Aaron4osu

I have a form that sends a confirmation email to the user who filled out the form. It worked for a while then suddenly stopped working for my email address which I had been using to test the form. It still forms for some email address, just not one with a specific domain. Could their be any kind of spam filter issues for how I have it configured? $mail->SetFrom('emailAddress', 'Name');$mail->AddReplyTo('emailAddress', 'Name'); $mail->AddAddress('emailAddress', 'Name');$mail->Subject = $subject;$mail->AltBody = "To view the message, please use an HTML compatible email viewer!"; // optional, comment out and test $mail->MsgHTML($requester_message);if(!$mail->Send()) { echo "Mailer Error: " . $mail->ErrorInfo;} else { //echo "Message sent!";} // end else

I'm working on a site that allows users to vote on different items. Right now I have it so after they vote the buttons are disabled until the page refreshes. I want to come up with a solution so I can keep users from voting more than once. Currently, I'm storing votes in a mysql database, but I'm not storing who voted. I have 3 different types of users. 1) members logged in 2)members not logged in 3)site visitors with no user account I'm trying to think of the best ways to handle each type. I know for my logged in users I can use their user_id and check if they have voted before for that specific element. But how should I handle the other two (2 & 3). Here is what I'm thinking... a) Should I use set a cookie for each of their votes. Then, loop through all of their vote cookies each time they vote before adding the vote to the database? There could be hundreds of votes. If they have cookie turned off this would not work. b)maybe grab their ipaddress somehow and store that in the database. Then query and loop through all votes looking for a duplicate ipaddress/item combo? c)something more efficient?

I'm trying to build a query that joins 3 tables but I can't seem to figure it out. I want pull all records from the cast table where cast.cast_video_id=1. It should return several users and I want to list their info which is listed in the users table and their position name which is in the position table. here is what I have thus far which returns this error: ERROR: Please separate SQL statements with the Statement Delimiter Preference value - currently ; - when using Execute All SELECT users.first_name, users.last_name, users.city, users.state, cast.cast_pos_id, cast.cast_video_id, positions.pos_name, cast.cast_user_id, users.user_pic_path FROM users INNER JOIN (positions INNER JOIN [cast] ON positions.pos_id = cast.cast_pos_id) ON users.user_id = cast.cast_user_id where cast.cast_video_id=1; here are my mysql tables used in query CREATE TABLE `cast` ( `cast_id` int(11) NOT NULL AUTO_INCREMENT, `cast_user_id` int(11) DEFAULT NULL, `cast_video_id` int(11) DEFAULT NULL, `cast_pos_id` int(11) DEFAULT NULL, `cast_detail` varchar(200) DEFAULT NULL, PRIMARY KEY (`cast_id`), FOREIGN KEY(cast_user_id) REFERENCES users(user_id), FOREIGN KEY(cast_video_id) REFERENCES videos(vid_id), FOREIGN KEY(cast_pos_id) REFERENCES positions(pos_id) ); CREATE TABLE `users` ( `user_id` int(11) NOT NULL AUTO_INCREMENT, `user_name` varchar(20) COLLATE utf8_unicode_ci DEFAULT NULL, `password` varchar(20) COLLATE utf8_unicode_ci DEFAULT NULL, `first_name` varchar(20) COLLATE utf8_unicode_ci NOT NULL, `last_name` varchar(30) COLLATE utf8_unicode_ci NOT NULL, `city` varchar(20) COLLATE utf8_unicode_ci DEFAULT NULL, `state` varchar(20) COLLATE utf8_unicode_ci DEFAULT NULL, `zip` int(5) DEFAULT NULL, `email` varchar(50) COLLATE utf8_unicode_ci NOT NULL, `user_pic_path` varchar(200) COLLATE utf8_unicode_ci DEFAULT NULL, PRIMARY KEY (`user_id`), FOREIGN KEY(user_pos_id) REFERENCES positions(pos_id) ); CREATE TABLE `positions` ( `pos_id` int(11) NOT NULL AUTO_INCREMENT, `pos_name` varchar(30) COLLATE utf8_unicode_ci NOT NULL, `pos_desc` varchar(300) COLLATE utf8_unicode_ci DEFAULT NULL, PRIMARY KEY (`pos_id`) );

Actually I think the initial post fixed it, but for some reason my ftp client wasn't copying over the previous version so I kept running the same script. Very sorry and thanks for the help. This is what I ended up using which is working perfect. <?php session_start(); session_unset(); session_destroy(); header("Location: http://aaronhaas.com/pitchshark6/index.php?vid_id=1"); ?>

Also, I just noticed something else that is weird. I accidentally took out the header function that redirects back to the index page but it is still redirecting there without it. I've tried clear the cache and removing the file and replacing it again and still doing the same. Is that normal for a php script to return back to the previous page?

Thanks guys but I'm still trouble. Ive tried both of these and neither is working for me: I also tried taking out everything but the unset lines. <?php session_start(); session_unset(); session_destroy(); ?> and <?php session_start(); unset($_SESSION); session_destroy(); ?>

I'm trying to use this php to clear out the session variables and return to the index page. However it's not clearing them out. Any ideas? logout.php <?php session_start(); unset($_SESSION['user_id']); unset($_SESSION['username']); session_destroy(); header("Location: http://aaronhaas.com/pitchshark6/index.php?vid_id=1"); ?> then in my navigation I'm using this code to either display their username and a logout link to logout.php or if they are not logged in display a sign in link. <?php // if logged in if (isset($_SESSION['user_id'])) { // display echo "<a href='#'>".$_SESSION['username']."</a> "; echo "<a href='scripts/logout.php'>Log Out</a> "; } // if not logged in else { // display login link echo "<a href='login.php'>Sign In</a>"; } ?> here is my super simple login script $username = $_POST['username']; $password = $_POST['password']; //Check if the username or password boxes were not filled in if(!$username || !$password){ //if not display an error message echo "<center>You need to fill in a <b>Username</b> and a <b>Password</b>!</center>"; }else{ // find user by username and password - working $userQuery = 'SELECT * FROM users WHERE user_name ='.'"'. $username.'" AND password='.'"'. $password.'"' ; $users = mysql_query($userQuery); $user = mysql_fetch_array($users); $_SESSION['user_id'] = $user['user_id']; $_SESSION['username'] = $user['username']; header("Location: http://aaronhaas.com/pitchshark6/index.php?vid_id=1"); }

sorry just saw the post on headers

I was hoping I could get someone to look at my ERD and let me know if it looks ok. Below is an explanation and I attached a pdf of my erd. I'm creating a website where a user can: 1)create a profile 2)then have the ability to create many projects. each project along with the other fields will have drop down menus for: -Genre -Format -Status 3) The project creator will assign positions needed for the project. The project crew table will have a drop down menu (crew positions) with types of crew positions (producer, director, etc...). Until a position is filled it will have an open to apply status. When the position is filled the users name will appear for that position. 4) Each Project Crew Position(in the project crew table) will have many users apply to that position. They will be stored in the audition table. Anyway, I hope this makes sense. If you have any questions I'll be monitoring this closely and get respond quickly. Thanks 17588_.pdf

I'm trying to create a database driven poll to allow users to "like" or "dislike" each video on my site. But instead of radio buttons and a submit button I want to use images with jquery handling the submit when a choice is made. I have found 2 tutorials that each accomplish part of what I want. This tutorial has a poll which is perfect because I can pass in a poll id (which will be the same value as my video id) and it will load that video's poll. http://www.webresourcesdepot.com/ajax-poll-script-with-php-mysql-jquery/ This tutorial replaces the radio buttons with images. http://www.weblee.co.uk/2009/05/30/radio-button-replacement-with-style/ I have them both working separately on this page http://aaronhaas.com/poll2/ I can't seem to figure out how to combine the two together. The poll script generates its html inside of a php function. In the code below I have commented out the line ( getPoll(1) ) that calls the function and replaced it with the html it generates to make it easier to see what is going on. Another problem is each form has a different action so both actions somehow need to be combined together into a function. I was hoping someone might want to see if they can combine the two together and post it as a tutorial or demo. <?php require("inc/db.php"); require("inc/functions.php"); ?> <!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>Ajax Poll Script - Demo</title> <!-- styles and js for poll --> <script src="inc/js/jquery-1.4.2.min.js" type="text/javascript"></script> <script src="inc/js/poll.js" type="text/javascript"></script> <style> body { font-family: Arial, Helvetica, sans-serif; font-size: 1em; } #pollWrap{ width: 550px; margin-left:40px; } #pollWrap h3 { display:none; color:red; font-size: 1em; margin-bottom: 5px; display:none; } #pollWrap ul { margin: 0; padding: 0 0 0 5px; } #pollWrap li { padding: 0; /*overflow:hidden;*/ /*for our lovely friend IE6 to behave nicely*/ font-size: 0.8em; display: inline; } #pollWrap li span { font-size: 0.7em; } .pollChart { margin-left: 25px; height: 10px; width:1px; /*Adding rounded corners to the graphs - Optional - START*/ -moz-border-radius-topright: 4px; -moz-border-radius-bottomright: 4px; -webkit-border-top-right-radius: 4px; -webkit-border-bottom-right-radius: 4px; /*Adding rounded corners to the graphs - Optional - END*/ } #pollSubmit { margin-top: 5px; } #pollMessage { color:#C00; font-size: 0.8em; font-weight: bold; } </style> <!-- styles and js for image radio buttons --> <link rel="stylesheet" type="text/css" href="css/radio.css"> <script type="text/javascript" src="js/jquery-1.3.2.min.js"></script> <script type="text/javascript" src="js/radio-demo.js"></script> </head> <body> <div id="greenlight"> <div id="options"> <ul id="list"> <li class="active"><a href="#" class="option1 active" id="link1"><span>Option</span></a></li> <li><a href="#" class="option2" id="link2"><span>Option</span></a></li> </ul> </div> <!-- close options --> <form action="step2.html" method="post" id="radioform"> Green Light: <input name="option1" type="radio" value="option1" id="option1" checked="checked" /><br /> Cancel: <input name="option1" type="radio" value="option2" id="option2" /><br /> </form> <!-- <p><a href="#" class="toggleform">Show Hidden Form Radion Buttons</a></p> --> <?php // getPoll(1); //$pollID ?> <div id="pollWrap"> <form name="pollForm" method="post" action="inc/functions.php?action=vote"> <h3>Poll Question 1</h3> <ul> <li> <input name="pollAnswerID" id="pollRadioButton1" value="1" type="radio"> Answer1 for Poll1 <span id="pollAnswer1"></span> </li> <li class="pollChart pollChart1"></li> <li> <input name="pollAnswerID" id="pollRadioButton2" value="2" type="radio"> Answer2 for Poll1 <span id="pollAnswer2"></span> </li> <li class="pollChart pollChart2"></li> </ul> <input name="pollSubmit" id="pollSubmit" value="Vote" type="submit"> <span id="pollMessage"></span> <img src="ajaxLoader.gif" alt="Ajax Loader" id="pollAjaxLoader"> </form> </div> </div><!-- close greenlight --> </body> </html>

I'm getting the following error after running a query when I run mysql_fetch_array(): Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in public_html/pitchShark2/Poll/inc/functions.php on line 24 Any help will be greatly Line 24 is: while($row = mysql_fetch_array($result)) // app_config.php defines: DATABASE_HOST, DATABASE_USERNAME, DATABASE_PASSWORD <?php require_once '../scripts/app_config.php'; require_once '../scripts/database_connection.php'; //require("db.php"); //GETTING VARIABLES START if (isset($_POST['action'])) { $action = mysql_real_escape_string($_POST['action']); } if (isset($_POST['pollAnswerID'])) { $pollAnswerID = mysql_real_escape_string($_POST['pollAnswerID']); } //GETTING VARIABLES END function getPoll($pollID){ $query = "SELECT * FROM polls LEFT JOIN pollAnswers ON polls.pollID = pollAnswers.pollID WHERE polls.pollID = " . $pollID . " ORDER By pollAnswerListing ASC"; $result = mysql_query($query); //echo $query;jquery $pollStartHtml = ''; $pollAnswersHtml = ''; while($row = mysql_fetch_array($result)) { $pollQuestion = $row['pollQuestion']; $pollAnswerID = $row['pollAnswerID']; $pollAnswerValue = $row['pollAnswerValue']; if ($pollStartHtml == '') { $pollStartHtml = '<div id="pollWrap"><form name="pollForm" method="post" action="inc/functions.php?action=vote"><h3>' . $pollQuestion .'</h3><ul>'; $pollEndHtml = '</ul><input type="submit" name="pollSubmit" id="pollSubmit" value="Vote" /> <span id="pollMessage"></span><img src="ajaxLoader.gif" alt="Ajax Loader" id="pollAjaxLoader" /></form></div>'; } $pollAnswersHtml = $pollAnswersHtml . '<li><input name="pollAnswerID" id="pollRadioButton' . $pollAnswerID . '" type="radio" value="' . $pollAnswerID . '" /> ' . $pollAnswerValue .'<span id="pollAnswer' . $pollAnswerID . '"></span></li>'; $pollAnswersHtml = $pollAnswersHtml . '<li class="pollChart pollChart' . $pollAnswerID . '"></li>'; } echo $pollStartHtml . $pollAnswersHtml . $pollEndHtml; } function getPollID($pollAnswerID){ $query = "SELECT pollID FROM pollAnswers WHERE pollAnswerID = ".$pollAnswerID." LIMIT 1"; $result = mysql_query($query); $row = mysql_fetch_array($result); return $row['pollID']; } function getPollResults($pollID){ $colorArray = array(1 => "#ffcc00", "#00ff00", "#cc0000", "#0066cc", "#ff0099", "#ffcc00", "#00ff00", "#cc0000", "#0066cc", "#ff0099"); $colorCounter = 1; $query = "SELECT pollAnswerID, pollAnswerPoints FROM pollAnswers WHERE pollID = ".$pollID.""; $result = mysql_query($query); while($row = mysql_fetch_array($result)) { if ($pollResults == "") { $pollResults = $row['pollAnswerID'] . "|" . $row['pollAnswerPoints'] . "|" . $colorArray[$colorCounter]; } else { $pollResults = $pollResults . "-" . $row['pollAnswerID'] . "|" . $row['pollAnswerPoints'] . "|" . $colorArray[$colorCounter]; } $colorCounter = $colorCounter + 1; } $query = "SELECT SUM(pollAnswerPoints) FROM pollAnswers WHERE pollID = ".$pollID.""; $result = mysql_query($query); $row = mysql_fetch_array( $result ); $pollResults = $pollResults . "-" . $row['SUM(pollAnswerPoints)']; echo $pollResults; } //VOTE START if ($action == "vote"){ if (isset($_COOKIE["poll" . getPollID($pollAnswerID)])) { echo "voted"; } else { $query = "UPDATE pollAnswers SET pollAnswerPoints = pollAnswerPoints + 1 WHERE pollAnswerID = ".$pollAnswerID.""; mysql_query($query) or die('Error, insert query failed'); setcookie("poll" . getPollID($pollAnswerID), 1, time()+259200, "/", ".webresourcesdepot.com"); getPollResults(1); } } //VOTE END if (mysql_real_escape_string($_GET['cleanCookie']) == 1){ setcookie("poll1", "", time()-3600, "/", ".webresourcesdepot.com"); header('Location: http://webresourcesdepot.com/wp-content/uploads/file/ajax-poll-script/'); } ?> database_connection.php <?php require_once 'app_config.php'; mysql_connect(DATABASE_HOST, DATABASE_USERNAME, DATABASE_PASSWORD) or handle_error("there was a problem connecting to the database that holds the information we need to get you connected.", mysql_error()); mysql_select_db(DATABASE_NAME) or handle_error("there's a configuration problem with our database.", mysql_error());

I'm trying to manually insert a url into a video table. But it is a long url with all kinds of special characters Here is an example: <object classid="clsid:D27CDB6E-AE6D-11cf-96B8-444553540000" width="640" height="320" id="viddler_gridweb_2"><param name="movie" value="//www.viddler.com/player/35da7dee/"/><param name="allowScriptAccess" value="always"/><param name="allowNetworking" value="all"/><param name="allowFullScreen"value="true"/><param name="flashVars" value="f=1&autoplay=f&loop=0&nologo=0&hd=0"/><embed src="//www.viddler.com/player/35da7dee/" width="640" height="320" type="application/x-shockwave-flash" allowScriptAccess="always" allowFullScreen="true" allowNetworking="all" name="viddler_gridweb_2" flashVars="f=1&autoplay=f&loop=0&nologo=0&hd=0"></embed></object> Is there a way without trying to escape all of the characters?

I'm working on a script that take information from a form and enters it into a mysql database. Everything was working except that the way it is written you must select an image to upload to your profile. I wanted to make the image upload part optional so I added an if/else to the script. If (user supplies image) { process image } else { use missing image icon } I'm getting the following error :Parse error: syntax error, unexpected T_ELSE ... on line 70 (which is where I have my else statement. From reading the forums it sounds like I'm missing a curly brace, but there I think I have the braces right. Here is the part of the code described. Line 70 is :else { <?php // if user provided image if (!empty($_FILES[$image_fieldname])) { // Make sure we didn't have an error uploading the image ($_FILES[$image_fieldname]['error'] == 0) or handle_error("the server couldn't upload the image you selected.", $php_errors[$_FILES[$image_fieldname]['error']]); // Is this file the result of a valid upload? // The @ supresses PHP's errorfor this and uses our custom version instead. @is_uploaded_file($_FILES[$image_fieldname]['tmp_name']) or handle_error("you were trying to do something naughty. Shame on you!", "Uploaded request: file named '{$_FILES[$image_fieldname]['tmp_name']}'"); // Is this actually an image? @getimagesize($_FILES[$image_fieldname]['tmp_name']) or handle_error("you selected a file for your picture that isn't an image.", "{$_FILES[$image_fieldname]['tmp_name']} isn't a valid image file."); // Name the file uniquely $now = time(); while (file_exists($upload_filename = $upload_dir . $now . '-' . $_FILES[$image_fieldname]['name'])) { $now++; // Finally, move the file to its permanent location @move_uploaded_file($_FILES[$image_fieldname]['tmp_name'], $upload_filename) or handle_error("we had a problem saving your image to its permanent location.", "permissions or related error moving file to {$upload_filename}"); } // end if user provided image else { // set file name equal to missing picture icon $upload_filename = "uploads/profile_pics/missing_user.png"; } ?> I also inclused the entire file if you need to see that. <?php require_once 'scripts/app_config.php'; require_once 'scripts/database_connection.php'; $upload_dir = SITE_ROOT . "uploads/profile_pics/"; $image_fieldname = "user_pic"; // Potential PHP upload errors $php_errors = array(1 => 'Maximum file size in php.ini exceeded', 2 => 'Maximum file size in HTML form exceeded', 3 => 'Only part of the file was uploaded', 4 => 'No file was selected to upload.'); $first_name = trim($_REQUEST['first_name']); $last_name = trim($_REQUEST['last_name']); $username = trim($_REQUEST['username']); $password = trim($_REQUEST['password']); $email = trim($_REQUEST['email']); $bio = trim($_REQUEST['bio']); $facebook_url = str_replace("facebook.org", "facebook.com", trim($_REQUEST['facebook_url'])); $position = strpos($facebook_url, "facebook.com"); if ($position === false) { $facebook_url = "http://www.facebook.com/" . $facebook_url; } $twitter_handle = trim($_REQUEST['twitter_handle']); $twitter_url = "http://www.twitter.com/"; $position = strpos($twitter_handle, "@"); if ($position === false) { $twitter_url = $twitter_url . $twitter_handle; } else { $twitter_url = $twitter_url . substr($twitter_handle, $position + 1); } // if user provided image if (!empty($_FILES[$image_fieldname])) { // Make sure we didn't have an error uploading the image ($_FILES[$image_fieldname]['error'] == 0) or handle_error("the server couldn't upload the image you selected.", $php_errors[$_FILES[$image_fieldname]['error']]); // Is this file the result of a valid upload? // The @ supresses PHP's errorfor this and uses our custom version instead. @is_uploaded_file($_FILES[$image_fieldname]['tmp_name']) or handle_error("you were trying to do something naughty. Shame on you!", "Uploaded request: file named '{$_FILES[$image_fieldname]['tmp_name']}'"); // Is this actually an image? @getimagesize($_FILES[$image_fieldname]['tmp_name']) or handle_error("you selected a file for your picture that isn't an image.", "{$_FILES[$image_fieldname]['tmp_name']} isn't a valid image file."); // Name the file uniquely $now = time(); while (file_exists($upload_filename = $upload_dir . $now . '-' . $_FILES[$image_fieldname]['name'])) { $now++; // Finally, move the file to its permanent location @move_uploaded_file($_FILES[$image_fieldname]['tmp_name'], $upload_filename) or handle_error("we had a problem saving your image to its permanent location.", "permissions or related error moving file to {$upload_filename}"); } // end if user provided image else { // set file name equal to missing picture icon $upload_filename = "uploads/profile_pics/missing_user.png"; } $insert_sql = sprintf("INSERT INTO users " . "(first_name, last_name, username, " . "password, email, " . "bio, facebook_url, twitter_handle, " . "user_pic_path) " . "VALUES ('%s', '%s', '%s', '%s', '%s', '%s', '%s', '%s', '%s');", mysql_real_escape_string($first_name), mysql_real_escape_string($last_name), mysql_real_escape_string($username), mysql_real_escape_string(crypt($password, $username)), mysql_real_escape_string($email), mysql_real_escape_string($bio), mysql_real_escape_string($facebook_url), mysql_real_escape_string($twitter_handle), mysql_real_escape_string($upload_filename)); // Insert the user into the database mysql_query($insert_sql) or handle_error(mysql_error()); // Redirect the user to the page that displays user information header("Location: show_user.php?user_id=" . mysql_insert_id()); ?>

I'm trying to check the number of results returned in a query. Currently there is one result being returned but i'm getting this error when I try to run line 38: $num_rows = mysql_num_rows($ratings); Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /home/haas12/public_html/login/rateVideo.php on line 38 Rows $ratingsQuery = "SELECT * FROM haas12_test.ratings "; $ratings = mysqli_query($conn, $ratingsQuery) or die ("Couldn't execute query."); $num_rows = mysql_num_rows($ratings); echo "$num_rows Rows\n";

Thanks! That fixed that part, but I tried to do the same think here and it isn't working. $query = "SELECT * FROM haas12_test.videos where username = \"$username\""; It works if I substitute $username for an actual username from the database. I read the link you posted on the heredoc method but it didn't understand it.