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PHP - Php Variable Error, With This Cookie Code.
Hi a get a server error with this code, this worked for awhile, now it doesn't work. I use the first part to extract the user name from the url. then parse it to the IF cookie function as a variable for the cookie name. The cookie is placed on another html page and i placed it like this:
<script language="javascript"> document.cookie = "{PROFILE_USERNAME}= 1"; </script> // this code works perfectly i used php to parser the variable to the javascript function, when i check the browser for cookies, it is placed correctly. this code i use to check the cookie's name to see if it exists. the variable below is taken from the url, and matches the username above, but when i execute the code i get this error: Illegal variable _files or _env or _get or _post or _cookie or _server or _session or globals passed to script. can anyone see the problem? <?php $url = $_SERVER['SCRIPT_FILENAME']; $newurl = explode("/", $url); $profmanager=$newurl[6]; $cook= $profmanager; if(!isset($_COOKIE['$cook'])) { $url2 = $_SERVER['HTTP_HOST']; $myurls = 'http://'.$url2.'/'; echo "<META HTTP-EQUIV=Refresh CONTENT=\"0;URL=$myurls\">"; } else { my page code here }?> // to note when i echo the $newurl[6] it does equal the username. the first code in on a html page/the javascript above, // the second code piece is AT THE TOP of a php page. // if there is a better way, liike matching values instead of the cookie name, i would be keen to see it in action. Similar Tutorials<? $name = $_GET['name']; $game2 = "2": setcookie('game_".$name."',$game2, time()+3600); ?> would this work? if not how can it be done? cheers matt Hi girls and boys I am trying to set a variable if a session OR a cookie has been set, but am unsure on how to write the statement... if (isset($_SESSION['name'])||isset($_COOKIE['name'])) {$variable = $_SESSION['name']||$_COOKIE['name'];} Obviously not working there, but just need a pointer here. any help is appreciated... Trying to figure out where my syntax error is. Also to know I'm using code igniter but I don't think that changes anything. if (isset(cookie('xtrcook')) AND (!empty(cookie('xtrcook'))) || trim(cookie('xtrcook') !== '') || (cookie('xtrcook') !== NULL) || (!is_numeric(cookie('xtrcook')))) http://codeigniter.com/user_guide/helpers/cookie_helper.html Hello All I wrote a tiny piece of cookie code (to set and record cookies and hits) that works pretty well but seems to have a bug in it that I cannot figure out. First, here is the code: Code: [Select] <?php $my_cookie = (isset($_COOKIE['my_cookie'])) ? true : false; $countFile = "./data/count.txt"; $cookieFile = "./data/cookie.txt"; $dateTime = date('Y-m-d\,H:i:s\,U\,T\,O\G\M\T'); $remoteIPAddress .= $_SERVER['REMOTE_ADDR']; $remoteUserAgent .= $_SERVER['HTTP_USER_AGENT']; $cookieVal = $dateTime; $cookieVal .= ","; $cookieVal .= $remoteIPAddress; $cookieVal .= ","; $cookieVal .= $remoteUserAgent; $cookieVal .= "\r\n"; if(!$my_cookie){ // read the visit counter $countHandle = fopen($countFile, 'r'); $visitCount = intval(trim(@fread($countHandle, filesize($countFile)))); fclose($countHandle); // incremenet the visit counter $visitCount++; // write the new visit count and cookie value to their respective files $countHandle = fopen($countFile, 'w'); $cookieHandle = fopen($cookieFile, "a+b"); if(!preg_match('/bot/',$remoteUserAgent)){ fwrite($countHandle, $visitCount); fwrite($cookieHandle, $cookieVal); } fclose($countHandle); fclose($cookieHandle); setcookie("my_cookie", $cookieVal, time()+31556926); } else{ // read the visit counter $countHandle = fopen($countFile, 'r'); $visitCount = fread($countHandle, filesize($countFile)); fclose($countHandle); } ?> What is happening is that sometimes, not every time, but, sometimes my cookie.txt count gets a 0 (zero) written back to it and I can't figure out why. Can anybody see anything in there that would have this effect? Like I said, it only gets zeroed out from time to time. Thanks to all for taking the time to read. Ok I have the following as part of a login script which is supposed to set a cookie upon successful login [php] if (mysql_num_rows($LoginResult) > 0) { $UserID = mysql_result($LoginResult, 0, 'user_id'); $Id = uniqid(); $IdRes = mysql_query("UPDATE ".MEMBER_LOGIN_TABLE." SET `unique_id`={$Id} WHERE user_id='{$UserID}", $db); setcookie('RAYTH_MEMBER_ID', $Id, time()+2592000); Echo "Logged In. Click <a href='index.php?act=idx'>Here</a> to Continue.<br>Note: If you click continue and you are not logged in please ensure cookies are enabled!"; } [/cookie] However when I log in I get: Code: [Select] Warning: Cannot modify header information - headers already sent by (output started at /home/rayth/public_html/style.php:1) in /home/rayth/public_html/login.php on line 38 Logged In. Click Here to Continue. Note: If you click continue and you are not logged in please ensure cookies are enabled! Line 38 is the setcookie() line. Can someone shed light on where I am going wrong here? Hi, I am very new to PHP and I am trying to execute the below code but getting these errors : Notice: Undefined variable: mysql_query in C:\wamp\www\process.php on line 16 Fatal error: Function name must be a string in C:\wamp\www\process.php on line 16 Code: <html><body> <?php mysql_connect("localhost","root",""); mysql_select_db("encryption") or die(mysql_error()); $username = $_POST['username']; $password = $_POST['password']; $mysql_query("INSERT INTO login (username,password) VALUES ('$username','$password')") or die(mysql_error()); ?> </body></html> ` $_POST['username'] & $_POST['password'] come from a previous page. I have no problem with that. Please help.. Thanks in advance. Everything seems to be working like it should iam getting Undegined variable Error : return in /Applications/MAMP/htdocs/modernCMS/_class/cms_class.php on line 37. cms_class.php file Code: [Select] <?php class modernCMS { var $host; var $username; var $password; var $db; function connect() { $con = mysql_connect($this->host, $this->username, $this->password) or die(mysql_error()); mysql_select_db($this->db, $con) or die(mysql_error()); } function get_content($id = '') { if($id != ""): $id = mysql_real_escape_string($id); $sql = "SELECT * FROM cms_content WHERE id = '$id'"; $return = '<p><a href="index.php">Go Back To Content</a></p>'; else: $sql = "SELECT * FROM cms_content ORDER BY id DESC"; endif; $res = mysql_query($sql) or die(mysql_error()); if(mysql_num_rows($res) != 0): while($row = mysql_fetch_assoc($res)) { echo '<h1><a href="index.php?id=' . $row['id'] . '">' . $row['title'] . '</a></h1>'; echo '<p>' . $row['body'] . '</p>'; } else: echo '<p>Uh Oh!, this doesn\'t exist!</p>'; endif; echo $return; } } //Class ends here index.php ?> Code: [Select] <?php include '_class/cms_class.php'; $obj = new modernCMS(); $obj->host = 'localhost'; $obj->username = 'root'; $obj->password = 'root'; $obj->db = 'modernCMS'; $obj->connect(); ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=UTF-8" /> <title>Untitled Document</title> <link rel="stylesheet" href="style.css" type="text/css" media="screen" title="no title" charset="utf-8"> </head> <body> <div id="page-wrap"> <?php if(isset($_GET['id'])): $obj->get_content($_GET['id']); else: $obj->get_content(); endif; ?> </div> </body> </html> if anyone could help would be great. thanks. Hi Guys I have a $_GET[] variable that is not echoing out correctly. If I look at $GET in the uri then I have upload-swf.php?s=C2YuzOj+FCPieffLIEYdrtPAAQDVeg+yT+P2+N4Echw= But when I echo <?php echo $_GET['s']; //echo decrypt($_GET['s']); ?> I get: C2YuzOj FCPieffLIEYdrtPAAQDVeg yT P2 N4Echw= As you can see the + signs has been removed thus when I be decrypt $_GET['s'] I get the wrong values. Hey Everyone. I've been using a piece of code several times before and its worked fine but now (after updating my php version) I get an error. Heres the code: Code: [Select] if(isset($_GET['id']) == TRUE){ if(is_numeric($_GET['id'])==FALSE) { $error = 1; } if($error == 1){ header("Location: " . $config_basedir); } else{ $validentry = $_GET['id']; } } else{ $validentry = 1; } The line with "if($error == 1){" gets flagged with an error saying: Notice: Undefined variable: error in C:\xampp\htdocs\sites\smd\index.php on line 8 I've used the code before with no problems and uploaded to my web server I also have no errors. What did I do wrong. I hope it isn't a seriously stupid error. Thanks in advance. Hi... I'm a little bit new to php and it seems I am getting a small issue that I am not able to resolve... Basicly I just have a page in wich you see the account file of a customer and you can add notes into it. Here's the account file script Code: [Select] <?php mysql_connect("localhost", "user", "pass") or die(mysql_error()); mysql_select_db("db") or die(mysql_error()); if( ($_GET['cmd']=="view") && (is_numeric($_GET['id'])) ) { $id=$_GET['id']; $data = mysql_query("select * from client_table WHERE id_client='$id'") or die(mysql_error()); while($info = mysql_fetch_array( $data )) { echo " <table border=\"0\" celppadding=\"0\" cellspacing=\"0\" style=\"font-weight: bold; width: 820px;\"> <tr> <td colspan=\"2\" style=\"height: 50px;\"><a href=\"../dashboard.php\">Panneau de controle</a> > <a href=\"clients.php\">clients</a> > <a href=\"view_client.php?cmd=view&id=".$info['id_client']."\">Fiche client</a> > Notes > ".$info['entreprise']."</td> </tr> <tr> <td style=\"padding-left: 10px\"> NEQ : ".$info['neq']."<br /> <h2>Nom de l'entreprise : ".$info['entreprise']."<br /></h2> <h2>Telephone : ".$info['client_phone']."<br /></h2> </td> </tr> </table> "; } } ?> <form enctype="multipart/form-data" action="process/insert_note.php" method="POST" target="main"> <input type="hidden" name="id_client" value="<?php echo $id; ?>"> <input type="hidden" name="note_date" value="<?php echo date("Y-m-d"); ?>"> <input type="hidden" name="note_time" value="<?php echo date("G:i:s"); ?>"> <table border="0" celppadding="0" cellspacing="0" style="font-weight: bold; width: 780px;"> <tr> <td colspan="0" style="padding-left: 10px">Ajouter une note : <br /><textarea name="note" style="height: 80px;" cols="90"></textarea><br /> <input type="submit" value="Ajouter une note"> </td> </tr> </table> </form> So when I add the note....the form data is passed to a php script to insert the note into the MySQL DB and I just want to be redirected back to the acount file page with the note on it.... Here's the insert script : Code: [Select] <?php header('location:../view_notes.php?cmd=view&id=$id_client'); require_once('../../auth.php'); $id_client = $_POST['id_client']; $con = mysql_connect("localhost","user","pass"); $note_date = $_POST['note_date']; $note_time = mysql_real_escape_string($_POST['note_time']); $note = mysql_real_escape_string($_POST['note']); if (!$con) { die('Could not connect: ' . mysql_error()); }mysql_select_db("db", $con); $sql="INSERT INTO notes_table ( id_client, note_date, note_time, note) VALUES ('$id_client','$note_date','$note_time','$note')";if (!mysql_query($sql,$con)) { die('Error: ' . mysql_error()); } ?> The problem I have is that the HEADER part is working but the value : header('location:../view_notes.php?cmd=view&id=$id_client'); is not passed so I am redirected to a blank page..... I tried to find the info on google but can't seems to find what wrong... Any help will be appreciated! Thanks having a problem using get variables to include certain pages. I have a management script for a news feed that uses get variables to display certain parts of the script. here is the template/ main part of the script <?php session_start(); require_once("vars.php"); $username = $_SESSION['username']; $is_admin = $_SESSION['is_admin']; $referer = $_SERVER['HTTP_REFERER']; //check to see if user is logged in and is an administrator if((isset($_SESSION['username'])) && ($_SESSION['is_admin'] == 1)){ //if all session variables are set, display admin panel ?> <html> <head> <title>Administration</title> <link rel="stylesheet" type="text/css" href="/cameo/css/admin.css" media="screen" /> </head> <body> <table cellspacing="0" cellpadding="3" class="layout"> <tr> <td colspan="2"><h3>Administration</h3></td> </tr> <tr> <td valign="top"> <a href="index.php?pg=news">Manage News Feed</a> </td> <td> <?php $id = $_GET['pg']; switch($id){ case news: include("manage_news.php"); case editpost: include('edit_post.php'); break; } ?> </td> </tr> </table> </body> </html> <?php } else { echo("You do not have permission to view this page. <br />"); echo("<a href=\"".$referer."\">Go Back</a>"); } ?> and heres the edit post page that the script refers to <?php require_once("vars.php"); $id = $_GET['id']; $query = "SELECT * FROM news_feed WHERE id=".$id; $result = mysqli_query($dbc,$query); $row = mysqli_fetch_array($result); ?> <form method="post" action="update.php"> <textarea><?php echo $row['post_body']; ?></textarea> </form> now, for some reason, the edit post form is showing up under the main news feed listing, even though the get id does not match the one in the switch statement. thanks in advance for any help Still new to PHP, trying to get around an error! Scenario is pretty simple, i'm trying to do a sample form validation via php, this is the code, Code: [Select] <?php if($_POST){ $t_name=trim($_POST["name"]); $t_email=trim($_POST["email"]); if($t_name==""){ $n_err=1; } elseif($t_email==""){ $e_err=1; } } ?> then towards the form fields i do this, Code: [Select] <?php if($n_err)echo "Please enter correct name"."</br"; ?> <?php if($e_err)echo "Please enter correct email"."</br>"; ?> But this gives an error Quote Notice: Undefined variable: n_err in C:\wamp\www\php\form\index.php on line 30 As far i see you can use php vars defined in section on others, only var inside function are local in nature if they are not using "global" prefix. Please help me figure out why im getting this error, i know it might be something simple, so kindly help, thanks! Hi, I am testing out some simple code while trying out PHP OOP to connect to the MySQL DB. I am getting an undefined variable error, but not sure why. the test code is: <?PHP error_reporting(E_ALL); include('includes/db.php'); $db = new db(); $db->query("SELECT title FROM blog_posts WHERE id = '7'"); if($sql) { while($r = mysql_fetch_array($sql)) { echo $r['title']; } } ?> and the code to connect to the DB is: <?PHP class db { private $hostname; private $username; private $password; private $database; private $connect; private $select_db; public function db() { $this->hostname = ""; $this->username = ""; $this->password = ""; $this->database = ""; } public function open_connection() { try { $this->connect = mysql_connect($this->hostname,$this->username,$this->password); $this->select_db = mysql_select_db($this->database); } catch(exception $e) { return $e; } } public function close_connection() { try { mysql_close($this->connect); } catch(exception $e) { return $e; } } public function query($sql) { try { $this->open_connection(); $sql = mysql_query($sql); } catch(exception $e) { return $e; } $this->close_connection(); return $sql; } } ?> the specific error is... Code: [Select] Undefined variable: sql in E:\DBTest.php on line 9 After recent updates to a wordpress site (updated non-related plugins, changed theme), the php code snippets are shooting back an error at the end of the feeds -- which were previously working fine for years. no other changes were made.
Notice: Undefined variable: response in /nas/content/live/usafact/wp-content/plugins/insert-php-code-snippet/shortcode-handler.php(26) : eval()'d code on line 8
URL: PHP code:
function do_post($url, $params)
{
$ch = curl_init();
curl_setopt($ch, CURLOPT_URL, $url);
curl_setopt($ch, CURLOPT_POSTFIELDS, $params);
$result = curl_exec($ch);
curl_close($ch);
return $response;
}
echo do_post("https://content.newbenefits.com/Feednocss.aspx", "hash=hDSJYjIb56KfGtxWE0I3OQ&Section=short_b2c");
The plugin developer said to replace $response with $result, but that did not work. Same error shows. Can someone please assist? I am trying to display some text "THIS IS A TEST" on the html page. I am getting this error: Notice: Undefined variable: text1 in /home/public_html/themes/videos/content.html on line 23
line 23 is this: <div class="test"><font color="#696969" font size="3" face="Arial">HELLO<?php echo $text1;?></font> The php file, related to the html page shows this (partially):
$text1 = "THIS IS A TEST";
$cateogry_id = '';
$videos = array();
if ($page == 'trending') {
$title = $lang->trending;
$db->where('privacy', 0);
$videos = $db->where('time', time() - 172800, '>')->orderBy('views', 'DESC')->get(T_VIDEOS, $limit);
}
else if ($page == 'latest') {
$title = $lang->latest_videos;
echo "$text1";
$db->where('privacy', 0);
$videos = $db->orderBy('id', 'DESC')->get(T_VIDEOS, $limit);
}
else if ($page == 'top') {
$title = $lang->top_videos;
$db->where('privacy', 0);
$videos = $db->orderBy('views', 'DESC')->get(T_VIDEOS, $limit);
}
else if ($page == 'category') {
if (!empty($_GET['id'])) {
if (in_array($_GET['id'], array_keys($categories))) {
$cateogry = PT_Secure($_GET['id']);
$title = $categories[$cateogry];
$cateogry_id = "data-category='$cateogry'";
$db->where('privacy', 0);
$videos = $db->where('category_id', $cateogry)->orderBy('id', 'DESC')->get(T_VIDEOS, $limit);
} else {
header("Location: " . PT_Link('404'));
exit();
}
}
}
what do I need to correct to remedy the error? Any help is appreciated. Hello i am new to this forum and coding. I am trying to fix a auto torrent downloading script which is php and curl. I have been stucked with Undefined offset error. Your help will be very appretiated, thanks. Starting download: Mp3-data1 Notice: Undefined variable: temp in /var/www/html/torrents/bgrahul2.php on line 338 Array ( => HTTP/1.1 200 OK Server: Transmission Content-Type: application/json; charset=UTF-8 Date: Wed, 21 Sep 2011 19:58:44 GMT Content-Length: 49 {"arguments":{"torrents":[]},"result":"success"} [1] => ) Marked those lines to bold Quote $result = callTransmission(array("filename" => "/var/www/html/donofwarez/xxx/torrents/$torrentfile", "download-dir" => "/home/Downloads/$title/"), "torrent-add", $arr[1]); callTransmission(array("seedRatioLimit" => 50, "seedRatioMode" => 1), "torrent-set", $arr[1]); /* echo "<pre>"; print_r($result); echo "</pre>"; */ if (preg_match('/"percentDone":[0-9.]+/i', $result[0], $matches)) $temp = str_replace('"id":', "", $matches[0]); // $temp="hlDr1wqu2ONtM3McVGHoVzfrkp2UgUcGwDt66IuxAe0LBkru"; $id[0] = $temp; //***OMG! We need the ids as integers! *** foreach ($id as $key => $value) { $id[$key] = (int) $value; } This is code from a voting script which does work, yet I am getting a notice that the variables inside the array are undefined, here's a showcase: Code: [Select] // POST BUTTONS inside the table if (isset($_POST['likes'])) $likes = $_POST['likes']; if (isset($_POST['dislikes'])) $dislikes = $_POST['dislikes']; if (isset($_POST['hidden_con_id'])) { $con_id = $_POST['hidden_con_id']; //$favorite = $_POST['favorite']; } $array = array ($likes, $dislikes, $con_id, $user_id); The error message: Code: [Select] Notice: Undefined variable: likes Notice: Undefined variable: dislikes How can I solve this one? Hi, this works: if(file_exists("myFile.txt")){etc.} this doesn't: $myFile = "myFile.txt"; if(file_exists($myFile)){etc.} Can anybody explain this to me? I'd really like to pass the filename as a variable... Kind regards, R. Hi i have created a query where i want some columns to be retrieved. i get these 2 errors and i dont understand why.
Notice: Undefined variable: db
Fatal error: Call to a member function query() on a non-object in
this is my query
function get_products_all() {
$connect = get_connected_db();
try {
$results = $db->query(" ***** error on this line****
SELECT id, product, description, price, picc
FROM products ORDER BY id DESC
");
} catch (Exception $e) {
echo "Data could not be retrieved from the database.";
exit;
}
$result = $results->fetchALL(PDO::FETCH_ASSOC);
return $result;
}
what am i doing wrong?
Have a ton of pop-up javascript links in my page. To tidy up the code, I have decided to declare all links at top of the page to keep them together and simply use the variable within the "a" brackets. Let me show you: Code: [Select] <? $brand_edit = 'javascript:void(0)"onclick="window.open('/admin/form.php?db=outlet&tbl=brands&auto_increment=<? echo $auto_increment; ?>&action=edtamp;step=1','none','width=750,height=250,menubar=no,status=no,resizable=no,location=no,toolbar=no,scrollbars=yes,left=50,top=50,titlebar=no')'; ?> <span class="float" style="width: 100px;"><a href="<? echo $brand_edit; ?>"><? echo $auto_increment; ?></a></span> However, something things to go wrong as I'm getting an "Parse error: syntax error, unexpected '=' in ... on line 12" Where am I going wrong with my syntax? |
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