PHP - How To Avoid Duplicate/multiple Sql Queries By Refreshing Page?
I'm using paypal PDT, so when the user has made payment he gets directed back to my site, on that page it executes a special sql query. I noticed I can refresh the page to make duplicate sql queries. How can I avoid this??
Similar TutorialsI have a sorting functionality, the code is he <?php $select_category = $_REQUEST['sort_category']; $sort_date_var = $_REQUEST['sort_date']; $sort_submit = $_POST['sortSubmit']; // Connect to the database $dbc = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME); if (($select_category == 'All') || (!isset($select_category)) && (!isset($sort_date_var))) { // Retrieve the chosen category from MySQL $query = "SELECT * FROM con"; $data = mysqli_query($dbc, $query); //Loop through the array of data while ($row = mysqli_fetch_array($data)) { echo "<table class='knuffixTable'>"; // Display the score data echo "<tr><td class='knuffix_name'>"; echo "<strong>" . htmlentities($row['name']) . "</strong><br /></td></tr>"; echo "<tr><td class='knuffix_contribution'><pre>" . $row['contribution'] . "</pre><br /></td></tr>"; echo "<tr><td class='knuffix_categoryDate'>" . $row['category'] . " | " . date('M d, Y', strtotime($row['contributed_date'])) . " </td></tr>"; echo "</table>"; } mysqli_close($dbc); } elseif (isset($select_category) && !isset($sort_date_var)) { // Retrieve the chosen category from MySQL $query = "SELECT * FROM con WHERE category = '$select_category'"; $data = mysqli_query($dbc, $query) or die (mysqli_error($dbc)); //Loop through the array of data while ($row = mysqli_fetch_array($data)) { echo "<table class='knuffixTable'>"; // Display the score data echo "<tr><td class='knuffix_name'>"; echo "<strong>" . htmlentities($row['name']) . "</strong><br /></td></tr>"; echo "<tr><td class='knuffix_contribution'><pre>" . $row['contribution'] . "</pre><br /></td></tr>"; echo "<tr><td class='knuffix_categoryDate'>" . htmlentities($row['category']) . " | " . date('M d, Y', strtotime($row['contributed_date'])) . " </td></tr>"; echo "</table>"; } mysqli_close($dbc); } elseif (!isset($select_category) && isset($sort_date_var)) { // Retrieve the chosen category from MySQL $query = "SELECT * FROM con ORDER BY contributed_date $sort_date_var"; $data = mysqli_query($dbc, $query) or die (mysqli_error($dbc)); //Loop through the array of data while ($row = mysqli_fetch_array($data)) { echo "<table class='knuffixTable'>"; // Display the score data echo "<tr><td class='knuffix_name'>"; echo "<strong>" . htmlentities($row['name']) . "</strong><br /></td></tr>"; echo "<tr><td class='knuffix_contribution'><pre>" . $row['contribution'] . "</pre><br /></td></tr>"; echo "<tr><td class='knuffix_categoryDate'>" . htmlentities($row['category']) . " | " . date('M d, Y', strtotime($row['contributed_date'])) . " </td></tr>"; echo "</table>"; } mysqli_close($dbc); } elseif (isset($select_category) && isset($sort_date_var)) { // Retrieve the chosen category from MySQL $query = "SELECT * FROM con WHERE category = '$select_category' ORDER BY contributed_date $sort_date_var"; $data = mysqli_query($dbc, $query) or die (mysqli_error($dbc)); //Loop through the array of data while ($row = mysqli_fetch_array($data)) { echo "<table class='knuffixTable'>"; // Display the score data echo "<tr><td class='knuffix_name'>"; echo "<strong>" . htmlentities($row['name']) . "</strong><br /></td></tr>"; echo "<tr><td class='knuffix_contribution'><pre>" . $row['contribution'] . "</pre><br /></td></tr>"; echo "<tr><td class='knuffix_categoryDate'>" . htmlentities($row['category']) . " | " . date('M d, Y', strtotime($row['contributed_date'])) . " </td></tr>"; echo "</table>"; } mysqli_close($dbc); } ?> With this code I'm able to: - showcase every category by DEFAULT when someone comes to the page. - sort only by category, - sort only by date, - sort by category AND date, The problem is as you can see I have a lot of duplicate code, if I for example want to change the table that is being printed out, I have to change it on all of them. I'd like to ask how I could avoid this duplicate code. Can I for example just have the table ONE TIME at a separate place and then insert the REFERENCE _after_ the query in each if statement? Like this: if (($select_category == 'All') || (!isset($select_category)) && (!isset($sort_date_var))) { // Retrieve the chosen category from MySQL $query = "SELECT * FROM con"; $data = mysqli_query($dbc, $query); table_here(); } elseif (isset($select_category) && !isset($sort_date_var)) { // Retrieve the chosen category from MySQL $query = "SELECT * FROM con WHERE category = '$select_category'"; $data = mysqli_query($dbc, $query) or die (mysqli_error($dbc)); table_here(); } I'm just coding since 2 month so I don't know how do it, maybe I can do it with functions? I've tried doing it with functions, but I got an error since the function doesn't contain any query statement and only the table because it was trying to fetch. it looked like this: function table_here () { //Loop through the array of data while ($row = mysqli_fetch_array($data)) { echo "<table class='knuffixTable'>"; // Display the score data echo "<tr><td class='knuffix_name'>"; echo "<strong>" . htmlentities($row['name']) . "</strong><br /></td></tr>"; echo "<tr><td class='knuffix_contribution'><pre>" . $row['contribution'] . "</pre><br /></td></tr>"; echo "<tr><td class='knuffix_categoryDate'>" . $row['category'] . " | " . date('M d, Y', strtotime($row['contributed_date'])) . " </td></tr>"; echo "</table>"; } mysqli_close($dbc); } Hi everyone I have the following PHP code $sql.= "INSERT INTO `data` (`info`, `write`, `date`) VALUES"; $sql.= "("; for ($c=0; $c < $num; $c++) { $sql.= '"'.str_replace('"', "", $data[$c]).'",'; } $sql.= "'".date('Y-m-d')."');"; If I print $sql; I get Code: [Select] INSERT INTO `data` (`info`, `write`, `date`) VALUES("data99","n",'2010-10-05'); INSERT INTO `data` (`info`, `write`, `date`) VALUES("data101","y",'2010-10-05'); INSERT INTO `data` (`info`, `write`, `date`) VALUES("data876","n",'2010-10-05'); what would I need to do to my PHP code in order for it to process each QUERY either all at once or one at a time Whatever I try, it either inserts nothing, or just does the first INSERT Any ideas? Thanks The whole code is below, followed by the part that isn't producing. It all works extremely well until the last part, and I have two instances of setting a variable to mysql_fetch_assoc that work. Not sure why the third attempt doesn't. I'm just trying to have a separate area that outputs 'msg'. Code: [Select] $sql = "SELECT * FROM players as p INNER JOIN schools as s WHERE p.tid = s.id ORDER BY status, playerLast"; $results = mysql_query($sql); echo '<div class="roster">'; $team = mysql_fetch_assoc($results); echo '<div class="team_info"> <div class="school">' . $team['school'] . '</div> <div class="coach">Coach ' .$team['coachFirst'] . ' ' . $team['coachLast'] .'</div> <div>Sectional: ' . $team['sectional'] . '</div> <div>Class: ' . $team['class'] . 'A</div> '; echo '</div>'; $currentStatus = false; //Flag to detect change in status while($player = mysql_fetch_assoc($results)) { if($currentStatus != $player['status']) { //Status has changed, display status header $currentStatus = $player['status']; echo '<br><b>'; if ($currentStatus == '1') {echo 'Returning Starters';} elseif ($currentStatus == '2') {echo 'Key Returners';} elseif ($currentStatus == '3') {echo 'Varsity Newcomers';} elseif ($currentStatus == '4') {echo 'Key Freshmen';} echo '</b><br>'; // echo "<div><b>{$currentStatus}</b></div>\n"; } //Display player info echo $player['playerFirst'] . ' ' . $player['playerLast'] . ', ' . $player['feet'] . '\'' . $player['inches'] . '",' . $player['position'] . ', ' . $player['year'] . ';<br>'; } //Player comments echo '<hr>'; echo '<div class="coach_comments"><span class="player_name">hello'; while($comments = mysql_fetch_assoc($results)) { echo $comments['playerFisrt'] . ' ' . $comments['playerLast'] . ': </span>' . $comments['msg']; } echo '</div>'; echo '</div>'; It's all working until what's below. All I get out of it is the "hello", which is just a test to make sure it wasn't a CSS issue. Quote //Player comments echo '<hr>'; echo '<div class="coach_comments"><span class="player_name">hello'; while($comments = mysql_fetch_assoc($results)) { echo $comments['playerFisrt'] . ' ' . $comments['playerLast'] . ': </span>' . $comments['msg']; } echo '</div>'; echo '</div>'; I'm working on a Stats page that will need to run multiple queries to access the information needed to dislpay on the page. Rather than running 30+ queries to get the user's Stats, is there a better method? Any tips that you suggest I further research? Thanks Hi all, I would like to perform two queries in a mysqli statement before closing the connection. I'm using prepared statements and I've read on the php.net you can perform a multi-query using mysqli->multi_query. My problem is that I'm only finding useful examples on how to select and fetch results not using prepared statements. I'm a bit lost with how to do this to update two tables. Currently i'm using two seperate connections to do the queries but want to combine these together: Query one (updates their password and resets number of logins to force password change on next login): Code: [Select] //reset logins to force the user to change password upon next login $loggedin = '0'; // connect to db for mysqli require('../db/db.php'); // updates password and number of logins $insert_stmt = $mysqli->stmt_init(); if ($insert_stmt->prepare("UPDATE users SET their_password=?, loggedin=? WHERE their_username=?")) { $insert_stmt->bind_param('sss', $their_password, $time, $user_email); $insert_stmt->execute(); $insert_stmt->close(); } // if query errors sends an email if ($mysqli->error) { try { throw new Exception("MySQL error $mysqli->error <br> Query:<br> $query", $mysqli->errno); } catch(Exception $e ) { $mess = "Error No: ".$e->getCode(). " - ". $e->getMessage() . "<br >"; $mess .= nl2br($e->getTraceAsString()); $contact_email = "webmaster@website.com"; $message_sub = "Mysqli Forgotten Password Query Error [UPASSWORD01]"; $hdrs = "From: " . $contact_email . "\r\n"; $hdrs .= "Reply-To: ". $contact_email . "\r\n"; $hdrs .= "MIME-Version: 1.0\r\n"; $hdrs .= "Content-Type: text/html; charset=UTF-8\r\n"; mail($contact_email, $message_sub, $mess, $hdrs); } header("refresh: 10; forgotpass.php"); die('ERROR: Unable to reset password. Please check you details and try again or report this error to us using our contact us form.<br><br>We will redirect you back to the forgotten password form in 10 seconds.'); exit(); } $mysqli->close(); Query two (updates a reset log - to keep track on our user password resets): Code: [Select] // change status $resetstatus = "Successful Reset"; // connect to db for mysqli require('../db/db.php'); // inserts a new record $null = NULL; $insert_stmt = $mysqli->stmt_init(); if ($insert_stmt->prepare("INSERT INTO passwordresets VALUES (?, ?, ?, ?, ?)")) { $insert_stmt->bind_param('issss', $null, $user_email, $time, $userip, $resetstatus); $insert_stmt->execute(); $insert_stmt->close(); } // if query errors sends an email if ($mysqli->error) { try { throw new Exception("MySQL error $mysqli->error <br> Query:<br> $query", $mysqli->errno); } catch(Exception $e ) { $mess = "Error No: ".$e->getCode(). " - ". $e->getMessage() . "<br >"; $mess .= nl2br($e->getTraceAsString()); $contact_email = "webmaster@website.com"; $message_sub = "Mysqli Forgotten Password Query Error [UARESETLOG01]"; $hdrs = "From: " . $contact_email . "\r\n"; $hdrs .= "Reply-To: ". $contact_email . "\r\n"; $hdrs .= "MIME-Version: 1.0\r\n"; $hdrs .= "Content-Type: text/html; charset=UTF-8\r\n"; mail($contact_email, $message_sub, $mess, $hdrs); } header("refresh: 10; forgotpass.php"); die('ERROR: Unable to reset password. Please check you details and try again or report this error to us using our contact us form.<br><br>We will redirect you back to the forgotten password form in 10 seconds.'); exit(); } $mysqli->close(); Any ideas would be greatly appreciated. Thank you Not sure where to begin describing ths one?! On my website I have Articles. Beneath each Article, Users can leave one or more Comments. Next to each User's Comment, I have the following in the left margin... - Username - User Online Status - User Photo - User Location - User # of Posts At the top of my file, I have my PHP code and I start off by running a Prepared Statement looking for the Article the User is requesting in the URL. If the Article is found, I store the Results, and then run another Prepared Statement looking for Comments. If Comments are found, I store the Results. Then down in the HTML section, I display the Article and beneath it I loop through the Article's Comments like this... Code: [Select] // ******************************** // Display Comments on Article. * // ******************************** while (mysqli_stmt_fetch($stmt2)){ // Display User Info. * // Display User Comments. * } This seems to be working nicely, except that there is one last piece of User Info that is messing things up... I can get all of the User and Comments data in this query... Code: [Select] // ************************ // Build Comments Query. * // ************************ // Build query. $q2 = 'SELECT m.first_name, m.username, m.photo_name, m.photo_label, m.location, m.created_on, m.logged_in, m.last_activity, c.created_on, c.body, c.status FROM member AS m INNER JOIN comment AS c ON m.id = c.member_id WHERE c.status="Approved" AND c.article_id=? ORDER BY c.created_on'; However, in order to get the User's "Number of Posts" I need another query. And I'm not sure how to get the "Post Count" for each User and then mesh it back into what I currently have?! Does that make sense?! Also, it is conceivable that later I might add additional User Info that comes from yet another Query/Source. (If I was using OOP and PDO and had a clear separation of "Presentation" from "Business Logic" I'm sure would be easier, but for now I need to work with the structure I have since any of those three things are way beyond the scope of this question and my current abilities!!!!) Thanks, Debbie Hi, I am currently working on a very detailed personal project during my vacation to keep me busy, but I have a problem, I am trying to create a unique id, without having the chance of a repetition of the id, even if its infinitely small. So I found the flickr ticket id system to be a good choice, since it was sequential and effective. My problem now is, how can I execute 2 queries simultaneously, given the hypothetical change that if I execute them with two separate queries (mysql_query), that there is the chance that another query could be executed in between the break, causing two id's to be the same. Since mysql_query can only execute a single query I am a bit stuck. I thought about possibly using CGI, but then again I have never used it, so I don't know its limitations. Can anyone suggestion anything to me for this problem, as to how I can execute two query strings in the same query. Hello I have a question. I'm trying to perform multiple queries based off the results from a query. Is this possible ? $result = mysql_query("SELECT id FROM sometable"); for each id returned $result1 = mysql_query("SELECT * FROM someothertable WHERE id=result from $result query"); Any help would be GREAT This topic has been moved to MySQL Help. http://www.phpfreaks.com/forums/index.php?topic=330871.0 Main script gets class information from a database and prints them so long as the class start date or end date is after today (actually includes today).
Main script calls "instructors.php". It queries another database based on the instructor name ($chef) and then prints the bio information for that instructor.
"instructors.php" works fine on it's own, when I add "$chef = "Chef Name" ("Chef Name" is in the Instructors database). When it's called from the main script, nothing shows up in that area - even though "Chef Name" is in the database. All of the other data is printed fine, just not anything from instructors.php. I verified that it's actually including the file, as I can add "echo "test";" to the top of instructors.php and it prints fine in the main script.
Any ideas of what I'm missing?
Main Script
<?php // Get required login info include "/path/to/login/info/file.php"; // Get required login info - changed for this post. $db = new mysqli('localhost', $username, $password, $database); // Connect to DB using required login info if($db->connect_errno > 0){ die('Unable to connect to database [' . $db->connect_error . ']'); } unset($username);// put these variables back to null unset($password);// put these variables back to null unset($database);// put these variables back to null //query db $sql = <<<SQL SELECT * FROM `ft_form_7` WHERE DATE(class_start_date) >= CURDATE() OR DATE(class_end_date) >= CURDATE() ORDER BY class_start_date ASC SQL; if(!$result = $db->query($sql)){ // if there is an error in running the query, show error message. die('There was an error running the query [' . $db->error . ']'); } while($row = $result->fetch_assoc()){ // Get start date information $start_date = $row['class_start_date']; // Get event_start_date for conversion and call it $start_date $start_date_formatted = date("l M d, Y", strtotime($start_date)); // Convert start_date $end_date = $row['class_end_date']; // Get event_end_date for conversion and call it $start_date $end_date_formatted = date("M d, Y", strtotime($end_date)); // Convert start_date // Get time information. $start_time = $row['class_start_time']; // Get event_start_time for conversion and call it $start_time $start_time_formatted = date("h:i A", strtotime($start_time)); // Convert start_time $end_time = $row['class_end_time']; // Get event_end_time for conversion and call it $end_time $end_time_formatted = date("h:i A", strtotime($end_time)); // Convert end_time // echo information... echo "<h2>" , $row['class_name'],"</h2>" ; // echo event name echo "<p><strong>",$start_date_formatted; // echo the start date if (empty($start_time)) { echo ''; } else { echo " (", $start_time; } // echo start time if (empty($end_date)) { echo ''; } else { echo " -","<br />", $end_date_formatted; } // echo end date if (empty($end_time)) { echo ')'; } else { echo " - ", $end_time, ")"; } // echo end time // if there is no start time, echo nothing. (otherwise it seems to echo 4pm). If it does contain a time, echo the time. echo "</strong><br />"; $chef = $row['Instructor']; global $chef; if ($chef != NULL) { require ('instructors.php'); } echo $row['class_description'], "<br />"; echo "<strong>" , $row['type'], " - Cost: $",$row['cost'] , " - #" , $row['course_number'] , "</strong><br />" , "</p>"; // echo class type and cost } $db->close(); $result->free(); ?>instructors.php <?php include "/path/to/login/info/file.php"; // Get required login info - changed for this post. $db_instructors = new mysqli('localhost', $username, $password, $database); // Connect to DB using required login info if($db_instructors->connect_errno > 0){ die('Unable to connect to database [' . $db_instructors->connect_error . ']'); } unset($username);// put these variables back to null unset($password);// put these variables back to null unset($database);// put these variables back to null //query db $sql_instructors = <<<SQL SELECT * FROM ft_form_8 WHERE chef_name = '$chef' SQL; if(!$result_instructors = $db_instructors->query($sql_instructors)) { // if there is an error in running the query, show error message. die('There was an error running the query [' . $db_instructors->error . ']'); } while($row_instructors = $result_instructors->fetch_assoc()) { $chef_full_name = $row_instructors['chef_name']; $chef_id = $row_instructors['submission_id']; $full_bio = $row_instructors['full_bio']; echo "<a href=\"#\" class=\"clickme\">" , $chef_full_name , "</a>"; echo "<div class=\"box\">"; echo $full_bio , "</div>"; } $db_instructors->close(); $result_instructors->free(); ?> Hi, I need help to check multiple rows for many columns, so that the script won't re-insert a duplicate entry. For eg: $isRowExist=mysql_query("SELECT (field1,field2) from table1 WHERE field1='$f1' AND field2='$f2' "); if(mysql_num_rows($isRowExist)==1) print "already in table1!"; else //proceed to insert into table table1 *BUT I tried this and it says mysql_num_rows expects resource, and this warning only shows when I have multipel columns I want to compare (it works just fine with one column in the WHERE clause above) Please help! This topic has been moved to MySQL Help. http://www.phpfreaks.com/forums/index.php?topic=327017.0 First of all excuse me if this topic is inappropriate in this forum. But I think it's rather a PHP problem. I can't figure out multiple duplicate database records on submitting a form. The database table have two columns: the first one 'Id' with AUTO_INCREMENT and the second one 'Name'. Here's the php code for database insertion and the form: ------------------------------------------------------------------ <?php if($_GET['add_name']){ $host = *******; $user = *******'; $pass = *******; $db = *******; $con = mysql_connect($host,$user,$pass) or die; mysql_select_db($db,$con); $name = $_GET['add_name']; $sql = "INSERT INTO names (Name) VALUES ('$name')"; mysql_query($sql); } ?> <form action="<?php echo $_SERVER['PHP_SELF'] ?>" method="GET"> Your Name: <input name="add_name" type="text" /> <input type="submit" value="Submit" /> </form> ------------------------------------------------------------------ After submitting the form to itself once I have multiple Name entries with different Ids. The curious thing is that with Chrome browser I get two duplicate records, with Mozilla - three of them. Seems like mysql_query runs multiple times. It works fine when submitting the form to a separate script and not to itself. Do I miss something? It must be very basic. The Script:
<h1>Do Add a Message to the MySQL Database</h1> <form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="POST"> <textarea name="message"></textarea> <br/> <input type="submit" name="submit"/> </form> <?php // The Connection to the Database // Taken Out ?> <?php // To insert the text data into the MySQL database. if(isset($_POST['submit'])){ $tqs = "INSERT INTO messages (`message`) VALUES ('{$_POST['message']}')"; $tqr = mysqli_query($dbc, $tqs) or die(mysqli_error($dbc)); } ?> <?php // To select the text data from the MySQL database. $tqs = "SELECT * FROM messages"; $tqr = mysqli_query($dbc, $tqs); // To print out the text data inside of table on the page. echo "<h1>This Is Where the Messages Gets Printed on Screen</h1>"; echo "<table><tr><td>ID</td><td>The Message</td></tr>"; while($row = mysqli_fetch_assoc($tqr)){ echo "<tr><td>" . $row['id'] . "</td><td>" . $row['message'] . "</td></tr>"; } echo "</table>"; ?>1. When I have added text with the form to the MySQL database... 2. ... and I have clicked on "page reload" in Firefox to reload the page... 3. ... then the before submitted text gets submitted again to the MySQL database. So basically, add text with the form to the MySQL database, reload the page in Firefox, and the before added text will get submitted to the MySQL database again. My Question Is: What is the proper way to avoid this? Edited by glassfish, 06 October 2014 - 10:18 AM. MySQL returns an error in the form of a number, a state, and a message. Without parsing the message you will not be able to determine what column is duplicated.While parsing the error code, I have also notice that, if you have multiple unique fields as duplicates, only the first duplicate encountered will be returned in the message. This is not very helpful to the end user.
Is there any way to parse the returned error code to reflect all duplicate fields, please see sample code below?
$error=array(); $sql = 'INSERT INTO staff(username, email, phone) VALUES (?, ?, ?)'; $stmt = $conn->stmt_init(); $stmt = $conn->prepare($sql); // bind parameters and insert the details into the database $stmt->bind_param('sss', $username, $email, $phone); $stmt->execute(); if ($stmt->errno == 1062) { $errors[] = "One of the fields is already in use."; } Hi, In the following code I'm using a combo box to select a relay and activate it. The code works but if I refresh the page, the last selected relay is activated again. How can I reset the page to default to value '0' which has no relay control? TIA
<!DOCTYPE html> <html> <head> <title>Relay control</title> <meta content="text/html; charset=utf-8" http-equiv="Content-Type" /> <meta name="apple-mobile-web-app-capable" content="yes" /> <meta name="apple-mobile-web-app-status-bar-style" content="black-translucent" /> <link rel="stylesheet" type="text/css" href="style.css" /> </head> <body> <h2><center>Status y control de relés</center></h2> <?php $arrleds = array(); if ($_SERVER["REQUEST_METHOD"] == "POST") { $selected_val = $_POST['sel']; if ($selected_val == '1'){ $cmd = exec("sudo ./trigger.py b 0"); } elseif ($selected_val == '2'){ $cmd = exec("sudo ./trigger.py b 1"); } elseif ($selected_val == '3'){ $cmd = exec("sudo ./trigger.py b 2"); } .... else {} } ?> <table class="center"> <tr> <td></td> <td><h1>Activación de relé</h1> <form action="<?php echo htmlspecialchars($_SERVER["PHP_SELF"]); ?>" method="post"> <select name='sel' onchange='submit();'> <option value='0'>Sin activar</option> <!-- no action --> <option value='1'>Secadora 1</option> <option value='2'>Secadora 2</option> <option value='3'>Lavadora S</option> .... </select> </form> </td> <td></td> </tr> <tr> <td></td> <td><a href="index.php">Home</a> <td></td> </tr> </table> </body> </html>
i have a form with two buttons, one is for submitting the form while button is for generating a random a code and placing it in a textfield. this works fine but the problem is anytime i click on the generate button, it refreshes the page thereby validating the form , which i don't want. How do i make it perform the function without refreshing the page.... I need Delete Duplicate Email Records That Are Attached To One Account But Can Be Found In Multiple Accounts I have a table, consumer_mgmt. It collects consumer information from various forms. These forms are available through different pages that are part of a business package. A business can offer these signups to gather names and emails from consumers for various types of specials they may offer. So a consumer my be in the consumer_mgmt table 5, 10, 15 times for that particular business. But, that consumer may be in the consumer_mgmt table multiple times for a different business. So multiple times for multiple businesses. I need to remove duplicates for each business account so the consumer is only the consumer_mgmt only once for each business. There are approximately 15,000 rows currently in the consumer_mgmt table. I'm not sure where to begin on the logic. Since there are multiple business accounts that the emails are attached to, would one have to build a case for each loop? Hi, I have a "user profile" page and it has a profile image upload form. I have it so that the old profile image is deleted, and the new image is uploaded and the new image is echoed out using the name of the new file from the database. The problem is, when I click submit to add the new image, the old image still stays there and the new image does not show up. Only when I manually click refresh on my browser does the new image show up. That is going to be bad for my users. Please if anyone can help, I'd greatly appreciate it. Below is the code I am using to delete the old image, and code to upload the image as well, when the photo upload form is used. Code: [Select] if(isset($_POST['photoUpload'])) { if(file_exists("images/".$currentUser.".jpg")){ unlink("images/".$currentUser.".jpg"); clearstatcache(); } //reads the name of the file the user submitted for uploading $image=$_FILES['image']['name']; //if it is not empty if ($image) { //get the original name of the file from the clients machine $filename = stripslashes($_FILES['image']['name']); //get the extension of the file in a lower case format $extension = getExtension($filename); $extension = strtolower($extension); //if it is not a known extension, we will suppose it is an error and will not upload the file, //otherwise we will do more tests if (($extension != "jpg") && ($extension != "jpeg") && ($extension != "png") && ($extension != "gif")) { //print error message echo '<h1>Unknown extension!</h1>'; $errors=1; } } //end if here //was else here //get the size of the image in bytes //$_FILES['image']['tmp_name'] is the temporary filename of the file //in which the uploaded file was stored on the server $size=filesize($_FILES['image']['tmp_name']); //compare the size with the maxim size we defined and print error if bigger if ($size > MAX_SIZE*1024) { echo '<h1>You have exceeded the size limit!</h1>'; $errors=1; } //we will give an unique name, for example the time in unix time format $image_name= $currentUser .'.'.$extension; //$image_name=time().'.'.$extension; //the new name will be containing the full path where will be stored (images folder) $newname=$image_name; //"images/". // Connects to your Database mysql_connect("host", "user", "pass") or die(mysql_error()); mysql_select_db("database") or die(mysql_error()) ; //we verify if the image has been uploaded, and print error instead $copied = copy($_FILES['image']['tmp_name'], "images/".$newname); // update the photo name in the database $result = mysql_query("UPDATE users SET profilePhoto='$newname' WHERE username='$currentUser'") or die(mysql_error()); if (!$copied) { echo '<h1>Copy unsuccessful!</h1>'; $errors=1; } else{ $dir="images/"; echo "<p>Profile Picture Change Successful!</p>"; echo "<img src='{$dir}{$newname}' alt='{$newname}' height='200' width='200' />"; echo "<p></p>"; } } Here is my code if(isset($_POST['pass_btn'])) { $numchars = 8; $chars = explode(',','2,3,4,5,6,7,8,9,a,b,c,d,e,f,g,h,i,j,k,m,n,p,q,r,s,t,u,v,w,x,y,z,2,3,4,5,6,7,8,9,A,B,C,D,E,F,G,H,J,K,L,M,N,P,Q,R,S,T,U,V,W,X,Y,Z,2,3,4,5,6,7,8,9'); $password=''; for($i=0; $i<$numchars;$i++) { $password.=$chars[rand(0,count($chars)-1)]; } } This is set up so when a person pressing the pass_btn, a password is provided. This works just fine. However, it refreshes the page and clears all other data that user may have entered. Suggestions? |