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  • HTML5 development in PHP projects

    - by Tomas Mysik
    Today, we would like to show you how you can in NetBeans 7.4 develop your HTML5 applications directly in your PHP projects. And because everything has already been described on the NetBeans Web Client blog, we will just provide a link to this great blog post: HTML5 development in Java EE and PHP projects. Enjoy it! :) That's all for today, as always, please test it and report all the issues or enhancements you find in NetBeans Bugzilla. Also, please do not forget that all the comments here are moderated.

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  • How to set up an rsync backup to Ubuntu securely?

    - by ws_e_c421
    I have been following various other tutorials and blog posts on setting up a Ubuntu machine as a backup "server" (I'll call it a server, but it's just running Ubuntu desktop) that I push new files to with rsync. Right now, I am able to connect to the server from my laptop using rsync and ssh with an RSA key that I created and no password prompt when my laptop is connected to my home router that the server is also connected to. I would like to be able to send files from my laptop when I am away from home. Some of the tutorials I have looked at had some brief suggestions about security, but they didn't focus on them. What do I need to do to let my laptop with send files to the server without making it too easy for someone else to hack into the server? Here is what I have done so far: Ran ssh-keygen and ssh-copy-id to create a key pair for my laptop and server. Created a script on the server to write its public ip address to a file, encrypt the file, and upload to an ftp server I have access to (I know I could sign up for a free dynamic DNS account for this part, but since I have the ftp account and don't really need to make the ip publicly accessible I thought this might be better). Here are the things I have seen suggested: Port forwarding: I know I need to assign the server a fixed ip address on the router and then tell the router to forward a port or ports to it. Should I just use port 22 or choose a random port and use that? Turn on the firewall (ufw). Will this do anything, or will my router already block everything except the port I want? Run fail2ban. Are all of those things worth doing? Should I do anything else? Could I set up the server to allow connections with the RSA key only (and not with a password), or will fail2ban provide enough protection against malicious connection attempts? Is it possible to limit the kinds of connections the server allows (e.g. only ssh)? I hope this isn't too many questions. I am pretty new to Ubuntu (but use the shell and bash scripts on OSX). I don't need to have the absolute most secure set up. I'd like something that is reasonably secure without being so complicated that it could easily break in a way that would be hard for me to fix.

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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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  • SSL + Jquery + Ajax

    - by chobo2
    Hi I starting too look at a bit of security into my site. My site I would consider a very low security risk as it has really no personal information from the user other than email. However the security risk will go up a bit as I am partnering with a company and the initial password for this companies users will be the same password they use essentially to get onto the network and every piece of software. So I have up my security( what is fine by me...I wanted to get around to this anyways). So one of my security concerns is this. A user logs in. form submit(non ajax is done). Password is hashed & Salted and compared to one in the database. Reject or let them proceed. So this uses no jquery or ajax but is just asp.net mvc and C#. Still if my understanding is right the password is sent in clear text. So if a use SSL and I would not need to worry about that is this correct? If that is true is that all I need? Second the user can change their password at anytime. This is done through ajax. So when the password is sent it is sent in clear text( and I can verify this by looking at firebug). So if I have SSL enabled on this page is that all I need or do I need to do more? So I am just kinda confused of what I need to make the password being sent to the server(both ajax and full post ways secure). I am not sure if I need to do more then SSL or if that is enough and if it is not enough what is the next layer of security?

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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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