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PHP - Unable To Show Session Name
I have written my index.php script where after verifying the password the script displays the user's name.
But the user name is not displayed by it. Here is my script Similar TutorialsCode: [Select] <html> <head> <script type="text/javascript"> function a() { var x = document.register.username.value; var y = document.register.pass.value; var z = document.register.pass2.value; if(x==""&& y==""&& z=="") { alert("Please insert all message!"); return false; } if(x=="") { alert("Please insert an username!"); return false; } if(y=="") { alert("Please insert an password!"); return false; } if(z=="") { alert("Please insert an password2!"); return false; } if (y!=z) { alert("Your passwords did not match"); return false; } } </script> </head> <?php mysql_connect("localhost","root") or die(mysql_error()); mysql_select_db("cute") or die(mysql_error()); if (isset($_POST["sub"])) { $_POST['pass'] = md5($_POST['pass']); if (!get_magic_quotes_gpc()) { $_POST['username'] = addslashes($_POST['username']); $_POST['pass'] = addslashes($_POST['pass']); } $usercheck = $_POST["username"]; $check = mysql_query("SELECT username FROM regis WHERE username = '$usercheck'") or die(mysql_error()); $check2 = mysql_num_rows($check); //if the name exists it gives an error if ($check2 != 0) { echo("<SCRIPT LANGUAGE='JavaScript'>window.alert('Sorry, the username" ." ".$usercheck." ". "is already in use.')</SCRIPT>"); echo ("<SCRIPT LANGUAGE='JavaScript'>setTimeOut(window.location = 'registration.php',1)</script>"); } else if($_POST['username'] && $_POST['pass'] && $_POST['pass2'] ) { $insert = "INSERT INTO regis(username, password) VALUES ('".$_POST['username']."', '".$_POST['pass']."')"; $add_member = mysql_query($insert); echo("<SCRIPT LANGUAGE='JavaScript'>window.alert('Registration had been succesfully added :)')</SCRIPT>"); } } ?> <body> <form name="register" method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>" onsubmit="return a()"> <table border='0'> <tr><td>Username:</td><td><input type="text"name="username" maxlength="60"></td></tr> <tr><td>Password:</td><td><input type="password" name="pass" maxlength="10"></td></tr> <tr><td>Confirm Password:</td><td><input type="password" name="pass2" maxlength="10"></td></tr> <tr><th colspan=2><input type="submit" name="sub" value="Register"></th></tr></table> </form> </body> </html> This my registration page can you show how to redirect to login.php using $SESSION and also it kept the username information in session if the session is true $_SESSION['frommonth'] i want it to show the selected session Code: [Select] <select name="from_month" id="from_month"> <option value="<?php echo $_SESSION['frommonth'];?>" selected="selected">mm</dd> <?for ($i=1;$i<=12;$i++):?> <option value="<?=$i?>"><?=$months[$i]?></dd> <?endfor;?> </select> how could i do this? show list of files uploaded by current session user to the database. I want to show different users when they log in to the website...they can see a list of old files that they have uploaded. can anyone tell me the code/script to this.....please, ty I am having trouble resolving an error. Warning: session_start() [function.session-start]: Cannot send session cache limiter - headers already sent (output started at /home/s519970/public_html/header.php:27) in /home/s519970/public_html/admin/login.php on line 2 What I can gather is I can't use "header (Location: 'admin.php')" after i've used session_start(). I have tried to replace the header (Location: 'admin.php') with this: echo "<script>document.location.href='admin.php'</script>"; echo "<script>'Content-type: application/octet-stream'</script>"; I've been trying to read up on solutions but haven't been able to get it sorted. If anyone can offer some advice that would be greatly appreciated as im new to php. Code: [Select] <?php session_start(); if(isset($_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! 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 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. I want to show a cookie of a referral's username on a sign up page. The link is like this, www.mysite.com/signup?ref=johnsmith. The cookie doesn't show if I go to that url page. But it does show up once I reload the page. So I'm wondering if it's possible to show the cookie the first time around, instead of reloading the page? Here is my code.
// This is in the header
$url_ref_name = (!empty($_GET['ref']) ? $_GET['ref'] : null);
if(!empty($url_ref_name)) {
$number_of_days = 365;
$date_of_expiry = time() + 60 * 60 * 24 * $number_of_days;
setcookie( "ref", $url_ref_name, $date_of_expiry,"/");
} else if(empty($url_ref_name)) {
if(isset($_COOKIE['ref'])) {
$user_cookie = $_COOKIE['ref'];
}
} else {}
// This is for the sign up form
if(isset($_COOKIE['ref'])) {
$user_cookie = $_COOKIE['ref'];
?>
<fieldset>
<label>Referred By</label>
<div id="ref-one"><span><?php if(!empty($user_cookie)){echo $user_cookie;} ?></span></div>
<input type="hidden" name="ref" value="<?php if(!empty($user_cookie)){echo $user_cookie;} ?>" maxlength="20" placeholder="Referrer's username" readonly onfocus="this.removeAttribute('readonly');" />
</fieldset>
<?php
}
I am displaying rows from a database onto a page using: while($row=mysql_fetch_array($query)){ echo $row['name']; } I need to figure out how to limit the rows shown to the page to 100. And if there are 100 rows on the page, a link will be displayed at the bottom, that says "Next 100". Then this will display the next 100 rows. Can you give an example how to do this please? Thanks Hi , I know my code sucks but i'm learning fast!! I'm trying to show a form if the qty value in a database == 10 or a different form if the value ==20. I tried but failed. Any help really appreciated. Code: [Select] <?php require_once('Connections/book.php'); ?> <?php $colname_cardpayment = "-1"; if (isset($_GET['orderid'])) { $colname_cardpayment = (get_magic_quotes_gpc()) ? $_GET['orderid'] : addslashes($_GET['orderid']); } mysql_select_db($database_book, $book); $query_cardpayment = sprintf("SELECT * FROM cards WHERE orderid = '%s' ORDER BY qty ASC", $colname_cardpayment); $cardpayment = mysql_query($query_cardpayment, $book) or die(mysql_error()); $row_cardpayment = mysql_fetch_assoc($cardpayment); $totalRows_cardpayment = mysql_num_rows($cardpayment); // Database connect $con = mysql_connect("mysql1.myhost.ie","admin_book","root123"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("book_test", $con); //Parse Values from Coupon.php Form $orderid = mysql_real_escape_string(trim($_POST['orderid'])); $name = mysql_real_escape_string(trim($_POST['name'])); $surname = mysql_real_escape_string(trim($_POST['surname'])); $add1 = mysql_real_escape_string(trim($_POST['add1'])); $add2 = mysql_real_escape_string(trim($_POST['add2'])); $town = mysql_real_escape_string(trim($_POST['town'])); $county = mysql_real_escape_string(trim($_POST['county'])); $postcode = mysql_real_escape_string(trim($_POST['postcode'])); $phone = mysql_real_escape_string(trim($_POST['phone'])); $email = mysql_real_escape_string(trim($_POST['email'])); $letterstyle = mysql_real_escape_string(trim($_POST['letterstyle'])); $sql="INSERT INTO custdetails (orderid, name, surname, add1, add2, town, county, postcode, phone, email, letterstyle) VALUES ('$orderid','$name','$surname','$add1','$add2','$town','$county','$postcode','phone','$email','$letterstyle')"; if (!mysql_query($sql)) { die('Error: ' . mysql_error()); } ?> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=UTF-8" /> <title>Digital Scribe Books</title> <link href="style.css" rel="stylesheet" type="text/css" /> <script type="text/javascript"> function MM_preloadImages() { //v3.0 var d=document; if(d.images){ if(!d.MM_p) d.MM_p=new Array(); var i,j=d.MM_p.length,a=MM_preloadImages.arguments; for(i=0; i<a.length; i++) if (a[i].indexOf("#")!=0){ d.MM_p[j]=new Image; d.MM_p[j++].src=a[i];}} } function MM_swapImgRestore() { //v3.0 var i,x,a=document.MM_sr; for(i=0;a&&i<a.length&&(x=a[i])&&x.oSrc;i++) x.src=x.oSrc; } function MM_findObj(n, d) { //v4.01 var p,i,x; if(!d) d=document; if((p=n.indexOf("?"))>0&&parent.frames.length) { d=parent.frames[n.substring(p+1)].document; n=n.substring(0,p);} if(!(x=d[n])&&d.all) x=d.all[n]; for (i=0;!x&&i<d.forms.length;i++) x=d.forms[i][n]; for(i=0;!x&&d.layers&&i<d.layers.length;i++) x=MM_findObj(n,d.layers[i].document); if(!x && d.getElementById) x=d.getElementById(n); return x; } function MM_swapImage() { //v3.0 var i,j=0,x,a=MM_swapImage.arguments; document.MM_sr=new Array; for(i=0;i<(a.length-2);i+=3) if ((x=MM_findObj(a[i]))!=null){document.MM_sr[j++]=x; if(!x.oSrc) x.oSrc=x.src; x.src=a[i+2];} } </script> </head> <body onload="MM_preloadImages('images/buttons/home_over.png','images/buttons/books_over.png','images/buttons/cards_over.png','images/buttons/letters_over.png')"> <div id="snow"> <div id="wrapper"> <div id="header"> <div id="logo"><img src="images/digital_scripe.png" width="218" height="91" /></div> <div id="menu"><a href="index.php" onmouseout="MM_swapImgRestore()" onmouseover="MM_swapImage('Home','','images/buttons/home_over.png',1)"><img src="images/buttons/home_act.png" name="Home" width="131" height="132" border="0" id="Home" /></a><a href="books.php" onmouseout="MM_swapImgRestore()" onmouseover="MM_swapImage('Books','','images/buttons/books_over.png',1)"><img src="images/buttons/books_act.png" name="Books" width="131" height="132" border="0" id="Books" /></a><a href="cards.php" onmouseout="MM_swapImgRestore()" onmouseover="MM_swapImage('Cards','','images/buttons/cards_over.png',1)"><img src="images/buttons/cards_act.png" name="Cards" width="131" height="132" border="0" id="Cards" /></a><a href="letters.php" onmouseout="MM_swapImgRestore()" onmouseover="MM_swapImage('Letters','','images/buttons/letters_over.png',1)"><img src="images/buttons/letters_act.png" name="Letters" width="131" height="132" border="0" id="Letters" /></a></div> </div> <div id="content"> <?php echo 'Order ID is : '. $orderid . '.<br />'; if ($row2['qty'] == 10) echo "<div> <form action="https://www.paypal.com/cgi-bin/webscr" method="post"> <input type="hidden" name="cmd" value="_xclick"> <input type="hidden" name="business" value="accounts@agraphics.ie"> <input type="hidden" name="lc" value="IE"> <input type="hidden" name="item_name" value="10 Christmas Cards"> <input type="hidden" name="item_number" value="<? echo $orderid; ?>"> <input type="hidden" name="amount" value="12.99"> <input type="hidden" name="currency_code" value="EUR"> <input type="hidden" name="button_subtype" value="services"> <input type="hidden" name="shipping" value="2.99"> <input type="hidden" name="return" value="http://www.digitalscribe/thanks.php"> <input type="hidden" name="bn" value="PP-BuyNowBF:btn_buynowCC_LG.gif:NonHosted"> <input type="image" src="https://www.paypalobjects.com/en_US/i/btn/btn_buynowCC_LG.gif" border="0" name="submit" alt="PayPal - The safer, easier way to pay online!"> <img alt="" border="0" src="https://www.paypalobjects.com/en_US/i/scr/pixel.gif" width="1" height="1"> </form> </div>"; if ($row2['qty'] == 20) echo "<div> <form action="https://www.paypal.com/cgi-bin/webscr" method="post"> <input type="hidden" name="cmd" value="_xclick"> <input type="hidden" name="business" value="accounts@agraphics.ie"> <input type="hidden" name="lc" value="IE"> <input type="hidden" name="item_name" value="20 Christmas Cards"> <input type="hidden" name="item_number" value="<? echo $orderid; ?>"> <input type="hidden" name="amount" value="21.99"> <input type="hidden" name="currency_code" value="EUR"> <input type="hidden" name="button_subtype" value="services"> <input type="hidden" name="shipping" value="2.99"> <input type="hidden" name="return" value="http://www.digitalscribe/thanks.php"> <input type="hidden" name="bn" value="PP-BuyNowBF:btn_buynowCC_LG.gif:NonHosted"> <input type="image" src="https://www.paypalobjects.com/en_US/i/btn/btn_buynowCC_LG.gif" border="0" name="submit" alt="PayPal - The safer, easier way to pay online!"> <img alt="" border="0" src="https://www.paypalobjects.com/en_US/i/scr/pixel.gif" width="1" height="1"> </form> </div>"; ?> </div> <div id="footer" class="clear"><div id="sign"><div id="sign_text">Personalised<br /> Books</div> </div></div> </div></div> </body> </html> <?php mysql_free_result($cardpayment); ?> Hi, I would like to do the following but not sure how. If the you/user is on index.php of http://www.domain.com/ show one page If not show another How would I do this? Thanks Hello, I've never really used the update command before for mysql and I'm attempting to use it and struggling a little bit. I'm trying to use mysqli prepared statements.. here's the code that I have thus far: if($query = $database->connection->prepare("UPDATE videos SET comments=?, views=?, uploader=? WHERE title = ?")) { $query->bind_param('iiss', $comments, $views, $uploader, $title); $query->execute(); $result = $query->affected_rows; $query->close(); } For some reason I cannot get this working. I have created a modification page for the administrators to be able change any of the values and wanting to update the database to reflect the changes. When using the MySQL UPDATE command do all of the values have to get changed or modified, or am I able to pass back some of the same values? Like with the above code.. if I only wanted to update the views, would I still be able to just pass in the same values for comments and uploader and it would just replace the values? Hi all,
I am not sure why my header is not displaying the header image after using the CSS
I have a png file that repeats horizotally.
Please help
<!DOCTYPE html> <html> <head> <meta charset="UTF-8"> <title>Wikigets The online store</title> <style type="text/css"> body { margin:0 px;} #pageTop { background: url(style/headerline1.png); height:110 px; } </style> </head> <body> <div id="pageTop"> </div> <div id="pageMiddle"></div> <div id="pageBottom"></div> </body> </html> headerline1.png 2.82KB 0 downloads headerline1.png 2.82KB 0 downloads Code: [Select] <?php function checking_out() { $conn = db_connect(); $nickname=$_SESSION['valid_user']; $query="select sum(price) from preorders where name='".$nickname."'"; $result = $conn->query($query); if ($result) { echo '<h1>'.$result.'</h1>'; } } ?> This is not working, there is no result in the browser, any idea ? I get the unable to jump to row zero mysql error. Code: [Select] function is_admin($uid, $cid) { $uid = (int)$uid; $cid = (int)$cid; $sql = "SELECT `users`.`id` AS `uid`, `companies`.`companyid` AS `cid`, `companies`.`adminid` AS `aid` FROM `companies` LEFT JOIN `users` ON `users`.`id` = companies.adminid WHERE `users`.`id` = {$uid} AND `companies`.`companyid` = {$cid}"; $user = mysql_query($sql); return (mysql_result($user, 0) == '1') ? true : false; } Hello....I am using ajax,jquery with php to insert data in db.....but the script is not working. form.html: Code: [Select] <!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> <style type="text/css"> label{ display:block; } </style> <script type="text/javascript" src="jquery-1.5.min.js"> </script> <script type="text/javascript"> $(function() { $('#submit').click(function(){ $('#container').append('<img src="ajax-loader.gif" id="loading" alt="image" />'); var name=$('#name').val(); var email=$('#email').val(); var d=$('#d').val(); var m=$('#m').val(); var y=$('#y').val(); var add=$('#add').val(); var phone=$('#phone').val(); $.ajax({ url: 'process.php', type: 'POST', data: 'name=' + name + '&email=' + email + '&d=' + d + '&m=' + m + '&y=' + y + '&add=' + add + '&phone=' + phone, success: function(result){ $('#response').remove(); $('#container').append('<p id="response">' + result + '</p>'); $('#loading').fadeOut(500,function(){ $(this).remove(); }); } }); return false; }); }); </script> </head> <body> <h2>User Registeration</h2> <form action="process.php" method="POST"> <div id="container"> Name:<br> <input type="text" name="name" id="name" /><br> Email:<br> <input type="text" name="email" id="email" /><br> Date Of Birth:<br> <input type="text" id="d" name="d" size="2" /> <input type="text" id="m" name="m" size="2" /> <input type="text" id="y" name="y" size="4" /><br> Address:<br> <input type="text" id="add" name="add" /><br> Phone:<br> <input type="text" id="phone" name="phone" /><br> <input type="submit" name="submit" id="submit" value="GO!" /> </div> </form> </body> </html> process.php My db id named "jquery" and table is "tab" Code: [Select] <?php $conns=mysql_connect("localhost","root",""); if(!$conns) echo "error in connection"; mysql_select_db("jquery", $conns); $name=$_POST['name']; $email=$_POST['email']; $m=$_POST['m']; $d=$_POST['d']; $y=$_POST['y']; $add=$_POST['add']; $phone=$_POST['phone']; echo $name,"<br>"; echo $email,"<br>"; echo $m, "<br>"; echo $d,"<br>"; echo $y,"<br>"; echo $add, "<br>"; echo $phone,"<br>"; $query="INSERT INTO tab (name,email,d,m,y,add,phone) VALUES ('$name',$email','$d','$m','$y','$add','$phone')"; if(!mysql_query($query,$conns)) echo "Error"; else echo "DATA inserted"; ?> The output shown is: pulkit pulkit@gmail.com 1 26 1991 lucknow 987576787 Error For some values i entered. Hi I have got a database table for logging in. One user will not log in even though the data is valid Code: [Select] $user = ValidateKey($_SESSION["Name"] , $_SESSION["PWD1"] , $_SESSION["PWD2"]); // User if($Error == 1 || $user == 0 || $user == -1) { // header('Location: index.php'); echo $Error . " , " . $user; // result is 0, caf9eba77c55ab5ae81a01c25d1987d3 exit; } All other user are OK! Strange Desmond. |
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