PHP - Need Help With Session In Script
Hallo my problems is something with the session in my script as it redirects even when i logged in can someone help out please..
the code <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html> <head> <title> </title> </head> <body> <form action="<?php echo $_SERVER['PHP_SELF']; ?>" method ="post"> Username:<input type="text" name="username" /> Password:<input type="password" name="password" /> <input type="submit" name="submit" value="submit" /> </form> <?php $host = '---'; $user = '---'; $pass = '---'; $db = '---'; $connect = mysql_connect($host, $user, $pass) or die ('Could not connect'); $database = mysql_select_db($db) or die ('Could not connect'); $username = $_POST['username']; $password = $_POST['password']; $username = stripslashes($username); $password = stripslashes($password); $username = mysql_real_escape_string($username); $password = mysql_real_escape_string($password); $sql= " SELECT * FROM Users WHERE username= '".$username."' AND password = '".$password."' "; $result = mysql_query($sql); if(mysql_num_rows($result) == 1){ $_SESSION['username'] = $username; $_SESSION['password'] = $password; header("location: login_succes.php"); } ?> </body> </html> code login_succes.php <?php session_start(); if(!isset($_SESSION['username'])){ header('location: login.php'); } ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html> <head> <title> </title> <link rel="stylesheet" href="stylesheet.css" type="text/css" /> </head> <body> login succesfull </body> </html> Similar TutorialsSET UP: Windows vista # XAMPP 1.7.3, # Apache 2.2.14 (IPv6 enabled) + OpenSSL 0.9.8l # MySQL 5.1.41 + PBXT engine # PHP 5.3.1 # phpMyAdmin So I'm getting this error message: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '", )' at line 8. I can't see where the '", )' code is in that script. I've looked and contrl + find it but no joy. I've looked through the previous pages and mysql database to ensure the variables and the tables match up. Any help would be appreciated. Maybe some good mysql debug advice. Is there a good debug open source program out there? Any help would be appreciated. <?php //let's start our session, so we have access to stored data session_start(); include 'db.inc.php'; $db = mysql_connect('localhost', 'root', '') or die ('Unable to connect. Check your connection parameters.'); mysql_select_db('ourgallery', $db) or die(mysql_error($db)); //let's create the query $query = sprintf("INSERT INTO subscriptions ( name, email_address, membership_type, terms_and_conditions, name_on_card, credit_card_number, credit_card_expiration_data) VALUES ('%s','%s','%s','%s','%s','%s','%s')", $_SESSION['email_address'], $_SESSION['membership_type'], $_SESSION['terms_and_conditions'], $_POST['name_on_card'], $_POST['credit_card_number'], $_POST['credit_card_expiration_data']); //let's run the query mysql_query($insert_query); ?> Hi, I have an old code from 2004 and I would like to update it to use new Session object. That means instead of session_register using the $_SESSION super global variable. The main reason for this change is that wheneve I logged out from the software I get: Warning: Unknown: Your script possibly relies on a session side-effect which existed until PHP 4.2.3. Please be advised that the session extension does not consider global variables as a source of data, unless register_globals is enabled. You can disable this functionality and this warning by setting session.bug_compat_42 or session.bug_compat_warn to off, respectively in Unknown on line 0 Spo I replace my old code: <?php session_name("MySite"); session_start(); reset ($_GET); session_register("ADMIN"); session_register("ADMINNAME"); session_register("MAIL") ; $USERCOOKIE_FOR_TRACKING = array(); //to get all session variables foreach ($_SESSION as $key => $value) { $value=stripslashes(trim($value)); $$key=$value; } ?> with this new code: <?php session_name("MySite"); session_start(); reset ($_GET); $_SESSION['ADMIN']=""; $_SESSION['ADMINNAME']=""; $_SESSION['MAIL']=""; $USERCOOKIE_FOR_TRACKING = array(); //to get all session variables foreach ($_SESSION as $key => $value) { $value=stripslashes(trim($value)); $$key=$value; } ?> BUT now I cannot login to the software any more. looks like I am doing something wrong here. please tell me how do I upgrade my code. Thank you. I am trying to create an index page which contains registration and login field the problem that i get is on successful login a warning is displayed session_start() [function.session-start]: Cannot send session cookie - headers already sent by (output started at C:\xampp\htdocs\Eventz.com\index.php:116) in C:\xampp\htdocs\Eventz.com\index.php on line 235 This is the login part of my index.php this tag is inside an html table below the login form I also have a registration form and its php code above the login form Code: [Select] <?php if (isset($_REQUEST['pass'])) { $id=$_POST['id']; $pass=$_POST['pass']; $conn =mysql_connect("localhost","root",""); if (!$conn) { die('Could not connect: ' . mysql_error()); } /* checking connection....success! */ $e=mysql_select_db('test', $conn); if(!$e) { die(''.mysql_error()); } else { echo 'database selected successfully'; } if (isset($_REQUEST['id']) || (isset($_REQUEST['pass']))) { if($_REQUEST['id'] == "" || $_REQUEST['pass']=="") { echo "login fields cannot be empty"; } else { $sql=mysql_query("Select email,password from login where email='$id' AND password='$pass'"); $count=mysql_num_rows($sql); if($count==1) /* $count checks if username and password are in same row */ { session_start(); $_SESSION['id']=$id; echo "</br>Login Successful</br>"; } else { echo "</br>invalid</br>"; echo "please try to login again</br>"; } } } } ?> Any help or suggestion would be appreciated I am having trouble resolving an error. Warning: session_start() [function.session-start]: Cannot send session cache limiter - headers already sent (output started at /home/s519970/public_html/header.php:27) in /home/s519970/public_html/admin/login.php on line 2 What I can gather is I can't use "header (Location: 'admin.php')" after i've used session_start(). I have tried to replace the header (Location: 'admin.php') with this: echo "<script>document.location.href='admin.php'</script>"; echo "<script>'Content-type: application/octet-stream'</script>"; I've been trying to read up on solutions but haven't been able to get it sorted. If anyone can offer some advice that would be greatly appreciated as im new to php. Code: [Select] <?php session_start(); if(isset($_SESSION['user'])) echo "<script>document.location.href='admin.php'</script>"; echo "<script>'Content-type: application/octet-stream'</script>"; ?> <div id="loginform"> <form action="dologin.php" method="post"> <table> <tr> <td><span>Username:</span></td> <td><input type="text" name="username" /></td> </tr> <tr> <td><span>Password:</span></td> <td><input type="password" name="password" /></td> </tr> <tr> <td colspan="2" align="right"><input type="submit" name="login" value="Login" /></td> </tr> </table> </form> </div> I have tried using require_once('yourpage.php'); before my <head></head> tags in the header document where I've specified the html information but this doesn't seem to work. I've been advised to use ob_start("ob_gzhandler"); but I am not sure how to implement this. Any advice is greatly appreciated! in this page http://maximaart.com/newscp/ i have this problem Code: [Select] Warning: session_start() [function.session-start]: Cannot send session cache limiter - headers already sent (output started at /home/maximasy/public_html/newscp/index.php:1) in /home/maximasy/public_html/newscp/index.php on line 2 my source code is <?php session_start(); include_once("config.php"); include_once("functions.php"); $errorMessage = ''; if (isset($_POST['txtUserId']) && isset($_POST['txtPassword'])) { if ($_POST['txtUserId'] === "$user" && $_POST['txtPassword'] === "$pass") { // the user id and password match, $_SESSION['basic_is_logged_in'] = true; require("main.php"); exit;?> I'm making a simple login system with MySQL and PHP (very simple, I'm just starting with PHP). The MySQL portion is done, but I need to ensure only people who are logged in can see certain content. To check if people are logged in, my website checks that they have the $_SESSION['user'] variable set. If it is set, then it lets them continue through the website, if not, it tells them to login. Is that enough security, or can people simply inject a session cookie into their browser to spoof that they are logged in? My idea was to generate a session key cookie when they login (just a random string of letters and numbers) and store that in the database, then on every page, check to make sure their session key is the same thing that's in the database. Is this necessary? It seems expensive. hi everyone. i'm wondering what the best way is to create a session variable and pass it to an iframe. i need to do something along these lines, but it doesn't seem to pass the ID. Any hints on how i should accomplish this? Code: [Select] session_start(); $_SESSION['ID']=$_GET['ID']; // id from previous page $ID=session_id(); <iframe src="iframepage.php?ID=<?php echo $ID; ?>" style="width:680px; height:200px;" noresize="noresize" frameborder="0" border="0" scrolling="Yes" allowtransparency="true" /> </iframe> Evening! I've been iffing and ahhing over this and well im not too sure, hence the post. Code: [Select] // Redirects if there is no session id selected and echos the error on the previous page if(!isset($_GET['get']) || ($_GET['getget'])){ header("Location: #.php?error"); } So it should simply check if get is set if it isnt then see if getget is set? If not redirect and show the error. Now ive tried it and even when get/getget is set it still redirects, probably something silly. Care to share anyone? Harry. Just curious how other people feel about this. I am working on an application where a lot of info is pulled from MySQL and needed on multiple pages.
Would it make more sense to...
1. Pull all data ONCE and store it in SESSION variables to use on other pages
2. Pull the data from the database on each new page that needs it
I assume the preferred method is #1, but maybe there is some downside to using SESSION variables "too much"?
Side question that's kind of related: As far as URLs, is it preferable to have data stored in them (i.e. domain.com/somepage.php?somedata=something&otherdata=thisdata) or use SESSION variables to store that data so the URLs can stay general/clean (i.e. domain.com/somepage.php)?
Both are probably loaded questions but any possible insight would be appreciated.
Thanks!
Greg
Edited by galvin, 04 November 2014 - 10:30 AM. Hi everyone! I've been working on a php script to replace links that contain a query with direct links to the files they would redirect to. Hi i have this upload script which works fine it uploads image to a specified folder and sends the the details to the database. but now i am trying to instead make a modify script which is Update set so i tried to change insert to update but didnt work can someone help me out please this my insert image script which works fine but want to change to modify instead Code: [Select] <?php mysql_connect("localhost", "root", "") or die(mysql_error()) ; mysql_select_db("upload") or die(mysql_error()) ; // my file the name of the input area on the form type is the extension of the file //echo $_FILES["myfile"]["type"]; //myfile is the name of the input area on the form $name = $_FILES["image"] ["name"]; // name of the file $type = $_FILES["image"]["type"]; //type of the file $size = $_FILES["image"]["size"]; //the size of the file $temp = $_FILES["image"]["tmp_name"];//temporary file location when click upload it temporary stores on the computer and gives it a temporary name $error =array(); // this an empty array where you can then call on all of the error messages $allowed_exts = array('jpg', 'jpeg', 'png', 'gif'); // array with the following extension name values $image_type = array('image/jpg', 'image/jpeg', 'image/png', 'image/gif'); // array with the following image type values $location = 'images/'; //location of the file or directory where the file will be stored $appendic_name = "news".$name;//this append the word [news] before the name so the image would be news[nameofimage].gif // substr counts the number of carachters and then you the specify how how many you letters you want to cut off from the beginning of the word example drivers.jpg it would cut off dri, and would display vers.jpg //echo $extension = substr($name, 3); //using both substr and strpos, strpos it will delete anything before the dot in this case it finds the dot on the $name file deletes and + 1 says read after the last letter you delete because you want to display the letters after the dot. if remove the +1 it will display .gif which what we want is just gif $extension = strtolower(substr($name, strpos ($name, '.') +1));//strlower turn the extension non capital in case extension is capital example JPG will strtolower will make jpg // another way of doing is with explode // $image_ext strtolower(end(explode('.',$name))); will explode from where you want in this case from the dot adn end will display from the end after the explode $myfile = $_POST["myfile"]; if (isset($image)) // if you choose a file name do the if bellow { // if extension is not equal to any of the variables in the array $allowed_exts error appears if(in_array($extension, $allowed_exts) === false ) { $error[] = 'Extension not allowed! gif, jpg, jpeg, png only<br />'; // if no errror read next if line } // if file type is not equal to any of the variables in array $image_type error appears if(in_array($type, $image_type) === false) { $error[] = 'Type of file not allowed! only images allowed<br />'; } // if file bigger than the number bellow error message if($size > 2097152) { $error[] = 'File size must be under 2MB!'; } // check if folder exist in the server if(!file_exists ($location)) { $error[] = 'No directory ' . $location. ' on the server Please create a folder ' .$location; } } // if no error found do the move upload function if (empty($error)){ if (move_uploaded_file($temp, $location .$appendic_name)) { // insert data into database first are the field name teh values are the variables you want to insert into those fields appendic is the new name of the image mysql_query("INSERT INTO image (myfile ,image) VALUES ('$myfile', '$appendic_name')") ; exit(); } } else { foreach ($error as $error) { echo $error; } } //echo $type; ?> I'm having trouble echoing $year in my script. Listed below is the script, just below ,$result = mysql_query("SELECT * FROM $dbname WHERE class LIKE '%$search%'") or die(mysql_error());, in the script I try to echo $year. It doesn't show up in the table on the webpage. Everything else works fine. Any help wold be appreciated greatly. Thanks in advance. <?php include 'config2.php'; $search=$_GET["search"]; // Connect to server and select database. mysql_connect($dbhost, $dbuser, $dbpass)or die("cannot connect"); mysql_select_db("vetman")or die("cannot select DB"); $result = mysql_query("SELECT * FROM $dbname WHERE class LIKE '%$search%'") or die(mysql_error()); // store the record of the "" table into $row //$current = ''; echo "<table align=center border=1>"; echo "<br>"; echo "<tr>"; echo "<td align=center>"; ?> <div style="float: center;"><a><h1><?php echo $year; ?></h1></a></div> <?php echo "</td>"; echo "</tr>"; echo "</table>"; // keeps getting the next row until there are no more to get if($result && mysql_num_rows($result) > 0) { $i = 0; $max_columns = 2; echo "<table align=center>"; echo "<br>"; while($row = mysql_fetch_array($result)) { // make the variables easy to deal with extract($row); // open row if counter is zero if($i == 0) echo "<tr>"; echo "<td align=center>"; ?> <div style="float: left;"> <div><img src="<?php echo $image1; ?>"></div> </div> <?php echo "</td>"; // increment counter - if counter = max columns, reset counter and close row if(++$i == $max_columns) { echo "</tr>"; $i=0; } // end if } // end while } // end if results // clean up table - makes your code valid! if($i > 0) { for($j=$i; $j<$max_columns;$j++) echo "<td> </td>"; echo '</tr>'; } mysql_close(); ?> </table> I'm trying to use this script known as SimpleImage.php that can be found here <a href="http://www.white-hat-web-design.co.uk/articles/php-image-resizing.php">link</a> I'm trying to include what is on the bottom of the page to my existing script can anyone help me I've tried several ways but its not working. Code: [Select] <?php session_start(); error_reporting(E_ALL); ini_set('display_errors','On'); //error_reporting(E_ALL); // image upload folder $image_folder = 'images/classified/'; // fieldnames in form $all_file_fields = array('image1', 'image2' ,'image3', 'image4'); // allowed filetypes $file_types = array('jpg','gif','png'); // max filesize 5mb $max_size = 5000000; //echo'<pre>';print_r($_FILES);exit; $time = time(); $count = 1; foreach($all_file_fields as $fieldname){ if($_FILES[$fieldname]['name'] != ''){ $type = substr($_FILES[$fieldname]['name'], -3, 3); // check filetype if(in_array(strtolower($type), $file_types)){ //check filesize if($_FILES[$fieldname]['size']>$max_size){ $error = "File too big. Max filesize is ".$max_size." MB"; }else{ // new filename $filename = str_replace(' ','',$myusername).'_'.$time.'_'.$count.'.'.$type; // move/upload file $target_path = $image_folder.basename($filename); move_uploaded_file($_FILES[$fieldname]['tmp_name'], $target_path); //save array with filenames $images[$count] = $image_folder.$filename; $count = $count+1; }//end if }else{ $error = "Please use jpg, gif, png files"; }//end if }//end if }//end foreach if($error != ''){ echo $error; }else{ /* -------------------------------------------------------------------------------------------------- SAVE TO DATABASE ------------------------------------------------------------------------------------ -------------------------------------------------------------------------------------------------- */ ?> Hello, I stored a fsockopen function in a separate "called.php" file, in order to run it as another thread when it needs. The called script should return results to the "master.php" script. I'm able to run the script to get the socket working, and I'm able to get results from the called script. I tried for hours but I can't do the twice both My master.php script (with socket working): Code: [Select] <?php $command = "(/mnt/opt/www/called.php $_SERVER[REMOTE_ADDR] &) > /dev/null"; $result = exec($command); echo ("result = $result\r\n"); ?> and my called.php script Code: [Select] #!/mnt/opt/usr/bin/php-cli -q <?php $device = $_SERVER['argv'][1]; $port = "8080"; $fp = fsockopen($device, $port, $errno, $errstr, 5); fwrite($fp, "test"); fclose($fp); echo ("normal end of the called.php script"); ?> In the master script, if I use Code: [Select] $command = "(/mnt/opt/www/called.php $_SERVER[REMOTE_ADDR] &) > /dev/null"; the socket works, but I have nothing in $result (note also that I don't anderstand why the ( ... &) are needed!?) and if I use Code: [Select] $command = "/mnt/opt/www/called.php $_SERVER[REMOTE_ADDR]"; I have the correct text "normal end of the called.php script" in $result but the socket connection is not performed (no errors in php logs) Could you help me to find a way to let's work the two features correctly together? Thank you. hey guys im really just after a bit of help/information on 2 things (hope its in the right forum).
1. basically I'm wanting to make payments from one account to another online...like paypal does...im wondering what I would need to do to be able to do this if anyone can shine some light please?
2.as seen on google you type in a query in the search bar and it generates sentences/keywords from a database
example:
so if product "chair" was in the database
whilst typing "ch" it would show "chair" for a possible match
I know it would in tale sql & json but im after a good tutorial/script of some sort.
if anyone can help with some information/sites it would be much appreciated.
Thank you
hello all, What I want to do is, make the session ID clickable in a url here Code: [Select] Login Successful <a href="user.php">Conitnue</a> so when a user logs in, his ID gets in the link of Continue so he can only see his information so for example, if his id is 10, then the url would be ....user.php?id=10 Is there anyway to work out the age of a session, i.e the time that has elapsed since it was created? Cheers I am having trouble calling the session var "email" from the landing page. Here is the code that I am using. I am not even sure the session "email" is starting or registering. Code: [Select] $referer = $_SERVER['HTTP_REFERER']; $email = $_POST['email']; $sql="SELECT * FROM users WHERE email='$email' "; $result=mysql_query($sql); // Mysql_num_row is counting table row $count=mysql_num_rows($result); // If result matched $email, table row count must be 1 row if($count==1){ // Session Register email session_start(); $_SESSION['email'] = $email; header("location:".$referer2." "); exit(); } Hi, I have a PHP page that includes this JS: Code: [Select] <script type="text/javascript"> AC_FL_RunContent( 'codebase','http://download.macromedia.com/pub/shockwave/cabs/flash/swflash.cab#version=9,0,28,0','width','465','height','80','class','FlashBorder','src','img/skype_banner_eng','quality','high','pluginspage','http://www.adobe.com/shockwave/download/download.cgi?P1_Prod_Version=ShockwaveFlash','movie','img/skype_banner_eng' ); //end AC code </script> skype_banner_eng is a flash movie in English. I need to get this movie to change to a different languagdepending on the language variable loaded. I have done this with jpg images by doing this: Code: [Select] header_<?php echo ($_SESSION['session_idioma']);?>.jpg but whenI try to do the same within the JS it doesn't work....any ideas ? Thanks Firstly, I am new to the forum, so hello I am trying to code a login script for my website. I have got the login to work but how do I get it to create a session so the user stays logged in until they log out? Also how can I prevent access to success.php and fail.php so they cannot be accessed directly. I am new to PHP so please explain in detail for me. Here is my code... Code: [Select] <?php ob_start(); $host=""; // Host name $username=""; // Mysql username $password=""; // Mysql password $db_name=""; // Database name $tbl_name=""; // Table name mysql_connect("$host", "$username", "$password")or die("cannot connect"); mysql_select_db("$db_name")or die("cannot select DB"); $myusername=$_POST['myusername']; $mypassword=$_POST['mypassword']; $mypassword=md5($mypassword); $myusername = stripslashes($myusername); $mypassword = stripslashes($mypassword); $myusername = mysql_real_escape_string($myusername); $mypassword = mysql_real_escape_string($mypassword); $sql="SELECT * FROM $tbl_name WHERE email='$myusername' and password='$mypassword'"; $result=mysql_query($sql); $count=mysql_num_rows($result); if($count==1){ session_register("myusername"); session_register("mypassword"); header("location:success.php"); } else { header("location:fail.php"); } ob_end_flush(); ?> |