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  • Creating webpage on form submit?

    - by Joachim Mcdonald
    How is it possible to allow a user to create a webpage containing some html, based on their entries in a form? ie. I would want them to be able to input a name and when the button is clicked, a webpage called that name would be created. I imagine that this must be possible in php, but what functions/code would I be using? Thank you!

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  • How to get php form data to pdf

    - by Fero
    Hi I am displaying all the users in the form using php where the data are fetched from db. When i click on the icon all users data should be show in a pdf. How should this can be done. kindly advice. thanks in advance

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  • remove params from form request

    - by joe
    Hello folks, Upon user interaction, I need to remove certain input params from an HTML form before submission. Using javascript to remove the input fields from the DOM doesn't seem to actually remove the params from being sent through the request. Is there a way to delete or clear the actual request params?

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  • Copy information only, from an external webpage and import it into a form using PHP

    - by mpw5013
    I'm looking to extract key pieces of information from the following website to simplify the signup process. More or less I want to build a quick import script that will grab the following: 1) Username 2) Type (Model / Photographer) 3) Age 4) Sex 5) City, State, Zip 6) Email 7) Photos I will then take all of this data and add it to join form / push the photos through a manipulation script. Any ideas?

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  • HTML form to JSON without modern libraries

    - by Rok
    Does anyone know of a simple javascript library that serializes a form DOM object to JSON? I know jQuery has the serialize method but since I am writing an app for IE mobile, I can't use any of the new fancy libraries, must be oldschool simple javascript, as light as possible. Cheers

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  • checkbox checked with php form post?

    - by Patrick
    how do I check a checkbox? I've tried 1, On, Yes that doesn't work. putting the worked "checked" alone works but then how do I check with php after form post of the checkbox is checked? <input type="checkbox" class="inputcheckbox" id="newmsg" name=chk[newmsg2] value="1" />

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  • Is it possilbe to automatically submit a php form

    - by phpnoob
    I have a website in which I have a several php forms that I would like to fill out with auto generated content (for the purposes of exercising different content the user could submit). I would like to write a client side application that enables me to do so. Is there any way either using webtoolkit, java script etc of doing this? Thank all for your help, PHPNoob

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  • Not allow a href tags in form textarea

    - by saquib
    Hello friends, How can i prevent user to enter any url or link in contact form text area, i have tried it with this but its not working - if (!isset($_POST['submit']) && preg_match_all('/<a.*>.*<\/a>/', $_POST['query'])) { echo "<h1 style='color:red;'>HTML Tag Not allowed </h1>"; } else { //sendmail } Please help me

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  • Get user IP via comment Form

    - by jasmine
    I have inserted a hidden input in my comment form: $ip = $_SERVER['REMOTE_ADDR']; <input type="hidden" name="c-ip" value="<?php echo $ip; ?>"> With this input, ip column is empty in mysql. What is wrong in input. Thanks in advance

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  • PHP: Proper way of using a PDO database connection in a class

    - by Cortopasta
    Trying to organize all my code into classes, and I can't get the database queries to work inside a class. I tested it without the class wrapper, and it worked fine. Inside the class = no dice. What about my classes is messing this up? class ac { public function dbConnect() { global $dbcon; $dbInfo['server'] = "localhost"; $dbInfo['database'] = "sn"; $dbInfo['username'] = "sn"; $dbInfo['password'] = "password"; $con = "mysql:host=" . $dbInfo['server'] . "; dbname=" . $dbInfo['database']; $dbcon = new PDO($con, $dbInfo['username'], $dbInfo['password']); $dbcon->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION); $error = $dbcon->errorInfo(); if($error[0] != "") { print "<p>DATABASE CONNECTION ERROR:</p>"; print_r($error); } } public function authentication() { global $dbcon; $plain_username = $_POST['username']; $md5_password = md5($_POST['password']); $ac = new ac(); if (is_int($ac->check_credentials($plain_username, $md5_password))) { ?> <p>Welcome!</p> <!--go to account manager here--> <?php } else { ?> <p>Not a valid username and/or password. Please try again.</p> <?php unset($_POST['username']); unset($_POST['password']); $ui = new ui(); $ui->start(); } } private function check_credentials($plain_username, $md5_password) { global $dbcon; $userid = $dbcon->prepare('SELECT id FROM users WHERE username = :username AND password = :password LIMIT 1'); $userid->bindParam(':username', $plain_username); $userid->bindParam(':password', $md5_password); $userid->execute(); print_r($dbcon->errorInfo()); $id = $userid->fetch(); Return $id; } } And if it's any help, here's the class that's calling it: require_once("ac/acclass.php"); $ac = new ac(); $ac->dbconnect(); class ui { public function start() { if ((!isset($_POST['username'])) && (!isset($_POST['password']))) { $ui = new ui(); $ui->loginform(); } else { $ac = new ac(); $ac->authentication(); } } private function loginform() { ?> <form id="userlogin" action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post"> User:<input type="text" name="username"/><br/> Password:<input type="password" name="password"/><br/> <input type="submit" value="submit"/> </form> <?php } }

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  • HTML / PHP from

    - by user317128
    I am trying to code an all in one HTML/PHP contact from with error checking. When I load this file in my browser there is not HTML. I am a newb php programmer so most likely forgot something pretty obvious. <!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>All In One Feedback Form</title> </head> <body> <? $form_block = " <input type=\"hidden\" name=\"op\" value=\"ds\"> <form method=\"post\" action=\"$server[PHP_SELF]\"> <p>Your name<br /> <input name=\"sender_name\" type=\"text\" size=30 value=\"$_POST[sender_name]\" /></p> <p>Email<br /> <input name=\"sender_email\" type=\"text\" size=30 value=\"$_POST[sender_email]\"/></p> <p>Message<br /> <textarea name=\"message\" cols=30 rows=5 value=\"$_POST[message]\"></textarea></p> <input name=\"submit\" type=\"submit\" value=\"Send This Form\" /> </form>"; if ($_POST[op] != "ds") { //they see this form echo "$form_block"; } else if ($_POST[op] == "ds") { if ($_POST[sender_name] == "") { $name_err = "Please enter your name<br>"; $send = "no"; } if ($_POST[sender_email] == "ds") { $email_err = "Please enter your email<br>"; $send = "no"; } if ($_POST[message] == "ds") { $message_err = "please enter message<br>"; $send = "no"; } if ($send != "no") { //its ok to send $to = "[email protected]"; $subject = "All in one web site feed back"; $mailheaders = "From: website <some email [email protected]> \n"; $mailheaders .= "Reply-To: $_POST[sender_email]\n"; $msg = "Email sent from this site www.ccccc.com\n"; $msg .= "Senders name: $_POST[senders_name]\n"; $msg .= "Sender's E-Mail: $_POST[sender_email]\n"; $msg .= "Message: $_POST[message]\n\n"; mail($to, $subject, $msg, $mailheaders); echo "<p>Mail has been sent</p>"; } else if ($send == "no") { echo "$name_err"; echo "$email_err"; echo "$message_err"; echo "$form_block"; } } ?> </body> </html> FYI I am trying the example from a book named PHP 6 Fast and Easy Wed Development

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  • Is this PHP/MySQL login script secure?

    - by NightMICU
    Greetings, A site I designed was compromised today, working on damage control at the moment. Two user accounts, including the primary administrator, were accessed without authorization. Please take a look at the log-in script that was in use, any insight on security holes would be appreciated. I am not sure if this was an SQL injection or possibly breach on a computer that had been used to access this area in the past. Thanks <?php //Start session session_start(); //Include DB config require_once('config.php'); //Error message array $errmsg_arr = array(); $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: http://tapp-essexvfd.org/admin/index.php"); exit(); } //Create query $qry="SELECT * FROM user_control WHERE username='$login' AND password='".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(); //Collect details about user and assign session details $member = mysql_fetch_assoc($result); $_SESSION['SESS_MEMBER_ID'] = $member['user_id']; $_SESSION['SESS_USERNAME'] = $member['username']; $_SESSION['SESS_FIRST_NAME'] = $member['name_f']; $_SESSION['SESS_LAST_NAME'] = $member['name_l']; $_SESSION['SESS_STATUS'] = $member['status']; $_SESSION['SESS_LEVEL'] = $member['level']; //Get Last Login $_SESSION['SESS_LAST_LOGIN'] = $member['lastLogin']; //Set Last Login info $qry = "UPDATE user_control SET lastLogin = DATE_ADD(NOW(), INTERVAL 1 HOUR) WHERE user_id = $member[user_id]"; $login = mysql_query($qry) or die(mysql_error()); session_write_close(); if ($member['level'] != "3" || $member['status'] == "Suspended") { header("location: http://members.tapp-essexvfd.org"); //CHANGE!!! } else { header("location: http://tapp-essexvfd.org/admin/admin_main.php"); } exit(); }else { //Login failed header("location: http://tapp-essexvfd.org/admin/index.php"); exit(); } }else { die("Query failed"); } ?>

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  • php slideshow sample

    - by serhio
    I try to do a simple image slideshow in php (just cycle images, no links, no other effects). after some googling I found the following in net: <HTML> <HEAD> <TITLE>Php Slideshow</TITLE> <script language="javascript"> var speed = 4000; // time picture is displayed var delay = 3; // time it takes to blend to the next picture x = new Array; var y = 0; <?php $tel=0; $tst='.jpg'; $p= "./images"; $d = dir($p); $first = NULL; while (false !== ($entry = $d->read())) { if (stristr ($entry, $tst)) { $entry = $d->path."/".$entry; print ("x[$tel]='$entry';\n"); if ($first == NULL) { $first = $entry; } $tel++; } } $d->close(); ?> function show() { document.all.pic.filters.blendTrans.Apply(); document.all.pic.src = x[y++]; document.all.pic.filters.blendTrans.Play(delay); if (y > x.length - 1) y = 0; } function timeF() { setTimeout(show, speed); } </script> </HEAD> <BODY > <!-- add html code here --> <?php print ("<IMG src='$first' id='pic' onload='timeF()' style='filter:blendTrans()' >"); ?> <!-- add html code here --> </BODY> </HTML> but it displays only the first image from the cycle. Do I something wrong? the resulting HTML page is: <HTML> <HEAD> <TITLE>Php Slideshow</TITLE> <script language="javascript"> var speed = 4000; // time picture is displayed var delay = 3; // time it takes to blend to the next picture x = new Array; var y = 0; x[0]='./images/under_construction.jpg'; x[1]='./images/BuildingBanner.jpg'; x[2]='./images/littleLift.jpg'; x[3]='./images/msfp_smbus1_01.jpg'; x[4]='./images/escalator.jpg'; function show() { document.all.pic.filters.blendTrans.Apply(); document.all.pic.src = x[y++]; document.all.pic.filters.blendTrans.Play(delay); if (y > x.length - 1) y = 0; } function timeF() { setTimeout(show, speed); } </script> </HEAD> <BODY > <!-- add html code here --> <IMG src='./images/under_construction.jpg' id='pic' onload='timeF()' style='filter:blendTrans()' ><!-- add html code here --> </BODY> </HTML>

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  • Separation of presentation and business logic in PHP

    - by Markus Ossi
    I am programming my first real PHP website and am wondering how to make my code more readable to myself. The reference book I am using is PHP and MySQL Web Development 4th ed. The aforementioned book gives three approaches to separating logic and content: include files function or class API template system I haven't chosen any of these yet, as wrapping my brains around these concepts is taking some time. However, my code has become some hybrid of the first two as I am just copy-pasting away here and modifying as I go. On presentation side, all of my pages have these common elements: header, top navigation, sidebar navigation, content, right sidebar and footer. The function-based examples in the book suggest that I could have these display functions that handle all the presentation example. So, my page code will be like this: display_header(); display_navigation(); display_content(); display_footer(); However, I don't like this because the examples in the book have these print statements with HTML and PHP mixed up like this: echo "<tr bgcolor=\"".$color."\"><td><a href=\"".$url."\">" ... I would rather like to have HTML with some PHP in the middle, not the other way round. I am thinking of making my pages so that at the beginning of my page, I will fetch all the data from database and put it in arrays. I will also get the data for variables. If there are any errors in any of these processes, I will put them into error strings. Then, at the HTML code, I will loop through these arrays using foreach and display the content. In some cases, there will be some variables that will be shown. If there is an error variable that is set, I will display that at the proper position. (As a side note: The thing I do not understand is that in most example code, if some database query or whatnot gives an error, there is always: else echo 'Error'; This baffles me, because when the example code gives an error, it is sometimes echoed out even before the HTML has started...) For people who have used ASP.NET, I have gotten somewhat used to the code-behind files and lblError and I am trying to do something similar here. The thing I haven't figured out is how could I do this "do logic first, then presentation" thing so that I would not have to replicate for example the navigation logic and navigation presentation in all of the pages. Should I do some include files or could I use functions here but a little bit differently? Are there any good articles where these "styles" of separating presentation and logic are explained a little bit more thoroughly. The book I have only has one paragraph about this stuff. What I am thinking is that I am talking about some concepts or ways of doing PHP programming here, but I just don't know the terms for them yet. I know this isn't a straight forward question, I just need some help in organizing my thoughts.

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  • Converting PHP pagination to jQuery?

    - by ClarkSKent
    Hey, I have been trying to get this pagination class that I am using to be more ajaxy - meaning when I click on the page number like page [2] the data loads, but I want to load in the data without going to a different page (HTTP request in the background, with no page reloads). Being new to both php and jquery, I am a little unsure on how to achieve this result, especially while using a php class. This is what the main page looks like by the way: <?php $categoryId=$_GET['category']; echo $categoryId; ?> <script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.4.1/jquery.min.js"></script> <script type="text/javascript" src="jquery_page.js"></script> <?php //Include the PS_Pagination class include('ps_pagination.php'); //Connect to mysql db $conn = mysql_connect('localhost', 'root', 'root'); mysql_select_db('ajax_demo',$conn); $sql = "select * from explore where category='$categoryId'"; //Create a PS_Pagination object $pager = new PS_Pagination($conn, $sql, 3, 11, 'param1=value1&param2=value2'); //The paginate() function returns a mysql //result set for the current page $rs = $pager->paginate(); //Loop through the result set echo "<table width='800px'>"; while($row = mysql_fetch_assoc($rs)) { echo "<tr>"; echo"<td>"; echo $row['id']; echo"</td>"; echo"<td>"; echo $row['site_description']; echo"</td>"; echo"<td>"; echo $row['site_price']; echo"</td>"; echo "</tr>"; } echo "</table>"; echo "<ul id='pagination'>"; echo "<li>"; //Display the navigation echo $pager->renderFullNav(); echo "</li>"; echo "</ul>"; ?> <div id="loading" ></div> <div id="content" ></div> Would I need to do something with this part of the class?, as seen above: $pager = new PS_Pagination($conn, $sql, 3, 11, 'param1=value1&param2=value2'); Or this?: echo $pager->renderFullNav(); I don't no much about jquery,but i guess I would start it like: $("#pagination li").click(function() { Then load something maybe... I don't no. Any help on this would be great. Thanks.

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  • upload file on database through php code

    - by ruhit
    hi all I have made an application to upload files and its workingout well.now I want to upload my files on database and I also want to display the uploaded files names on my listby accessing the database....so kindly help me. my codes are given below- function uploadFile() { global $template; //$this->UM_index = $this->session->getUserId(); switch($_REQUEST['cmd']){ case 'upload': $filename = array(); //set upload directory //$target_path = "F:" . '/uploaded/'; for($i=0;$i<count($_FILES['ad']['name']);$i++){ if($_FILES["ad"]["name"]) { $filename = $_FILES["ad"]["name"][$i]; $source = $_FILES["ad"]["tmp_name"][$i]; $type = $_FILES["ad"]["type"]; $name = explode(".", $filename); $accepted_types = array('text/html','application/zip', 'application/x-zip-compressed', 'multipart/x-zip', 'application/x-compressed'); foreach($accepted_types as $mime_type) { if($mime_type == $type) { $okay = true; break; } } $continue = strtolower($name[1]) == 'zip' ? true : false; if(!$continue) { $message = "The file you are trying to upload is not a .zip file. Please try again."; } $target_path = "F:" . '/uploaded/'.$filename; // change this to the correct site path if(move_uploaded_file($source, $target_path )) { $zip = new ZipArchive(); $x = $zip->open($target_path); if ($x === true) { $zip->extractTo("F:" . '/uploaded/'); // change this to the correct site path $zip->close(); unlink($target_path); } $message = "Your .zip file was uploaded and unpacked."; } else { $message = "There was a problem with the upload. Please try again."; } } } echo "Your .zip file was uploaded and unpacked."; $template->main_content = $template->fetch(TEMPLATE_DIR . 'donna1.html'); break; default: $template->main_content = $template->fetch(TEMPLATE_DIR . 'donna1.html'); //$this->assign_values('cmd','uploads'); $this->assign_values('cmd','upload'); } } my html page is <html> <link href="css/style.css" rel="stylesheet" type="text/css"> <!--<form action="{$path_site}{$index_file}" method="post" enctype="multipart/form-data">--> <form action="index.php?menu=upload_file&cmd=upload" method="post" enctype="multipart/form-data"> <div id="main"> <div id="login"> <br /> <br /> Ad No 1: <input type="file" name="ad[]" id="ad1" size="10" />&nbsp;&nbsp;Image(.zip)<input type="file" name="ad[]" id="ad1" size="10" /> Sponsor By : <input type="text" name="ad3" id="ad1" size="25" /> <br /> <br /> </div> </div> </form> </html>

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  • PHP & MySQL pagination display problem.

    - by TaG
    I asked a similar question like this yesterday but after waiting for ever I figured out part of the problem but now I'm stuck again I'm trying to display ... when the search results are to long because my pagination links will keep on displaying and will not stop until every link is displayed on the page. For example I'm trying to achieve the following in the example below. Can some one help me fix my code so I can update my site. Thanks This is what I want to be able to do. First Previous 1 2 ... 5 6 7 8 9 10 11 12 13 ... 199 200 Next Last Here is my pagination code that displays the links. $display = 20; if (isset($_GET['p']) && is_numeric($_GET['p'])) { $pages = $_GET['p']; } else { $q = "SELECT COUNT(id) FROM comments WHERE user_id=3"; $r = mysqli_query ($mysqli, $q) or trigger_error("Query: $q\n<br />MySQL Error: " . mysqli_error($mysqli)); $row = mysqli_fetch_array ($r, MYSQLI_NUM); $records = $row[0]; if ($records > $display) { $pages = ceil ($records/$display); } else { $pages = 1; } } if (isset($_GET['s']) && is_numeric($_GET['s'])) { $start = $_GET['s']; } else { $start = 0; } //content goes here if ($pages > 1) { echo '<br /><p>'; $current_page = ($start/$display) + 1; if ($current_page != 1) { echo '<a href="index.php">First</a>'; } if ($current_page != 1) { echo '<a href="index.php?s=' . ($start - $display) . '&p=' . $pages . '">Previous</a> '; } for ($i = 1; $i <= $pages; $i++) { if ($i != $current_page) { echo '<a href="index.php?s=' . (($display * ($i - 1))) . '&p=' . $pages . '">' . $i . '</a> '; } else { echo '<span>' . $i . '</span> '; } } if ($current_page != $pages) { echo '<a href="index.php?s=' . ($start + $display) . '&p=' . $pages . '">Next</a>'; } if ($current_page != $pages) { echo '<a href="index.php?s=' . ($pages - 1) . '&p=' . $pages . '">Last</a>'; } echo '</p>'; }

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