PHP - Finding Url Path
I am trying to find the URL to a directory TWO levels above the script level (if you know what I mean).
For example.. the script sits at Code: [Select] http://www.domain.com/TestArea2/members/index.php BUT I need to find the path to Code: [Select] http://www.domain.com/TestArea2/ the nearest I get is... $domain = $_SERVER['HTTP_HOST']; $domain .= $_SERVER['REQUEST_URI']; echo $domain; // outputs www.domain.com/TestArea2/members/index.php Is there a simple method to do this or have I got to split the string and work backwards? Many thanks Similar TutorialsGiven a directory - folder1/subfolder1, I would like to loop through all subdirectories under the given directory and return the file path relative to the given directory. Example: Code: [Select] Given Directory FolderA FolderB SubFolderA SubFolderB SubSubFolderA File1 FolderC the file path for File1 should be FolderB/SubFolderB/SubSubFolderA/ Any ideas? Hi: I'm a newby regarding uploading files to MySQL, turning report output into an HREF, and getting MySQL data via a hyperlink. I have successfully uploaded, files to MySQL, I have also been able to display filename information as a hyperlink in report output, but when I click on the hyperlink, I get the following message format on a 404 page: The requested URL /current_dir_of_requesting_page/filename.filetype was not found on this server. My reporting page has teh following line of code in a report table to create the hyperlink: <td style="width:225px"><? echo "<a href=".$row['name'].">".$row['name']."</a>";?></td> Can anyone assist me with this? 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? Hi I have a file, that is copied from other files. All other files work perfectly. But, for some reason this one is throwing back an error. I've been over it so many time, but can't see whats wrong. Error: Fatal error: Call to undefined method stdClass::save() in /home/p/o/powtest/web/public_html/admin/lib/ajax_php/add_interests.php on line 28 add_interests.php Code: [Select] <?PHP require_once("../../../includes/initialize.php"); $flag = 0; $ID = $_POST['ID']; $type = $_POST['type']; $category = $_POST['category']; $interest = $_POST['interest']; $expInt = explode("\n", $interest); $DMOD = date('Y-m-d'); $TMOD = date('H:i:s'); $check_entry = Admin_interest::if_exists(clean_input_value($category)); if($check_entry == 0){ $new_category = Admin_interest::make($ID, clean_input_value($category), clean_input_value($type), $DMOD, $TMOD); if($new_category && $new_category->save()){ $CID = $new_category->id; $flag = 1; } } foreach($expInt as $expInts){ $check_entry = Admin_interests_sub::if_exists($CID, clean_input_value($expInts)); if($check_entry == 0){ $new_interest = Admin_interests_sub::make($ID, $CID, clean_input_value($expInts), $DMOD, $TMOD); if($new_interest && $new_interest->save()){ $message = 'Thank You: Your list Has Been Saved'; }else{ $message = "Sorry, There was an error"; } } } echo $message; ?> Class: Code: [Select] <?PHP require_once(LIB_PATH.DS.'database.php'); class Admin_interests_sub { protected static $table_name="admin_interests_sub"; protected static $db_fields = array('id', 'category_id', 'interest_sub', 'dateMod', 'timeMod'); public $id; public $category_id; public $interest_sub; public $dateMod; public $timeMod; public static function make($ID, $category_id, $interest_sub, $DMOD, $TMOD){ if(!empty($interest_sub)){ $interest = new Admin_interests_sub(); $interest->id = (int)$ID; $interest->category_id = (int)$category_id; $interest->interest_sub = $interest_sub; $kw->dateMod = $DMOD; $kw->timeMod = $TMOD; return $kw; }else{ return false; } } protected function attributes(){ $attributes = array(); foreach(self::$db_fields as $field){ if(property_exists($this, $field)){ $attributes[$field] = $this->$field; } } return $attributes; } protected function sanitized_attributes(){ global $database; $clean_attributes = array(); foreach($this->attributes() as $key => $value){ $clean_attributes[$key] = $database->escape_value($value); } return $clean_attributes; } public function save(){ return !empty($this->id) ? $this->update() : $this->create(); } public function create(){ global $database; $attributes = $this->sanitized_attributes(); $sql = "INSERT INTO ".self::$table_name." ("; $sql .= join(", ", array_keys($attributes)); $sql .= ") VALUES ('"; $sql .= join("', '", array_values($attributes)); $sql .= "')"; if($database->query($sql)){ $this->id = $database->insert_id(); return true; }else{ return false; } } public function update(){ global $database; $attributes = $this->sanitized_attributes(); $attribute_pairs = array(); foreach($attributes as $key => $value){ $attribute_pairs[] = "{$key}='{$value}'"; } $sql = "UPDATE ".self::$table_name." SET "; $sql .= join(", ", $attribute_pairs); $sql .= " WHERE id=".$database->escape_value($this->id); $database->query($sql); return ($database->affected_rows() == 1) ? true : false; } public function delete(){ global $database; $sql = "DELETE FROM ".self::$table_name." "; $sql .= "WHERE id=".$database->escape_value($this->id); $sql .= " LIMIT 1"; $database->query($sql); return ($database->affected_rows() == 1) ? true : false; } } ?> Any help finding this bug will be a big help. Thanks Okay I'm done with searching for answers, working on this preg_match for about 3 hours now.
I'm looking for "($0.01/$0.02 USD)" in a string. The needle might be slighty different. Possible strings that I might look for is: (€xx.xx/€xx.xx EUR) where the EUR can be changed into USD, including its signs. 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, User submits 1 or more refrerences via a form. On the action page I want to check if any id's on the page match any values submitted in $_POST. Code: [Select] $Ref = $_POST['Ref']; $source = file_get_contents( 'E:/wamp/www/Project/File.php' ); $document = new DOMDocument; $document->validateOnParse = true; $document->LoadHTML($source); for($i=0; $i<count($Ref); $i++) { $ID = $document->getElementById('$Ref[$i]'); } I have a table on the source page with the same id as a reference contained in $Ref, but getElementById is not seeing/recognsing it. No idea why not. My query is not finding the last recieptnum entry, it is finding the number 9 everytime for some odd reason. Im trying to incrementally increase this each time a reciept is created. $getreceiptnum = mysql_query("SELECT receiptnum FROM accounting WHERE agency = '$agency' ORDER BY receiptnum DESC LIMIT 1") or die(mysql_error()); $recieptarray = mysql_fetch_array($getreceiptnum); $recieptnum = $recieptarray['receiptnum']; echo $recieptnum; //calculate age $birthdate = "1978-04-26"; //birth date... actually being obtained from a database $today = date("Y-m-d H:i:s"); // The exact date $age = date_diff($str_birthday, $today); echo $age; I'd like a simple code to echo the age of someone with the mysql database information that's in their record. This doesn't work. I have no idea why. Nothing seems to work that I've found on the net. Please help. Thanks. What I want to be able to do is this: If I have a list of values in my database: 1 13 26 15 12 4 I do a query of this database and sort it asc so it looks like this: 1 4 12 13 15 26 I want to know how I can use php and/or sql query to find these values based off of "position" in the list. If I wanted the 3rd value in the list, that value would be 12. If i wanted the 2nd value in the list it would be 4. How would I go about doing this? Thank you in advance. hi there all i need the code that adds the url to a page like in the new forums cheers.
you get a image of that website and the url of the page
can anyone help please
thank you
like if i type www,google.com <<< you get a image and link automaticly..
Edited by php-real-degree, 15 September 2014 - 05:07 PM. Howdy Colleagues,
I was wondering whether you know ways/websites/channels/magic owls to how to help find a job in USA as a programmer as in for relocating there while the company pays for the Visa and the travel expenses?
My land-lord's son found a job as a programmer in Texas and the company was generous and kind enough to pay his travelling expenses, as well for his wife and kid.
I have an account in Monster & Indeed and I send CVs very often, but no luck so far.
And I want to go there to work and live, not like some people to rely on welfare, I have worked my whole life.
Thank you!
BR,
Stefany
Hello. I think I have a little unique and difficult to solve problem that I really need to post here. I, as a counter-strike portal administrator, am running some CS servers. With automatical script, the server uploads demos (videos of gameplay) on FTP in this format: 101229154428.dem and so on. The number is a date in this format: year, month, day, hours, minutes, seconds => for the example it's 29th December 2010, 15h, 44m and 28s. Then I have a "report" page. The important part on it is that, there is a date in unix timestamp in each row like 1336837680 (29th december 2010, 15h, 47m, 00s). And now what I need: Find a demo that corresponds to the unix timestamp. If there is a timestamp 1336837680, it should find 101229154428.dem. You know what I mean, there is not one demofile, there are plenty. Just find the demo, that contains a record of the time used in timestamp. Pretty difficult, isn't it? I would love to see if anybody came up with anything. Best regards! Im trying to find the time between the last database entry and the current time and then echo it so it says something like '2 minutes ago'. I tried doing: $last_post_gap = strtotime('NOW') - strtotime($topic_info->topic_last_post_time); $last_post_gap = date('i', strtotime($last_post_gap)); $last_post = date('F j, Y', strtotime($topic_info->topic_last_post_time)); if ($last_post <= '- 1 DAY') { $last_post = 'Yesterday at '.date("g:i a", strtotime($topic_info->topic_last_post_time)); } if ($last_post <= '- 1 MINUTE') { $last_post = 'Less Than 1 Minute Ago'; } else { $last_post = $last_post_gap.' Minutes Ago'; } but that just displays 0 minutes ago regardless of the time. Is there anything im missing? Hi , I'm new on this forum so don't judge too hard.. But I have a few questions. I am recently new to PHP as this is my first computer language I've learnt besides HTML. My first question is, what's the difference between !== and !=, and is it more secure to use !== when comparing two fields such as passwords? I've different things and this has confused me.. And my second question is, I have this piece of code Code: [Select] /* option error checking */ $field = "option"; // Use field name for option /* Check if the option picked is from the list of options to choose from*/ if($suboption !== "option1" || $suboption !== "option2" || $suboption !== "option3" || $suboption !== "option4" || $suboption !== "option5" || $suboption !== "option6" || $suboption !== "option7"){ $form->setError($field, "* Unexpected error with option"); } I'm guessing in the long run, this will slow down my script.. is there an easier way of checking without using || $suboption !== and is it okay if I've used !==? Like I said I'm new to this so please don't flame.. I've done some research but I can't seem to find some accurate answer. Thank you, ZT my problem is when i click on hyperlink it start copying it self insertrow=yes in address bar how to tackle this Code: [Select] $httpquery = "{$_SERVER['PHP_SELF']}?{$_SERVER['QUERY_STRING']}&insertrow=yes" ; <a href='{$httpquery}' >New</a> <br> if (isset($_GET['insertrow'])) { echo "<br> {$_GET['insertrow']} "; } Hello, I would like to know how to include a root file in a 4 deep folder using dots. Is it good : include "../../../config.php" or i can simple include "./config.php" ? How does this works? Is it better to write all the path? Thank you Hi, How can i show image using absolute path instead of virtual path?? Help please It's the simple things that frustrate me sometimes. My project has a one root folder. Under that are three separate folders (admin, img, and inc). "admin" contains admin pages, "img" contains images, and "inc" contains utility files to be included and used by other scripts. I'll just take one specific problem as an example. In my inc folder is a file called "send_email.php". It needs to include another file called "class.phpmailer.php" in the same directory (inc). Now, since these files are never called directly but included by scripts in the rest of the project, I cannot simply do "require('class.phpmailer.php')". I have a file in the base project directory called validation.php. It needs to include inc/send_email.php to use its functions to, obviously, send email. Now, that part is no problem. I just require('inc/send_email.php') since that is a valid path from validation.php. My problem comes when send_email.php tries to include class.phpmailer.php. As I said before, I cannot just directly include it without any path since it's not being called directly. It "inherits" the path of the file that called it so just a direct require to it. My "solution" was to add a config value called "path" to my config file in the root directory. It would ideally hold the path from the server root to my project folder. In my case I set it to "/danny/smswebalerts/" which is indeed the correct path from my web root. I use $config['path'] to link to my images and CSS files and it works correctly. I thought it would solve my problem when I did the following in send_email.php: require_once($config['path'] . 'inc/class.phpmailer.php'); That indeed correctly builds the full path to that file: /danny/smswebalerts/inc/class.phpmailer.php. When I open up a page that includes send_email, however, I get the following error: Code: [Select] Warning: require_once(/danny/smswebalerts/inc/class.phpmailer.php) [function.require-once]: failed to open stream: No such file or directory in /home/danny/workspace/smswebalerts/inc/send_email.php on line 20 I would appreciate if somebody could give me some nice solution before I just say "screw it" to organization and plop everything in the same directory. It's really frustrating that my project is being held up by some technicality like this. Hey guys! I just joined this forum today, and I'm wondering about something... I see a lot of sites, and in the code of the site, style sheets and javascript includes are references in a manner similar to this... "href='http://www.somesite.com/css/css.php/someDirectory/style.css'" How is this done? The path following the "css.php" is what confuses me. Thanks for any information you have! |