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PHP - Php And Writing To The The User's Hdd, Possible ?
I have a web app hosted on Just Host that I have nearly finished writing but still needs to have coded the ability to write a text file to user's computer. the user should be able to specify where on their computer they would like the file to be stored. Is this possible ? And can you tell me how it's done, or point me to a souce that can tell me
I rarely ever ask for help regarding programming, but this has flumoxed me. If it is not possible to do this, then would I have to generate this file on the Host's server (Just Host in my case), then download it ? If this can be done then can you please tell me how, as any info I have found related to file downloads seems a bit obscure thanks in advance Similar TutorialsUntil I switched to Plesk 9 I had no trouble writing a log file for an application that I wrote. It used PHP's FOPEN() command. It would write the file as the user. Now it does not write at all. If I run it manually it the file is owned by siteadmin but writes it as as root:root.. If I run it through a browser it is owned by apache:apache. What's changed? Code: [Select] <?php if (file_exists('test.txt')) { unlink('test.txt'); // if it already exists let's erase it as a workaround } $fp = fopen('test.txt', 'w') or die("Unable to open file."); fwrite($fp, strftime('%c')."\r\n"); fclose($fp); ?> A similar block of code would write a ticker file that later would display on our pages. It was written and saved as the main site user, such as "siteadmin" on Plesk. Now PHP reports that it is unable to open or save the file. Permissions. We've changed nothing. This script has run since 2002 using chmod 644. We worked around this by creating an external script and using a CRON to write the file and this works fine. This suggests that the user has permission to open and write the file! Imagine our confusion. Don't laugh, but this server still has PHP 4.3.9, which adds to the confusion. It's run this way for years. Quote PHP 4.3.9 (cgi) (built: Sep 27 2006 20:40:56) Copyright (c) 1997-2004 The PHP Group Zend Engine v1.3.0, Copyright (c) 1998-2004 Zend Technologies with the ionCube PHP Loader v3.1.16, Copyright (c) 2002-2006, by ionCube Ltd., and with Zend Extension Manager v1.0.10, Copyright (c) 2003-2006, by Zend Technologies with Zend Optimizer v3.0.1, Copyright (c) 1998-2006, by Zend Technologies I would appreciate your assistance, there are tons of login scripts and they work just fine. However I need my operators to login and then list their activities for the other operators who are logged in to see and if desired send their clients on the desired activity. I have the login working like a charm and the activities are listed just beautifully. How do I combine the two tables in the MySQL with PHP so the operator Logged in can only make changes to his listing but see the others. FIRST THE ONE script the member logges in here to the one table in MSQL: <?php session_start(); require_once('config.php'); $errmsg_arr = array(); $errflag = false; $link = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD); if(!$link) { die('Failed to connect to server: ' . mysql_error()); } $db = mysql_select_db(DB_DATABASE); if(!$db) { die("Unable to select database"); } function clean($str) { $str = @trim($str); if(get_magic_quotes_gpc()) { $str = stripslashes($str); } return mysql_real_escape_string($str); } $login = clean($_POST['login']); $password = clean($_POST['password']); if($login == '') { $errmsg_arr[] = 'Login ID missing'; $errflag = true; } if($password == '') { $errmsg_arr[] = 'Password missing'; $errflag = true; } if($errflag) { $_SESSION['ERRMSG_ARR'] = $errmsg_arr; session_write_close(); header("location: login-form.php"); exit(); } $qry="SELECT * FROM members WHERE login='$login' AND passwd='".md5($_POST['password'])."'"; $result=mysql_query($qry); if($result) { if(mysql_num_rows($result) == 1) { session_regenerate_id(); $member = mysql_fetch_assoc($result); $_SESSION['SESS_MEMBER_ID'] = $member['member_id']; $_SESSION['SESS_FIRST_NAME'] = $member['firstname']; $_SESSION['SESS_LAST_NAME'] = $member['lastname']; session_write_close(); header("location: member-index.php"); exit(); }else { header("location: login-failed.php"); exit(); } }else { die("Query failed"); } ?> ................................................. ................................ Now I need the person who logged in to the table above to be able to make multiple entries to the table below <? $ID=$_POST['ID']; $title=$_POST['title']; $cost=$_POST['cost']; $activity=$_POST['activity']; $ayear=$_POST['aday']; $aday=$_POST['ayear']; $seats=$_POST['special']; $special=$_POST['seats']; mysql_connect("xxxxxx", "xxx350234427", "========") or die(mysql_error()); mysql_select_db("xxxx") or die(mysql_error()); mysql_query("INSERT INTO `activity` VALUES ('ID','$title', '$cost','$activity', '$aday', '$ayear', '$special', '$seats')"); Print "Your information has been successfully added to the database!" ?> Click <a href="member-profile.php">HERE</a> to return to the main menu <?php ?> hi, i have made a website where people resgister their details of them and products. they have to enter the following details in form Name of company name of the product company address email id password mobile number contact and brief details about their company
user can then login with email id and pwd. now after login ..user will get a page where he can upload the photos of products images and their price, so now my question is that when he finishes uploading (|by clicking on upload button) the product images and price text box ..then on final uploaded webspage it should show all other things which he registerd before (company name , mobile number etc) along with images and price...hence the main question that user does not need to enter mobile and address while uploading images and filling proce ..but on the final page it should show mobile and address along with price and images..as user is not going to enter mobile and address again and again as he will have multiple products to upload.
Actually, what i want to do is to use the email to fetch the $email,$password and $randomnumber from database after Hi, so far I have managed to set up a somewhat basic login website with a mysql database backend. Once they have logged on they go to a "main menu" page. What I need to define is that user A sees button A but only that button, etc. (Then of course that same rule would have to apply if they tried to directly go to the page, but I am guessing I can do that in the same way that I currently do to force a login). If anyone has any tutorials or sample code I would much appreciate it. Thanks, Hi, I am getting frustrated beyond belief at the moment with trying to get a very simple script to run, I am using PHP 5.3.3 and MySQL 5.1 on a Win2k8 server with IIS7.5. Basically my script is connecting to a local database, running a single select query, returning those rows and building up a string from them. The problem is that I am receiving complete BS responses from PHP that the access is denied for the user being specified. This is complete rubbish since the user can connect via mysql, sqlyog, ASP.NET MVC without issue but for some bizarre reason it is not working via PHP. The code for the script is here : Code: [Select] <?php $mysql = mysql_connect('127.0.0.1:3306', 'myuser', 'mypass', 'mydatabase'); if (!$mysql) { die(mysql_error()); $content = "<nobr></nobr>"; } else { $result = mysql_query('SELECT * FROM tblEventGroup'); $content = "<nobr>"; if ($result) { while($row = mysql_fetch_assoc($result)) { $content .= "<span>"; $content .= $row['GroupName']; $content .= "</span>"; $content .= "<a href=\"../Event/EventSearch?groupid="; $content .= $row['GroupId']; $content .= "\" target=\"_blank\">Book here</a> "; } } mysql_close($mysql); $content .= "</nobr>"; } ?> I cannot for the life of me understand what the problem is, the return error is Access denied for user 'myuser'@'localhost' (using password: YES) Hi guys, I am trying to put together a little system that allows users to log onto my website and access there own personal page. I am creating each page myself and uploading content specific to them which cannot be viewed by anyone else. I have got the system to work up as far as: 1/ The user logs in 2/ Once logged in they are re-directed to their own page using 'theirusername.php' Thats all good and working how I need it too. The problem I have is this. If I log onto the website using USER A details - I get taken to USER A's page like I should but - If I then go to my browser and type in USERBdetails.php I can then access USER B's page. This cannot happen!! I need for USER A not to be able to access USER B profile - there is obviously no point in the login otherwise! If you are not logged in you obviously cannot access any secure page. That much is working! Please find below the code I am using: LOGIN <?php session_start(); function dbconnect() { $link = mysql_connect("localhost", "username", "password") or die ("Error: ".mysql_error()); } ?> <?php if(isset($_SESSION['loggedin'])) { header("Location:" . strtolower($username) . ".php"); if(isset($_POST['submit'])) { $username = mysql_real_escape_string($_POST['username']); $password = mysql_real_escape_string($_POST['password']); $mysql = mysql_query("SELECT * FROM clients WHERE username = '{$username}' AND password = '{$password}'"); if(mysql_num_rows($mysql) < 1) { die("Password or Username incorrect! Please <a href='login.php'>click here</a> to try again"); } $_SESSION['loggedin'] = "YES"; $_SESSION['username'] = $username; $_SESSION['name'] header("Location:" . strtolower($username) . ".php"); } ?> HEADER ON EACH PHP PAGE <?php session_start(); if(!isset($_SESSION['loggedin'])) { die(Access to this page is restricted without a valid username and password); ?> --------------------------------------------------- Am I right in thinking it is something to do with the "loggedin" part? The system I have here is adapted from a normal login system I have been using for years. The original just checks the details and then does a 'session start'. This one obviously has to re-direct to a user specific page. To do this I used the <<header("Location:" . strtolower($username) . ".php");>> line to redirect to a page such as "usera.php" or "userb.php" Any help would be greatly appreciated! Ta Hallo everybody,
i have the following code.
but i get allways this error while the user exist in the database.
User not found!
what do i do wrong?
thank you very much for your help
Rafal
<html>
<head>
<?php
$connection = mysql_connect("db.xyz.com", "username", "password")
or die ("connection fehler");
mysql_select_db("db0123456789")
or die ("database fehler");
$email = $_POST["inp_email"];
$pwd = $_POST["inp_pwd"];
if($email && $pwd)
{
$chkuser = mysql_query("SELECT email FROM gbook WHERE email = '($email)' ");
$chkuserare = mysql_num_rows($chkuser);
echo $email;
echo $pwd;
if ($chkuserare !=0)
{
$chkpwd = mysql_query("SELECT pwd FROM gbook WHERE email = '($email)' ");
$pwddb = mysql_fetch_assoc($chkpwd);
if ($pwd != $pwddb["pwd"])
{
echo "password is wrong!";
}
else
{
echo "login successed";
}
}
else
{
echo "User not found!";
}
}
else
{
echo "Pleas enter your email and password!";
}
mysql_close($connection);
?>
</head>
<body>
<form action="login.php" method="post">
Email <input type="text" name="inp_email"><br>
Password <input type="text" name="inp_pwd"><br>
<input type="submit" name="submit" value="login">
</form>
</body>
</html>
Edited by rafal, 21 September 2014 - 04:33 PM. Hallo everybody,
the user is in the table, but i get error (user not found!).
thank you very much for your help
Rafal
<!DOCTYPE html>
<html>
<head>
<title>index</title>
<meta http-EQUIV="CONTENT-LANGUAGE" content="en">
<?php
SESSION_START();
include("abc.php");
$link2 = mysqli_connect("$hoster", "$nameuser", "$password", "$basedata")
or die ("connection error" . mysqli_error($link2));
$email = $_POST["inp_email"];
$pwd = $_POST["inp_pwd"];
if($email && $pwd)
{
$chkuser = mysqli_query("SELECT email FROM $table2 WHERE email = '$email' ");
$chkuserare = mysqli_num_rows($chkuser);
if ($chkuserare !=0)
{
$chkpwd = mysqli_query("SELECT pwd FROM $table2 WHERE email = '$email'");
$pwddb = mysqli_fetch_assoc($chkpwd);
if (md5($pwd) != $pwddb["pwd"])
{
echo "Password is wrong!";
}
else
{
$_SESSION['username'] = $email;
header ('Location:list.php');
}
}
else
{
echo "user not found!";
}
}
else
{
echo "enter your Email and Password!";
}
mysqli_close($link2);
?>
</head>
<body style="font-family: arial;margin: 10; padding: 0" bgcolor="silver">
<font color="black">
<br>
<form action="index.php" method="post">
<b>Login</b><br><br>
<table width="100%">
<tr><td>
Email:<br><input type="text" name="inp_email" style="width:98%; padding: 4px;"><br>
Password:<br><input type="password" name="inp_pwd" style="width:98%; padding: 4px;"><br>
<br>
<input type="submit" name="submit" value="Login" style="width:100%; padding: 4px;">
</td></tr>
</table>
</form>
</font>
</body>
</html>
Hello, i've got some shop script which has 2 payment modules which i'd like to use for something else, the payment modules only work if the user is logged in though, i tried to make them standalone scripts but that didn't work out too well. So now i decided to go another way and just let everyone have the same session so everyone will be using the same username&password automatically. the index file looks like this: Code: [Select] <?php include('./inc/config.php'); include('./inc/functions.php'); include('./lang/'.$language.'.lng'); $id = addslashes($_REQUEST["id"]); $user = addslashes($_REQUEST["username"]); $pass = addslashes($_REQUEST["password"]); $language = strtolower($language); if(empty($id)) $id =1; $file = mysql_query('SELECT * FROM navi_'.$language.' WHERE id="'.$id.'"'); if(mysql_num_rows($file)>0) $file = mysql_fetch_array($file); else $file = mysql_fetch_array(mysql_query('SELECT * FROM navi_'.$language.' WHERE id="404"')); if(!empty($user) AND !empty($pass)) {$query = mysql_query('SELECT * FROM users WHERE username="'.$user.'" AND pass="'.md6($pass).'"'); if(mysql_num_rows($query) == 1) {$_SESSION[$session_prefix."user"] = ucfirst($user); echo'<meta http-equiv="refresh" content="0; url=index.php?id=8">';} else $error = 'Username oder Passwort ist falsch.';} include('./designe/'.$designe.'/head.tpl'); include('./designe/'.$designe.'/navi.php'); include('./designe/'.$designe.'/middle.tpl'); if(file_exists('./pages/'.$file["file"])) {echo'<h1>'.ucfirst($file["title"]).'</h1>'; include('./pages/'.$file["file"]);} if(!empty($error)) echo '<font color="red">'.$error.'</font>'; include('./designe/'.$designe.'/foot.tpl'); ?> Now i tried alot of things including adding: Code: [Select] session_start(); $_SESSION["username"] = "peter"; $_SESSION["user"] = "peter"; $_SESSION["id"] = "1"; $_SESSION["pass"] = "peter"; $_SESSION["password"] = "peter"; or Code: [Select] $id = "1"; $user = "peter"; $username = "peter"; $pass = "peter"; $password = "peter"; also a combination of both, nothing works, but i don't understand why ? Any help is appreciated. /Edit, i tried adding it to the paymentmodule .php aswell, but no luck. I have written some code to past information from one page to a diffrent page however all the variables are sent over but they are not writting to the database could anyone help me out please .... $ud_P_Id = $_POST['ud_P_Id']; $ud_LastName = $_POST['ud_LastName']; $ud_FirstName = $_POST['ud_FirstName']; $ud_GroupCode=$_POST['ud_GroupCode']; $ud_P_Unit1=$_POST['ud_P_Unit1']; $ud_P_Unit2=$_POST['ud_P_Unit2']; $ud_P_Unit3=$_POST['ud_P_Unit3']; $ud_P_Unit6=$_POST['ud_P_Unit6']; $ud_P_Unit14=$_POST['ud_P_Unit14']; $ud_P_Unit20=$_POST['ud_P_Unit20']; $ud_P_Unit27=$_POST['ud_P_Unit27']; $ud_P_Unit28=$_POST['ud_P_Unit28']; $ud_P_Unit42=$_POST['ud_P_Unit42']; $ud_P_Unit10=$_POST['ud_P_Unit10']; $ud_P_Unit11=$_POST['ud_P_Unit11']; $ud_P_Unit12=$_POST['ud_P_Unit12']; $ud_P_Unit13=$_POST['ud_P_Unit13']; $ud_P_Unit1454=$_POST['ud_P_Unit1454']; $ud_P_Unit15=$_POST['ud_P_Unit15']; $ud_P_Unit16=$_POST['ud_P_Unit16']; $ud_P_Unit17=$_POST['ud_P_Unit17']; $ud_P_Unit18=$_POST['ud_P_Unit18']; $con = mysql_connect('localhost', 'lccstude_progre', '********'); $db= "lccstude_pro"; if (! $con) die("Couldn't connect to MySQL"); mysql_select_db($db , $con) or die("Couldn't open $db: ".mysql_error()); mysql_query("UPDATE BTECL31113 SET FirstName='$ud_FirstName' , LastName='$ud_LastName' , GroupCode='$ud_GroupCode' , Unit1='$ud_P_Unit1' , Unit2='$ud_P_Unit2' , Unit3='$ud_P_Unit3' , Unit6='$ud_P_Unit6' , Unit14='$ud_P_Unit14' , Unit20='$ud_P_Unit20' , Unit27='$ud_P_Unit27' , Unit28='$ud_P_Unit28' , Unit42='$ud_P_Unit42' , Unit10='$ud_P_Unit10' , Unit11='$ud_P_Unit11' , Unit12='$ud_P_Unit12' , Unit13='$ud_P_Unit13' , Unit1454='$ud_P_Unit1454' , Unit15='$ud_P_Unit15' , Unit16= $ud_P_Unit16' , Unit17='$ud_P_Unit17' , Unit18='$ud_P_Unit18' WHERE P_Id='$ud_P_Id'"); echo "Record Updated"; mysql_close($con); Please any help would be great I am using a cache script, well it writes a array to a file like this: Code: [Select] fwrite($fh, '<?php'."\n\n".'define(\'PUN_LOTTERY_LOADED\', 1);'."\n\n".'$lottery = '.var_export($output2, true).';'."\n\n".'?>'); output2 is Code: [Select] $result2 = $db->query('MY QUERY '); $output2 = array(); while ($cur_donors = $db->fetch_assoc($result2)) $output2[] = $cur_donors; Now, I want to ditch the mysql and I want to use this script with 7 variables that I already have loaded, so I dont need to use the mysql, how do I add my 7 variables to my var_export function instead of using mysql to loop them? I am using a script I adapted from a tutorial to print the contents of a text box to a txt file. Basically, it's a really simple way of seeing who has logged in. I only have a handful of users. The problem is, although the text file is being created in the proper folder, it isn't being written to and just remains blank. <div align="center"> <table width="300" border="2" bordercolor="#FFFFFF" style="-moz-border-radius: 18px; -webkit-border-radius: 18px;" height="120" cellpadding="0" cellspacing="0"> <tr> <form name="form1" method="post" action="checklogin.php"> <td> <table width="100%" border="0" cellpadding="3" cellspacing="1" background="images/loginbg.jpg" style="-moz-border-radius: 15px; -webkit-border-radius: 15px;"> <tr align="center"> <td colspan="3"><font color="#FFFFFF"><strong>Family Login </strong></font></td> </tr> <tr> <td width="78"><font color="#000000">Username</font></td> <td width="6">:</td> <td width="294"><input name="myusername" type="text" id="myusername"> <?php $myusername = $_POST['myusername']; $data = "$myusername\n"; //open the file and choose the mode $fh = fopen("logs/login.txt", "a"); fwrite($fh, $data); fclose($fh); ?></td> </tr> <tr> <td><font color="#000000">Password</font></td> <td>:</td> <td><input name="mypassword" type="password" id="mypassword"></td> </tr> <tr> <td> </td> <td> </td> <td><input type="submit" name="Submit" value="Login"> </td> </tr> </table> </td> </form> </tr> </table> </div> I'm not sure what's going wrong but I'm guessing it's the placing of the php, or at least some of it. I'd quite like to add the time they logged in as well. Any idea's anyone? I need help! I cant get this to write the correct way! What i need is based on the value of what is posted to the script it has to write it in the config file and also make a folder! Please help Code: [Select] <?php $start = '$uploadpath=\''; $structure = '../banner/images/'.$_POST['FOLDER'].'\';\n'; $myFile = "PHP/confup.php"; $fh = fopen($myFile, 'w') or die("can't open file"); $stringData = "<?\n"; fwrite($fh, $stringData); $stringData = "$start $structure"; fwrite($fh, $stringData); $stringData = "?>\n"; fwrite($fh, $stringData); fclose($fh); // Desired folder structure // To create the nested structure, the $recursive parameter // to mkdir() must be specified. if (!mkdir($structure, 0777, true)) { die('Failed to create folders...'); } // ... ?> Greetings, I am trying to write XML using SimpleXML for a web service call. I was having success until the XML got a little more complicated. Here is the XML format where I am running into problems: Code: [Select] <Agent> <Person Last='Smith' First='John'> <Addresses /> <PhoneNumbers> <Phone Type='1' Number='888-555-1212'> </PhoneNumbers> </Person> </Agent> Here is my php: Code: [Select] $xmlOutput = new SimpleXMLElement('<?xml version="1.0"?><ReportRequest> </ReportRequest>'); $xmlOutput->addAttribute('CID','9200'); $xmlOutput->addAttribute('Diagram','1'); $xmlOutput->addAttribute('DueDate','2011-11-15'); $xmlOutput->addAttribute('NumPhotos','6'); $xmlOutput->addAttribute('InspectAfter',''); $xmlOutput->addAttribute('PolicyNumber','JTC0004425'); $reportType = $xmlOutput->addChild('ReportType'); $reportType->addAttribute('CPType','Commercial'); $reportType->addAttribute('SectionIDs',''); $reportType->addAttribute('Description',''); $reportType->addAttribute('ReportTypeID','123'); $locations = $xmlOutput->addChild('Locations')->addChild('Addresses')->addChild('Address'); $locations->addAttribute('Zip','91216'); $locations->addAttribute('City','Chatsworth'); $locations->addAttribute('Line1','123 Main St'); $locations->addAttribute('Line2','Suite 201'); $locations->addAttribute('State','CA'); $locations->addAttribute('Latitude',''); $locations->addAttribute('Longitude',''); $agent = $xmlOutput->addChild('Agent')->addChild('Person')->addChild('Addresses')->addChild('PhoneNumbers')->addChild('Phone'); $agent->addAttribute('Last','Smith'); $agent->addAttribute('Email',''); $agent->addAttribute('First','John'); $agent->addAttribute('Title',''); $agent->addAttribute('Type','1'); $agent->addAttribute('Number','888-555-1212'); $agent->addAttribute('TypeName','Office'); $agent->addAttribute('Extension',''); Header('Content-type: text/xml'); echo $xmlOutput->asXML(); Everything outputs as it should until the line that starts with $agent. I want the output to match the section above. I have tried several variations but I cannot seem to figure it out. I know the current PHP sample doesn't work. Help?? Thanks in advance, John I don't think the start of my code is right?! Code: [Select] echo '<button onclick=\"gohere(viewpub.php?PubID='.$row['PubID'].')" id="button" type="button" class="ui-button ui-widget ui-state-default ui-corner-all ui-button-text-only" role="button" aria-disabled="false"><span class="ui-button-text">View Pub</span></button>'; Please help?! Hi everyone. Still new to PHP and to Object-Oriented Programming. My goal for today is to write a few dummy Classes in PHP that do enough so that I can see something on my webpage. I am wondering if there is someone here who would be willing to help in either the forums or one-on-one. It may sound like a silly request, but this is all new to me as a Procedural Programmer who has actively programmed in about 10 years!! Thanks, TomTees Hey Guys. I am trying to write to the file depending on which condition is met. The code works fine on my local machiene but not on my remote server. I have also tried to output any error messages to see if it would output anything, and I don't get anyting on my browser. Can anyone help me with this issue? Thanks
<?php
if($_SERVER['REQUEST_METHOD'] == "POST") {
isset($_POST['interfax']) ? $option= "interfax" : $option= "";
isset($_POST['metrofax']) ? $option= "metrofax" : $option= "";
switch ($option) {
case 'interfax':
$file = "fax.php";
$fax_client = "interfax";
if(file_put_contents($file, "<?php ".'$fax_client = "' . $fax_client . '"'." ?>"))
{ echo "Successful"; } else {
die("Can't write file");
}
break;
// By defualt all the orders go to metrofax so by selecting the variable it resets it self
case 'metrofax':
$file = "fax.php";
$fax_client = "metrofax";
file_put_contents($file, "<?php ".'$fax_client = "' . NULL . '"'." ?>");
break;
}
}
?>
<form action="#" method="POST">
<input type="radio" name="interfax" value="interfax">Switch To Interfax<br>
<input type="radio" name="metrofax" value="metrofax">Switch To Metrofax<br>
<input type='submit' name="submit" >
im trying to make a program that changes some files from 0 to 1,but im having some trouble... heres my code, Code: [Select] $id = $_GET["id"]; $file1 = "http://mysite.co.cc/users/".$id."/file1.txt"; $fh = fopen($file1, 'w'); fwrite($fh, "1"); fclose($fh); I followed this tut: http://www.tizag.com/phpT/filewrite.php I cant find anything wrong with the code,but it will not make the file have 1 in it. I'm writing a script that takes user input from a html form and updates the database with the new data. In this example the user is updating the data about the university they attend. The problem with this is, the script to update the database with the new information appears to be not working - the new data is not being added to the database. The SQL queries used in the script work in phpMyadmin, and the variables - $uni and $username - contain the data the correct data. This has me stumped, could someone look over this for me please and tell me what is going wrong here. Code: [Select] <?php include 'connect.php'; session_start(); $_SESSION['username']; if(!(isset($_SESSION['login']) && $_SESSION['login']!= " ")){ header("Location: login.php"); } $username = $_SESSION['username']; $tablename = 'usr_test'; $uni = $_POST['uni']; $uni = stripslashes($uni); $uni = mysql_real_escape_string($uni); $uni = trim($uni); if(isset($uni)) { $username = $_SESSION['username']; $loc = mysql_query("SELECT * FROM usr_test WHERE usr = '$username'"); if (mysql_num_rows($loc) == 0) header("Location:notlogged.php"); else { extract(mysql_fetch_array($loc)); mysql_query("UPDATE usr_test SET uni = ('$uni') WHERE usr = '$username'") or die (mysql_error()); header("Location:profile.php"); } } ?> |
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