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  • PHP : Une vulnérabilité importante a été découverte dans le CMS open source [e107]

    Les équipes de PHP Sécurity ont révélé ce Mercredi dernier une vulnérabilité importante dans le gestionnaire de contenu open source e107. Cette vulnérabilité permettrait d'exécuter du code PHP à partir du BBCODE provenant des formulaires. [IMG]http://e107.org/e107_themes/imprint/images/i_logo.png[/IMG] Saisis par l'alerte l'équipe de développement de e107 n'ont pas tardé de sortir un patch qui n'est pas encore officiel. -> Annonce de la vulnérabilité -> Annonce de l'info sur le site officiel de e107 ->

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  • Missing Password check

    - by AAA
    I am using the code below, it checks for empty fields and verifies email, but even if the password is correct it won't login. the password has been inserted with md5 protection, below is the code. I am new to this so please bare with me. Thanks! PHP: session_start(); //Checks if there is a login cookie if(isset($_COOKIE['ID_my_site'])) //if there is, it logs you in and directes you to the members page { $email = $_COOKIE['ID_my_site']; $pass = $_COOKIE['Key_my_site']; $check = mysql_query("SELECT * FROM accounts WHERE email = '$email'")or die(mysql_error()); while($info = mysql_fetch_array( $check )) { if ($pass != $info['password']) { } else { header("Location: home.php"); } } } //if the login form is submitted if (isset($_POST['submit'])) { // if form has been submitted // makes sure they filled it in if(!$_POST['email'] | !$_POST['password']) { die('You did not fill in a required field.'); } // checks it against the database if (!get_magic_quotes_gpc()) { $_POST['email'] = addslashes($_POST['email']); } $check = mysql_query("SELECT * FROM accounts WHERE email = '".$_POST['email']."'")or die(mysql_error()); //Gives error if user dosen't exist $check2 = mysql_num_rows($check); if ($check2 == 0) { die('That user does not exist in our database. <a href=add.php>Click Here to Register</a>'); } while($info = mysql_fetch_array( $check )) { $_POST['password'] = stripslashes($_POST['password']); $info['password'] = stripslashes($info['password']); $_POST['password'] = md5($_POST['password']); //gives error if the password is wrong if ($_POST['password'] != $info['password']) { die('Incorrect password, please try again.'); } else { // if login is ok then we add a cookie $_POST['email'] = stripslashes($_POST['email']); $hour = time() + 3600; setcookie(ID_my_site, $_POST['email'], $hour); setcookie(Key_my_site, $_POST['password'], $hour); //then redirect them to the members area header("Location: home.php"); } } } else { // if they are not logged in <form action="<?php echo $_SERVER['PHP_SELF']?>" method="post"> <table border="0"> <tr><td colspan=2><h1>Login</h1></td></tr> <tr><td>email:</td><td> <input type="text" name="email" maxlength="40"> </td></tr> <tr><td>Password:</td><td> <input type="password" name="password" maxlength="50"> </td></tr> <tr><td colspan="2" align="right"> <input type="submit" name="submit" value="Login"> </td></tr> </table> </form> } Here is the registration code: PHP: // here we encrypt the password and add slashes if needed $_POST['password'] = md5($_POST['password']); if (!get_magic_quotes_gpc()) { $_POST['password'] = mysql_escape_string($_POST['password']); $_POST['email'] = mysql_escape_string($_POST['email']); $_POST['full_name'] = mysql_escape_string($_POST['full_name']); $_POST['user_url'] = mysql_escape_string($_POST['user_url']); } // now we insert it into the database $insert = "INSERT INTO accounts (Uniquer, Full_name, Email, Password, User_url) VALUES ('".$uniquer."','".$_POST['full_name']."', '".$_POST['email']."','".$_POST['password']."', '".$_POST['user_url']."')"; $add_member = mysql_query($insert); After using ini_set function i got to see the error, i am getting this message but not sure what it means: Notice: Undefined index: password in /var/www/domain.com/htdocs/login.php on line 103 Notice: Use of undefined constant password - assumed 'password' in /var/www/domain.com/htdocs/login.php on line 11

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  • Wordpress and Jquery slide

    - by kwek-kwek
    I am integrating a Jquery slider inside of wordpress here is the demo of the slider. I can see the div that is their but for some reason it is not showing up. View the working site here Now my problem is that this code: <script type="text/javascript"> var _siteRoot='index.php',_root='index.php';</script> <script type="text/javascript" src="<?php bloginfo('template_url'); ?>/js/jquery.js"></script> <script type="text/javascript" src="<?php bloginfo('template_url'); ?>/js/scripts.js"></script> represents and index.html, but in wordpress I enabled permalinks. Any clue what would be the _siteRoot is? here is the complete code HEADER <script type="text/javascript"> var _siteRoot='index.php',_root='index.php';</script> <script type="text/javascript" src="<?php bloginfo('template_url'); ?>/js/jquery.js"></script> <script type="text/javascript" src="<?php bloginfo('template_url'); ?>/js/scripts.js"></script> Here are the images: <div id="slide-holder"> <div id="slide-runner"> <a href=""><img id="slide-img-1" src="images/nature-photo.png" class="slide" alt="" /></a> <a href=""><img id="slide-img-2" src="images/nature-photo1.png" class="slide" alt="" /></a> <a href=""><img id="slide-img-3" src="images/nature-photo2.png" class="slide" alt="" /></a> <a href=""><img id="slide-img-4" src="images/nature-photo3.png" class="slide" alt="" /></a> <a href=""><img id="slide-img-5" src="images/nature-photo4.png" class="slide" alt="" /></a> <a href=""><img id="slide-img-6" src="images/nature-photo4.png" class="slide" alt="" /></a> <a href=""><img id="slide-img-7" src="images/nature-photo6.png" class="slide" alt="" /></a> <div id="slide-controls"> <p id="slide-client" class="text"><strong>post: </strong><span></span></p> <p id="slide-desc" class="text"></p> <p id="slide-nav"></p> </div> </div> <!--content featured gallery here --> </div> And the footer <script type="text/javascript"> if(!window.slider) var slider={};slider.data=[{"id":"slide-img-1"},{"id":"slide-img-2"},{"id":"slide-img-3"},{"id":"slide-img-4"},{"id":"slide-img-5"},{"id":"slide-img-6"},{"id":"slide-img-7"},{"id":"slide-img-8"}]; </script>

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  • Why does my Ajax function returns my entire code?

    - by JDelage
    I'm playing with sample code from the book "Head first Ajax". Here are the salient pieces of code: Index.php - html piece: <body> <div id="wrapper"> <div id="thumbnailPane"> <img src="images/itemGuitar.jpg" width="301" height="105" alt="guitar" title="itemGuitar" id="itemGuitar" onclick="getDetails(this)"/> <img src="images/itemShades.jpg" alt="sunglasses" width="301" height="88" title="itemShades" id="itemShades" onclick="getDetails(this)" /> <img src="images/itemCowbell.jpg" alt="cowbell" width="301" height="126" title="itemCowbell" id="itemCowbell" onclick="getDetails(this)" /> <img src="images/itemHat.jpg" alt="hat" width="300" height="152" title="itemHat" id="itemHat" onclick="getDetails(this)" /> </div> <div id="detailsPane"> <img src="images/blank-detail.jpg" width="346" height="153" id="itemDetail" /> <div id="description"></div> </div> </div> </body> Index.php - script: function getDetails(img){ var title = img.title; request = createRequest(); if (request == null) { alert("Unable to create request"); return; } var url= "getDetails.php?ImageID=" + escape(title); request.open("GET", url, true); request.onreadystatechange = displayDetails; request.send(null); } function displayDetails() { if (request.readyState == 4) { if (request.status == 200) { detailDiv = document.getElementById("description"); detailDiv.innerHTML = request.responseText; }else{ return; } }else{ return; } request.send(null); } And Index.php: <?php $details = array ( 'itemGuitar' => "<p>Pete Townshend once played this guitar while his own axe was in the shop having bits of drumkit removed from it.</p>", 'itemShades' => "<p>Yoko Ono's sunglasses. While perhaps not valued much by Beatles fans, this pair is rumored to have been licked by John Lennon.</p>", 'itemCowbell' => "<p>Remember the famous \"more cowbell\" skit from Saturday Night Live? Well, this is the actual cowbell.</p>", 'itemHat' => "<p>Michael Jackson's hat, as worn in the \"Billie Jean\" video. Not really rock memorabilia, but it smells better than Slash's tophat.</p>" ); if (isset($_REQUEST['ImageID'])){echo $details[$_REQUEST['ImageID']];} ?> All this code does is that when someone clicks on a thumbnail, a corresponding text description appears on the page. Here is my question. I have tried to bring the getDetails.php code inside Index.php, and modify the getDetails function so that the var url be "Index.php?ImageID="... . When I do that, I get the following problem: the function does not display the snippet of text in the array, as it should. Instead it reproduces the entire code - the webpage, etc - and then at the bottom the expected snippet of text. Why is that?

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  • Adding to database with multiple text boxes

    - by kira423
    What I am trying to do with this script is allow users to update a url for their websites, and since each user isn't going to have the same amount of websites is is hard for me to just add $_POST['website'] for each of these. Here is the script <?php include("config.php"); include("header.php"); include("functions.php"); if(!isset($_SESSION['username']) && !isset($_SESSION['password'])){ header("Location: pubs.php"); } $getmember = mysql_query("SELECT * FROM `publishers` WHERE username = '".$_SESSION['username']."'"); $info = mysql_fetch_array($getmember); $getsites = mysql_query("SELECT * FROM `websites` WHERE publisher = '".$info['username']."'"); $postback = $_POST['website']; $webname = $_POST['webid']; if($_POST['submit']){ var_dump( $_POST['website'] ); $update = mysql_query("UPDATE `websites` SET `postback` = '$postback' WHERE name = '$webname'"); } print" <div id='center'> <span id='tools_lander'><a href='export.php'>Export Campaigns</a></span> <div id='calendar_holder'> <h3>Please define a postback for each of your websites below. The following variables should be used when creating your postback.<br /> cid = Campaign ID<br /> sid = Sub ID<br /> rate = Campaign Rate<br /> status = Status of Lead. 1 means payable 2 mean reversed<br /> A sample postback URL would be <br /> http://www.example.com/postback.php?cid=#cid&sid=#sid&rate=#rate&status=#status</h3> <table class='balances' align='center'> <form method='POST' action=''>"; while($website = mysql_fetch_array($getsites)){ print" <tr> <input type ='hidden' name='webid' value='".$website['id']."' /> <td style='font-weight:bold;'>".$website['name']."'s Postback:</td> <td><input type='text' style='width:400px;' name='website[]' value='".$website['postback']."' /></td> </tr>"; } print" <td style='float:right;position:relative;left:150px;'><input type='submit' name='submit' style='font-size:15px;height:30px;width:100px;' value='Submit' /></td> </form> </table> </div>"; include("footer.php"); ?> What I am attempting to do insert the what is inputted in the text boxes to their corresponding websites, and I cannot think of any other way to do it, and this obviously does not works and returns a notice stating Array to string conversion If there is a more logical way to do this please let me know.

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  • Html LogIn form not functioning

    - by Tony C
    Ok, I have a login form that looks like this: <form id="loginForm" name="loginForm" method="post" action="login-exec.php"> <table width="300" border="0" align="center" cellpadding="2" cellspacing="0"> <tr> <td width="112"><b>Login</b></td> <td width="188"><input name="login" type="text" class="textfield" id="login" /></td> </tr> <tr> <td><b>Password</b></td> <td><input name="password" type="password" class="textfield" id="password" /></td> </tr> <tr> <td>&nbsp;</td> <td><input type="submit" name="Submit" value="Login" /></td> </tr> </table> </form> Now, This form is on a page in a directory called members. When i put it on a page in the home directory and change the action to "members/login-exec.php" When I try to logIn it just refreshes the page, but the name of the page in the browser changes to the actions taking place in the form. Any ideas on making this work guys? EDIT, heres the login-exec.php code: <?php //Start session session_start(); //Include database connection details require_once('config.php'); //Array to store validation errors $errmsg_arr = array(); //Validation error flag $errflag = false; //Connect to mysql server $link = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD); if(!$link) { die('Failed to connect to server: ' . mysql_error()); } //Select database $db = mysql_select_db(DB_DATABASE); if(!$db) { die("Unable to select database"); } //Function to sanitize values received from the form. Prevents SQL injection function clean($str) { $str = @trim($str); if(get_magic_quotes_gpc()) { $str = stripslashes($str); } return mysql_real_escape_string($str); } //Sanitize the POST values $login = clean($_POST['login']); $password = clean($_POST['password']); //Input Validations if($login == '') { $errmsg_arr[] = 'Login ID missing'; $errflag = true; } if($password == '') { $errmsg_arr[] = 'Password missing'; $errflag = true; } //If there are input validations, redirect back to the login form if($errflag) { $_SESSION['ERRMSG_ARR'] = $errmsg_arr; session_write_close(); header("location: login-form.php"); exit(); } //Create query $qry="SELECT * FROM members WHERE login='$login' AND passwd='".md5($_POST['password'])."'"; $result=mysql_query($qry); //Check whether the query was successful or not if($result) { if(mysql_num_rows($result) == 1) { //Login Successful 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: members.php"); exit(); }else { //Login failed header("location: login-failed.php"); exit(); } }else { die("Query failed"); } ?>

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  • Form Not Submitting

    - by John
    Hello, When I try to click on the "submit" button for the form below, nothing happens. Any ideas why not? Thanks in advance, John submit.php: <?php require_once "header.php"; $u = $_SESSION['username']; if (!isLoggedIn()) { // user is not logged in. if (isset($_POST['cmdlogin'])) { // retrieve the username and password sent from login form & check the login. if (checkLogin($_POST['username'], $_POST['password'])) { show_userbox2(); } else { echo "Incorrect Login information !"; show_loginform(); } } else { show_loginform(); } } else { . show_userbox2(); } echo '<div class="submittitle">Submit an item.</div>'; echo '<form action="http://www...com/.../submit2.php" method="post"> <input type="hidden" value="'.$_SESSION['loginid'].'" name="uid"> <div class="submissiontitle"><label for="title">Story Title:</label></div> <div class="submissionfield"><input name="title" type="title" id="title" maxlength="1000"></div> <div class="urltitle"><label for="url">Link:</label></div> <div class="urlfield"><input name="url" type="url" id="url" maxlength="500"></div> <div class="submissionbutton"><input name="submit" type="submit" value="Submit"></div> </form> '; ?> submit2.php: <?php //if($_SERVER['REQUEST_METHOD'] == "POST"){header('Location: http://www...com/.../submit2.php');} require_once "header.php"; if (isLoggedIn() == true) { $remove_array = array('http://www.', 'http://', 'https://', 'https://www.', 'www.'); $cleanurl = str_replace($remove_array, "", $_POST['url']); $cleanurl = strtolower($cleanurl); $cleanurl = preg_replace('/\/$/','',$cleanurl); $title = $_POST['title']; //$url = $_POST['url']; $uid = $_POST['uid']; $title = mysql_real_escape_string($title); $cleanurl = mysql_real_escape_string($cleanurl); $site1 = 'http://' . $cleanurl; $displayurl = parse_url($site1, PHP_URL_HOST); function isURL($url1 = NULL) { if($url1==NULL) return false; $protocol = '(http://|https://)'; $allowed = '[-a-z0-9]{1,63}'; $regex = "^". $protocol . // must include the protocol '(' . $allowed . '\.)'. // 1 or several sub domains with a max of 63 chars '[a-z]' . '{2,6}'; // followed by a TLD if(eregi($regex, $url1)==true) return true; else return false; } if(isURL($site1)==true) mysql_query("INSERT INTO submission VALUES (NULL, '$uid', '$title', '$cleanurl', '$displayurl', NULL)"); else echo "<p class=\"topicu\">Not a valid URL.</p>\n"; } else { // user is not loggedin show_loginform(); } if (!isLoggedIn()) { // user is not logged in. if (isset($_POST['cmdlogin'])) { // retrieve the username and password sent from login form & check the login. if (checkLogin($_POST['username'], $_POST['password'])) { show_userbox(); } else { echo "Incorrect Login information !"; show_loginform(); } } else { // User is not logged in and has not pressed the login button // so we show him the loginform show_loginform(); } } else { // The user is already loggedin, so we show the userbox. show_userbox(); } require_once "footer.php"; ?>

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  • Facebook iFrame APP not working in IE, works on every other browser

    - by Sean Ashmore
    So im getting a blank page when loading this page within an iFrame on Internet explorer, every other browser works fine.. I have also tried using p3p headers as other people have suggested, but to no avail. <?php require ("connect.php"); require ("config.php"); require ("fb_config.php"); ?> <!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01//EN" "http://www.w3.org/TR/html4/strict.dtd"> <html> <head> <title>Login handler</title> <meta http-equiv="content-type" content="text/html; charset=iso-8859-1"> <link rel="stylesheet" href="css/login.css" type="text/css"> </head> <body> <?//=$user?> <?php if($user == 0) { echo "You are not logged into facebook. Nice try."; }else{ $query = "SELECT id,fb_id,login_ip,login_count,activated,sitestate FROM login WHERE fb_id='".mysql_real_escape_string($user)."'"; $result = mysql_query($query) or die(mysql_error()); $row = mysql_fetch_array($result); if (mysql_num_rows($result) == 0) { $sql = "INSERT INTO login SET id = '', fb_id ='" .mysql_real_escape_string($user). "', name = '" .rand(10000000000000000,99999999999999999999). "', signup =NOW() , password = '" .mysql_real_escape_string($pass). "', state = '0', mail = '" .mysql_real_escape_string($_POST['mail']). "',location='".mysql_real_escape_string($randomlocation)."',location_start='".mysql_real_escape_string($randomlocation)."', signup_ip='".mysql_real_escape_string($_SERVER['REMOTE_ADDR'])."',ref='".mysql_real_escape_string($_POST['ref'])."', activation_id = '" .mysql_real_escape_string($activation_link). "',activated='2', killprotection = '$twodayprot',gender='" .mysql_real_escape_string($_POST["gender"]). "'"; $res = mysql_query($sql); } //if($row['fb_id'] != $user){ //echo "Your facebook ID: $user is NOT in the MW DB."; //exit(); //}else{ if(empty($row['login_ip'])){ $row['login_ip'] = $_SERVER['REMOTE_ADDR']; }else{ $ip_information = explode("-", $row['login_ip']); if (in_array($_SERVER['REMOTE_ADDR'], $ip_information)) { $row['login_ip'] = $row['login_ip']; }else{ $row['login_ip'] = $row['login_ip']."-".$_SERVER['REMOTE_ADDR']; } } $update_login = mysql_query("UPDATE login SET login_count=login_count+'1' WHERE name='".mysql_real_escape_string($_POST['username'])."'") or die(mysql_error()); $_SESSION['user_id'] = $row['id']; $result = mysql_query("UPDATE login SET userip='".mysql_real_escape_string($_SERVER['REMOTE_ADDR'])."',login_ip='".mysql_real_escape_string($row['login_ip'])."',login_count='0' WHERE id='".mysql_real_escape_string($_SESSION['user_id'])."'") or die(mysql_error()); if ($row['sitestate'] == 0){ header("location: home.php"); } elseif ($row['sitestate'] == 2) { header("location: killed.php?id={$row['id']}&encrypted={$row['password']}"); } else { header("location: banned.php?id={$row['id']}&encrypted={$row['password']}"); } }// id check. ?> </body> </html>

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  • JQuery Checkbox with Textbox Validation

    - by Volrath
    I am using Jorn's validation plugin. I have a a group of checkboxes beside a group of textboxes. The textboxes are disabled by default and will enable when the matching checkbox is checked. At least 1 checkbox has to be checked which is not a problem. However, when I check more than 2 checkboxes only 1 textbox validates. The form still submits even when the second checkbox is empty. $count = 0; while($row = mysql_fetch_array($rs)) { ?> <tr> <td> <label> <input type="checkbox" name="tDays[]" id="tDays<?php echo $count; ?>" value="<?php echo $row['promoDayID'];?>" onClick="enableTxt();" <?php if((isset($arrTDays) && in_array_THours($row['promoDayID'], $arrTDays)) || (!empty($arrSelectedTHours) && in_array_THours($row['promoDayID'], $arrSelectedTHours))) { echo "checked='checked'"; }?> validate="required:true" /> <?php echo $row['promoDay'];?>: </label> </td> <td align="right"> <input type="textbox" size="45" style="font-size:12px" name="tHours[]" id="tHours<?php echo $count; ?>" <?php if(isset($arrTDays) && in_array_THours($row['promoDayID'], $arrTDays)) { echo "value='" .getHours($row['promoDayID'], $arrTDays) ."'"; } elseif (!empty($arrSelectedTHours) && in_array_THours($row['promoDayID'], $arrSelectedTHours)) { echo "value='" .getHours($row['promoDayID'], $arrSelectedTHours). "'"; } else { echo "value='' disabled='disabled'"; }?> class="required" /> <label for="tHours[]" class="error" id="tHourserror<?php echo $count; ?>">Please enter the Trading Hour.</label> </td> </tr> <?php $count++; }//while ?> This is done using javascript: function enableTxt() { for (i = 0; i <= 7; i++) { if (document.getElementById("tDays" + i) != null && document.getElementById("tDays" + i).checked == true) { document.getElementById('tHours' + i).disabled = false; document.getElementById('tHourserror' + i).style.visibility = "visible"; } else if (document.getElementById("tDays" + i) != null) { document.getElementById('tHours' + i).disabled = "disabled"; document.getElementById('tHours' + i).value = ""; document.getElementById('tHourserror' + i).style.visibility = "hidden"; } } } Please kindly advise in detail as to how this problem can be solved. I am fairly weak in JQuery.

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  • I never really understood: what is CGI?

    - by claws
    CGI is a Comman Gateway Interface. As the name says, it is a "common" gateway interface for everything. It is so trivial and naive from the name. I feel that I understood this and I felt this every time I encountered this word. But frankly, I didn't. I'm still confused. I am a PHP programmer. I did lot of web development. user (client) request for page --- webserver(-embedded PHP interpreter) ---- Server side(PHP) Script --- MySQL Server. Now say my PHP Script can fetch results from MySQL Server && MATLAB Server && Some other server. So, now PHP Script is the CGI? because its interface for the between webserver & All other servers? I don't know. Sometimes they call CGI, a technology & othertimes they call CGI a program or someother server. What exactly is CGI? Whats the big deal with /cgi-bin/*.cgi? Whats up with this? I don't know what is this cgi-bin directory on the server for. I don't know why they have *.cgi extensions. Why does Perl always comes in the way. CGI & Perl (language). I also don't know whats up with these two. Almost all the time I keep hearing these two in combination "CGI & Perl". This book is another great example CGI Programming with Perl Why not "CGI Programming with PHP/JSP/ASP". I never saw such things. CGI Programming in C this confuses me a lot. in C?? Seriously?? I don't know what to say. I"m just confused. "in C"?? This changes everything. Program needs to be compiled and executed. This entirely changes my view of web programming. When do I compile? How does the program gets executed (because it will be a machine code, so it must execute as a independent process). How does it communicate with the web server? IPC? and interfacing with all the servers (in my example MATLAB & MySQL) using socket programming? I'm lost!! They say that CGI is depreciated. Its no more in use. Is it so? What is its latest update? Once, I ran into a situation where I had to give HTTP PUT request access to web server (Apache HTTPD). Its a long back. So, as far as I remember this is what I did: Edited the configuration file of Apache HTTPD to tell webserver to pass all HTTP PUT requests to some put.php ( I had to write this PHP script) Implement put.php to handle the request (save the file to the location mentioned) People said that I wrote a CGI Script. Seriously, I didn't have clue what they were talking about. Did I really write CGI Script? I hope you understood what my confusion is. (Because I myself don't know where I'm confused). I request you guys to keep your answer as simple as possible. I really can't understand any fancy technical terminology. At least not in this case. EDIT: I found this amazing tutorial "CGI Programming Is Simple!" - CGI Tutorial Which explains the concepts in simplest possible way. I've only have one complaint about this tutorial. Just to make what ever he explained complete he should have shown the C code he used for generating response for those GET / POST requests. I've also added link to this tutorial to Wikipedia's article : http://en.wikipedia.org/wiki/Common_Gateway_Interface

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  • Session caching problem

    - by Levani
    I have a strange problem with php sessions. I use them for authorization on my site. I store two variables - currently logged in user's id and username in session. When I log in with one username, than log out and log in again with another username the previous user's id is returned using the session variable instead of the current user. The most strange thing is that this happens only when it comes to insert some data into database. When I directly echo this variable the correct id is displayed, but when I insert new record into database this variable sends incorrect id. Here is the php code I use for sending data into database: <?php session_start(); //connect database require_once 'dbc.php'; $authorID = $_SESSION['user_id']; if ( $authorID != 0 ) { $content = htmlentities($_POST["answ_content"],ENT_COMPAT,'UTF-8'); $dro = date('Y-m-d H:i:s'); $qID = $_POST["question_ID"]; $author = 'avtori'; $sql="INSERT INTO comments (comment_ID, comment_post_ID, comment_author, comment_date, comment_content, user_id) VALUES (NULL, '$qID', '$author', '$dro', '$content', '$authorID')"; $result = mysql_query($sql); } else { echo 'error'; } ?> Can anyone please help? Here is the logout function: function logout() { global $db; session_start(); if(isset($_SESSION['user_id']) || isset($_COOKIE['user_id'])) { mysql_query("update `users` set `ckey`= '', `ctime`= '' where `id`='$_SESSION[user_id]' OR `id` = '$_COOKIE[user_id]'") or die(mysql_error()); } /************ Delete the sessions****************/ unset($_SESSION['user_id']); unset($_SESSION['user_name']); unset($_SESSION['user_level']); unset($_SESSION['HTTP_USER_AGENT']); session_unset(); session_destroy(); /* Delete the cookies*******************/ setcookie("user_id", '', time()-60*60*24*COOKIE_TIME_OUT, "/"); setcookie("user_name", '', time()-60*60*24*COOKIE_TIME_OUT, "/"); setcookie("user_key", '', time()-60*60*24*COOKIE_TIME_OUT, "/"); header("Location: index.php"); } Here is the authentication script: include 'dbc.php'; $err = array(); foreach($_GET as $key => $value) { $get[$key] = filter($value); //get variables are filtered. } if ($_POST['doLogin']=='Login') { foreach($_POST as $key => $value) { $data[$key] = filter($value); // post variables are filtered } $user_email = $data['usr_email']; $pass = $data['pwd']; if (strpos($user_email,'@') === false) { $user_cond = "user_name='$user_email'"; } else { $user_cond = "user_email='$user_email'"; } $result = mysql_query("SELECT `id`,`pwd`,`full_name`,`approved`,`user_level` FROM users WHERE $user_cond AND `banned` = '0' ") or die (mysql_error()); $num = mysql_num_rows($result); // Match row found with more than 1 results - the user is authenticated. if ( $num > 0 ) { list($id,$pwd,$full_name,$approved,$user_level) = mysql_fetch_row($result); if(!$approved) { //$msg = urlencode("Account not activated. Please check your email for activation code"); $err[] = "Account not activated. Please check your email for activation code"; //header("Location: login.php?msg=$msg"); //exit(); } //check against salt if ($pwd === PwdHash($pass,substr($pwd,0,9))) { // this sets session and logs user in session_start(); session_regenerate_id (true); //prevent against session fixation attacks. // this sets variables in the session $_SESSION['user_id']= $id; $_SESSION['user_name'] = $full_name; $_SESSION['user_level'] = $user_level; $_SESSION['HTTP_USER_AGENT'] = md5($_SERVER['HTTP_USER_AGENT']); //update the timestamp and key for cookie $stamp = time(); $ckey = GenKey(); mysql_query("update users set `ctime`='$stamp', `ckey` = '$ckey' where id='$id'") or die(mysql_error()); //set a cookie if(isset($_POST['remember'])){ setcookie("user_id", $_SESSION['user_id'], time()+60*60*24*COOKIE_TIME_OUT, "/"); setcookie("user_key", sha1($ckey), time()+60*60*24*COOKIE_TIME_OUT, "/"); setcookie("user_name",$_SESSION['user_name'], time()+60*60*24*COOKIE_TIME_OUT, "/"); } if(empty($err)){ header("Location: myaccount.php"); } } else { //$msg = urlencode("Invalid Login. Please try again with correct user email and password. "); $err[] = "Invalid Login. Please try again with correct user email and password."; //header("Location: login.php?msg=$msg"); } } else { $err[] = "Error - Invalid login. No such user exists"; } }

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  • Help! I got a runaway PHP script. My server is down.

    - by gAMBOOKa
    I got a PHP script that is looping and will continue to do so for about another hour. How do I stop it. The script explicitly overrides the time out and the memory buffer. It's on a shared hosting server with cPanel installed. The entire website is down until the script completes. I had added a usleep(100000) statement, but it doesn't appear to work.

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  • vhost.conf file in PLESK not working as intended

    - by Saif Bechan
    I have configured a vhost file for my domain but it does not seem to work. These are the steps I took, please correct me if I am wrong. First I made a file called vhost.conf in: /var/www/vhosts/*domain*/conf/vhost.conf The content of the vhost file looks like this: <Directory /var/www/vhosts/*domain*/httpdocs> php_admin_flag engine on php_admin_flag display_errors on </Directory> Now in my /etc/php.ini i set display_errors=Off After everything i rebuild with: /usr/local/psa/admin/sbin/websrvmng -a But I don't see the any errors in my page. When i turn on the display_errors in /etc/php.ini only then can I see the errors. I know for a fact that the vhost file is read, because when i type nonsense values i get an error when restarting apache saying there are errors in the vhost file. Anyone know what the problem can be. Should there be special settings in either the php.ini file or the httpd.conf file. The httpd.conf i edit is in /etc/httpd/conf/httpd.conf. Is this the file that PLESK uses or is there another, because the values i see there do not really reflect the http folders of my domain. The httpd file looks like this now. # The document root DocumentRoot "/var/www/html" # i guess this is the base directory <Directory /> Order Deny,Allow Deny from all Options None AllowOverride None </Directory> # And i guess here are all my domains located, but there aren't any here <Directory "/var/www/html"> Options None AllowOverride None Order allow,deny Allow from all </Directory> Only this directory /var/www/html is not used by me, I use the directory /var/www/vhosts. The only folder found in /var/www/html is a folder called awstats. Does plesk use other files, and where are they located. I hope this all makes sense to anyone, and i hope i can find a solution

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  • data is not posted in $_POST variable using AJAX [migrated]

    - by Oliver
    Im having a problem in one of my script. Server is running in php, and im using AJAX to post data. Here is my script. PHP script: 0){ echo "Search Result :"; for ($x=0;$xProject Name:   ".mysql_result($result,$x,"projname").""; echo "APMS ID:   ".mysql_result($result,$x,"apmsid").""; echo "Prefix/es:   ".mysql_result($result,$x,"projprefix").""; echo "Usage Type:   ".mysql_result($result,$x,"usagetype").""; echo "Rate:   ".mysql_result($result,$x,"projrate").""; echo "Offer Details:   ".mysql_result($result,$x,"offerdetails").""; } }else{ echo "No results found ..."; } }else{ echo "Problems encountered while processing the data ..."; } ? JS Script: function QueryPrefix() { var xmlhttp; var pStr = document.getElementById('Editbox2'); var htmlHolder = document.getElementById('Html1'); var butStr = document.getElementById('Button1'); if (pStr.value.length == 0){ alert("Please enter a value on the box provided!"); return; } pStr.value=""; if (window.XMLHttpRequest) {// code for IE7+, Firefox, Chrome, Opera, Safari xmlhttp=new XMLHttpRequest(); } else {// code for IE6, IE5 xmlhttp=new ActiveXObject("Microsoft.XMLHTTP"); } xmlhttp.onreadystatechange=function() { if (xmlhttp.readyState==4) { htmlHolder.innerHTML=xmlhttp.responseText; butStr.disabled=false; } } butStr.disabled=true; xmlhttp.open("POST","searchutype.php",false); xmlhttp.setRequestHeader("Content-type","application/x-www-form-urlencoded"); xmlhttp.send("pStr=" + pStr.value); }

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  • MemCache-repcached compile error

    - by Ramy Allam
    I'm trying to install [memcached-1.2.8-repcached-2.2.1]( http://sourceforge.net/projects/repcached/files/latest/download?source=files) And I have the following error after running the make command: make all-recursive make[1]: Entering directory `/usr/local/src/pro/memcached-1.2.8-repcached-2.2.1' Making all in doc make[2]: Entering directory `/usr/local/src/pro/memcached-1.2.8-repcached-2.2.1/doc' make[2]: Nothing to be done for `all'. make[2]: Leaving directory `/usr/local/src/pro/memcached-1.2.8-repcached-2.2.1/doc' make[2]: Entering directory `/usr/local/src/pro/memcached-1.2.8-repcached-2.2.1' gcc -DHAVE_CONFIG_H -I. -DNDEBUG -g -O2 -MT memcached-memcached.o -MD -MP -MF .d eps/memcached-memcached.Tpo -c -o memcached-memcached.o test -f 'memcached.c' || echo './'memcached.c memcached.c: In function ‘add_iov’: memcached.c:697: error: ‘IOV_MAX’ undeclared (first use in this function) memcached.c:697: error: (Each undeclared identifier is reported only once memcached.c:697: error: for each function it appears in.) make[2]: * [memcached-memcached.o] Error 1 make[2]: Leaving directory `/usr/local/src/pro/memcached-1.2.8-repcached-2.2.1' make[1]: * [all-recursive] Error 1 make[1]: Leaving directory `/usr/local/src/pro/memcached-1.2.8-repcached-2.2.1' make: * [all] Error 2 OS : Centos5.7 64bit gcc-4.1.2-51.el5 gcc-c++-4.1.2-51.el5 libgcc-4.1.2-51.el5 Note : Memcached and memcache extension for php are already installed root@server[~]# memcached -h memcached 1.4.5 php ext http://pecl.php.net/get/memcache-2.2.6.tgz

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  • POST Fail via AJAX Request?

    - by Jascha
    I can't for the life of me figure out why this is happening. This is kind of a repost (submitted to stackoverflow, but maybe a server issue?). I am running a javascript log out function called logOut() that has make a jQuery ajax call to a php script... function logOut(){ var data = new Object; data.log_out = true; $.ajax({ type: 'POST', url: 'http://www.mydomain.com/functions.php', data: data, success: function() { alert('done'); } }); } the php function it calls is here: if(isset($_POST['log_out'])){ $query = "INSERT INTO `token_manager` (`ip_address`) VALUES('logOutSuccess')"; $connection->runQuery($query); // <-- my own database class... // omitted code that clears session etc... die(); } Now, 18 hours out of the day this works, but for some reason, every once in a while, the POST data will not trigger my query. (this will last about an hour or so). I figured out the post data is not being set by adding this at the end of my script... $query = "INSERT INTO `token_manager` (`ip_address`) VALUES('POST FAIL')"; $connection->runQuery($query); So, now I know for certain my log out function is being skipped because in my database is the following data: if it were NOT being skipped, my data would show up like this: I know it is being skipped for two reasons, one the die() at the end of my first function, and two, if it were a success a "logOutSuccess" would be registered in the table. Any thoughts? One friend says it's a janky hosting company (hostgator.com). I personally like them because they are cheap and I'm a fan of cpanel. But, if that's the case??? Thanks in advance. -J

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