the deletion message echo $vv." deleted!<br>";}else{ // if $vv is a dir then scan it again for files scanr($vv); }} } scanr($dir); ?> I am trying to use the following code to include a directories contents and was wondering what im doing wrong, because its not working? Code: [Select] if ($handle = opendir('../configs')) { while (false !== ($FILEZ = readdir($handle))) { include "$FILEZ"; } closedir($handle); } I am trying to include a inc.php file within a normal .php file from a single directory as I am using it in several directories but cannot get it to work and have tried many different techniques. Any ideas on the coding that would work would be appreciated? Alternatively, I would be happy to set the constants in a mysql db but am unsure how I would unpack/read it so as to get it in the php program as code i.e. mysql db field 1 field 2 name $name Program <?php $name; ?> is there a way to do an include or require to a whole directory? I have only done includes and requires with single files in the past, but have a situation where this could save me a lot of time. When I use require('../config.php'); It does not works on my machine but it works on shared hosting Can someone help me what must be the issue? Thanks in advance CSJakharia Hi, I have a file structure for all my files on my website that works fine for the most part. The basic layout would be something similar to the following. - root folder - sub_folder1 - folder1 - folder2 -sub_folder2 - folder1 - folder2 Each of the folders would contain relative .php files carrying out specific functions etc. The issue I am having is as follows. If I am on a webpage that is located in either of the folder1/folder2 folders, and I need to redirect somebody to a file located in the root folder, I am not able to do this without the page not being displayed correctly. On the file located in the folder1 folder, I have a <? php include("page1.php");?> which works fine. Trying to access a file to redirect to from this folder, I have tried, <? php include("../page1.php");?>, but this only bring me up one level, <? php include("../../page1.php");?> which I think is wrong anyway, and doesn't work, and also <? php include("./page1.php");?>, again does not work. Does anyone know if it is possible to get the file from the subfolders to access a file located directly in the root? Sorry if I'm not being clear explaining the layout etc. Thanks. Hello, I've got what should be a simple solution here. I've tried many things but it just won't work for me. This is what I have. I've got a wysiwyg editor setup to write to a file called news.php. The file that has the editor is edit/index.php. It's just a simple textarea. The textarea is supposed to include news.php. For some reason, it won't let me include that file that is in a lower directory. The file that writes to news.php is called edit/write.php. edit/write.php doesn't write to news.php instead it writes to edit/news.php. The only reason I want to include the lower directory is so that I can put a .htaccess password protection on the edit directory. I'll include some code below. Code: (edit/index.php) [Select] <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.1//EN" "http://www.w3.org/TR/xhtml11/DTD/xhtml11.dtd"> <html xmlns="http://www.w3.org/1999/xhtml" dir="ltr"> <head> <title>Editing News</title> <meta http-equiv="content-type" content="text/html; charset=utf-8"/> <script type="text/javascript" src="nicEdit.js"></script> <script type="text/javascript"> bkLib.onDomLoaded(function() { nicEditors.allTextAreas() }); </script> <form method="post" action="write.php"> <p> <textarea name="content" style="width:100%;height:1000px;"><?php include '../news.php';?></textarea> <input type="submit" value="Save" /> </p> </form> </body> </html> Code: (edit/write.php) [Select] <?php $File = "../news.php"; $Handle = fopen($File, 'w'); $Data = $_POST['content']; fwrite($Handle, $Data); fclose($Handle); header("Location: index.php"); ?> index.php is just a simple include news.php. Am I missing an easy step? Any help would be great!! Thanks! Hi guys, I am using a MySQL table to store if a user has access permisisons to a file. The files are stored outside the webdirectory in drive F:\ So I have this code I have been playing around with in order to display the file I access and check permissions for: <?php $file = $_GET['file']; $myfile = 'F:\files\/'.$file.'.jpg'; echo "<img src='$myfile' />"; ?> So a sample case would be say on index.php <img src="http://somedomain.com/files?file=abcdefghijkl" /> and only if the current user had permissions to view that file would they see it. Any help on how I can do this is greatly appreciated. I do understand all my above code is probably totally wrong. Cheers I almost have it but can't figure why the code below only shows one file per folder. I think it is showing the last file. I need $ul somehow looping to show all the files in $file. <?php $year = date('Y'); if ($handle = opendir("results/$year/Boys")) { while (false !== ($file = readdir($handle))) { if ($file != "." && $file != "..") { if ($handle2 = opendir("results/$year/Boys/$file")) { while (false !== ($file2 = readdir($handle2))) { if ($file2 != "." && $file2 != "..") { $ul = "<li><a href=\"#\">$file2</a></li>"; $li = "<li><a target=\"_blank\" href=\"javascript:void(0)\">$file</a> <ul> $ul </ul> </li>"; } // close if file2 }echo $li; } } } }closedir($handle);closedir($handle2); ?> Hey all, I viewed this tutorial and the guy used $_SERVER['DOCUMENT_ROOT'] to reference a view layout for his specific index file. So I tried to imitate and did exactly what he did, but for me I got the following error: Code: [Select] Warning: include(/Users/jmerlino/Sites/mark/public/diet/views/layouts/shop.php) [function.include]: failed to open stream: No such file or directory in /Users/jmerlino/Sites/diet/index.php on line 22 Warning: include() [function.include]: Failed opening '/Users/jmerlino/Sites/mark/public/diet/views/layouts/shop.php' for inclusion (include_path='.:/usr/lib/php') in /Users/jmerlino/Sites/diet/index.php on line 22 This is the php code that is causing this error: include($_SERVER['DOCUMENT_ROOT'].'/'.'diet/views/layouts/'.$controller.'.php'); I'm not sure why it's going to: /Users/jmerlino/Sites/mark/public/diet/views/layouts/shop.php instead of: /Users/jmerlino/Sites/diet/views/layouts/shop.php Thanks for any response Hey guys, I've written a script that is suppose to Display the files within a specified directory /documents/ with check boxes to the right of them.. This worked out great. Then I wanted to devise a method to remove or unlink the selected files (via check box) when the user pressed a Submit button.. File Permissions on the server are 777 and I have hardcoded an unlink to test to see if it works and it does.. So the bug lies within my code.. Before I post the code let me explain my method in doing this. I created a for loop to name each check box after the corresponding file to the left and created a variable with the file name (to use later) Then on button press I made a for loop (it goes up to 50 because I wasn't sure how to make it the exact number of files and I didn't want to overload the server with say 1000000 :?) that goes through and checks the check box's values and if they equal 1, unlink the file in /documents/ Here's the code: Code: [Select] <html> <body> <form action="" method="post" class="delete"> <?php $dir = "documents/"; if ($handle = opendir($dir)) { echo '<h2>List of Files in '.$dir.'</h2>'; echo '<table>'; for ($i = -2; false !== ($file = readdir($handle)); $i++) { if ($file != "." && $file != "..") { echo '<tr><td><a href="'.$dir.''.$file.'">'.$file.'</a></td><td><input type="checkbox" name="Del['.$i.']" /></td</tr>'; $DelFile[$i] = $file; } } echo '</table>'; closedir($handle); } ?> <br /> <input type="submit" name="submit" value="Delete Checked Files" /> </form> <?php for ($i = 0; $i <= 50; $i++) if ('Del['.$i.']' == 1) unlink("'.$dir.''.$DelFile[$i].'"); ?> </body> </html> Help is extremely appreciated, and Thanks in advance guys! I have a directory full of pictures that I would like to display in a gallery. Originally I was going to write HTML like this...
<li> <img src=/photos/img_001.jpg"> </li>
And then simply copy and paste that for the number of photos I have, and then tweak the code to: img_002.jpg, img_003.jpg, and so on. I guess that is silly considering that I have over 500 photos to display! How could I use PHP to go to my /photos/ directory, scan all of the files in that directory, and then grab the file name so I could automate this process? Sorry, but I haven't written PHP in several years so all of this escapes me! ☹️ I am trying to find a script that will show just the file names of files in a specific directory. I tried this code posted previously on this forum and it works but shows the directory name. Code: [Select] <?php filesInDir('attachments'); function filesInDir($tdir) { $dirs = scandir($tdir); foreach($dirs as $file) { if (($file == '.')||($file == '..')) { } elseif (is_dir($tdir.'/'.$file)) { filesInDir($tdir.'/'.$file); } else { echo $tdir.'/'.$file."<br>"; } } } ?> Result is: attachments/Sedona_and_Slot_Canyon_links.docx Thanks for the help. Dear sirs, I need to upload files to FTP and To waste less time, I zip the files, but the problem that since it is Automation it zips with directory - like this C:/DIR.zip in zip file I have a DIR/files.... How to unzip all files to main directory on the server. Right now if the main dir is /main/html/ it unzips to /main/html/dir/files But I need /main/html/files Thank you. Hello everyone. I am having trouble trying to write some script that will unlink all files in a directory. Can anyone help me? I am not sure of the best way to do this? Thankyou. I am using the following code to list mp3 files from a directory and find out if there is a file for it in mysql if so not to include it in the list. The code below is listing each file twice I only would like them to be listed once. if($_POST['submit'] != 'select') { echo'Choose an mp3 to add information.<p><select name="file">'; $dir = "/uploads/"; $result = mysql_query("SELECT * FROM dj_pool") or die(mysql_error()); $num_rows = mysql_num_rows($result); // Open a known directory, and proceed to read its contents if (is_dir($dir)) { if ($dh = opendir($dir)) { while (($file = readdir($dh)) !== false) { $info = explode(".", $file); if($info[1] == "mp3") { $i = 0; while ($i < $num_rows) { $file2 = mysql_result($result,$i,mp3); if ($file != $file2) { echo "<option value='$info[0]'>$info[0]</option>"; } $i++; } } }closedir($dh);} } print "</select><p><input type=submit value=select name=submit>"; I have a function that reads all the picture files in a folder and displays them as thumbnails. It also displays a large version of the first image in the folder and clicking the thumbnail changes the large picture. I'm trying to use pictures of arrows as a button to allow users to click and have the next picture displayed large rather than having to click each individual thumbnail but don't really know how to get started. Can anyone suggest a method or give me a hint? This is my code so far if that helps: function displayPhotos(){ global $columns,$dir,$album; generateThumbnails(); $act = 0; $keyid = 0; $thumb_selected = (isset($_GET['thumb']) ? $_GET['thumb']:''); if ($thumb_selected !==""){ $dirName = substr($thumb_selected,0,strpos($thumb_selected,basename($thumb_selected))); $thumbName = basename($thumb_selected); $thumbFile = $dirName.$thumbName; $large = str_replace('_th.jpg','.jpg',$thumbFile); } else{ $picture_array = glob("$dir*"); $large = $picture_array[$keyid]; } echo "<tr><td colspan='3' height='300' width='400' align='center'><img src='$large' alt=''></td></tr> <tr><td><img src='gallery/icons/arrow-blue-rounded-left.jpg' alt='Previous'></td><td></td><td align='right'><img src='gallery/icons/arrow-blue-rounded-right.jpg' alt='Next'> </td></tr><tr><td height='50'></td></tr>"; if ($handle = opendir("$dir")) { while (false !== ($file = readdir($handle))) { $file=$dir.$file; if (is_file($file)){ if (strpos($file,'_th.jpg')){ ++$act; if ($act > $columns) { echo "</tr><tr><td width='160' height='120' align='center'><a href='?album=$album&thumb=$file'><img src='".$file."' alt='".$file."'/></a></td>"; $act = 1; } else { echo "<td width='160' height='120' align='center'><a href='?album=$album&thumb=$file'><img src='".$file."' alt='".$file."'/></a></td>"; } } } } } } Thanks very much to anyone having a look at this! Can you include a file and append info to be grabbed by using GET? For example, can you have an include file like this... Code: [Select] include('file.php?id=1'); And then in the file "file.php" I would use $_GET['id'] to do other things in the code. I tried it and it's not working. I'm getting this error... "Warning: include(file.php?id=1) [function.include]: failed to open stream: No such file or directory in..." Just want to find out if this should be doable before I continue testing why it's not working for me. If it should work, should I be using something other than include maybe? (like require or require_once?) Hi! I would appreciate your help... How can i include files with get and post data in order to avoid sql injections? The including data has utf8 encoding... Any suggestions??? Thanks in advanced!!! Displaying the number of files in a directory and its subdirectorys <?php $path = './images_cache'; if ($handle = opendir($path)) { while (false !== ($file = readdir($handle))) { if ($file != "." && $file != "..") { $fName = $file; $file = $path.'/'.$file; if(is_file($file)) $numfile++; if(is_dir($file)) $numdir++; }; }; closedir($handle); }; echo $numfiles.' files in a directory'; echo $numdir.' subdirectory in directory'; ?> |
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