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  • Active User Tracking, PHP Sessions

    - by Nik
    Alright, I'm trying to work on a function for active user counting in an AJAX based application. I want to express the below SQL Query in correct syntax but I'm not sure how to write it. The desired SQL Query is below: SELECT count(*) FROM active WHERE timestamp > time() - 1800 AND nick=(a string that doesn't contain [AFK]) Now, I do understand that time() - 1800 can be assigned to a variable, but how is timestamp > variable and nick that doesn't contain a string written in SQL?

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  • Database Error Handling: What if You have to Call Outside service and the Transaction Fails?

    - by Ngu Soon Hui
    We all know that we can always wrap our database call in transaction ( with or without a proper ORM), in a form like this: $con = Propel::getConnection(EventPeer::DATABASE_NAME); try { $con->begin(); // do your update, save, delete or whatever here. $con->commit(); } catch (PropelException $e) { $con->rollback(); throw $e; } This way would guarantee that if the transaction fails, the database is restored to the correct status. But the problem is that let's say when I do a transaction, in addition to that transaction, I need to update another database ( an example would be when I update an entry in a column in databaseA, another entry in a column in databaseB must be updated). How to handle this case? Let's say, this is my code, I have three databases that need to be updated ( dbA, dbB, dbc): $con = Propel::getConnection("dbA"); try { $con->begin(); // update to dbA // update to dbB //update to dbc $con->commit(); } catch (PropelException $e) { $con->rollback(); throw $e; } If dbc fails, I can rollback the dbA but I can't rollback dbb. I think this problem should be database independent. And since I am using ORM, this should be ORM independent as well. Update: Some of the database transactions are wrapped in ORM, some are using naked PDO, oledb ( or whatever bare minimum language provided database calls). So my solution has to take care this. Any idea?

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  • how to create a dynamic sql statement w/ python and mysqldb

    - by Elias Bachaalany
    I have the following code: def sql_exec(self, sql_stmt, args = tuple()): """ Executes an SQL statement and returns a cursor. An SQL exception might be raised on error @return: SQL cursor object """ cursor = self.conn.cursor() if self.__debug_sql: try: print "sql_exec: " % (sql_stmt % args) except: print "sql_exec: " % sql_stmt cursor.execute(sql_stmt, args) return cursor def test(self, limit = 0): result = sql_exec(""" SELECT * FROM table """ + ("LIMIT %s" if limit else ""), (limit, )) while True: row = result.fetchone() if not row: break print row result.close() How can I nicely write test() so it works with or without 'limit' without having to write two queries?

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  • How to handle multiple openIDs for the same user

    - by Sinan
    For my site I am using a login system much like the one on SO. A user can login with his Facebook, Google (Gmail openID), Twitter account. This question is not about specific oAuth or openID implementations. The question is how to know if the same user logins with different providers. Let me give an example: Bobo comes to site logins to site by clicking on "Login with Facebook". Because this is his first visit we create an account for him. Later Bobo comes to the site. This time he clicks on "Login with Google". So how do I know if this is the same person so I can add this provider to his account instead of creating a new (and duplicate) account. Can I trust solely on email? What is the best way to handle this. How does SO do it? Any ideas?

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  • Managing Foreign Keys

    - by jwzk
    So I have a database with a few tables. The first table contains the user ID, first name and last name. The second table contains the user ID, interest ID, and interest rating. There is another table that has all of the interest ID's. For every interest ID (even when new ones are added), I need to make sure that each user has an entry for that interest ID (even if its blank, or has defaults). Will foreign keys help with this scenario? or will I need to use PHP to update each and every record when I add a new key?

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  • Selecting a whole database over an individual table to output to file

    - by Daniel Wrigley
    :::::::: EDIT :::::::: New code for people to have a look at, one question I have with this is where do I set were the *.gz file is saved? $backupFile = $dbname . date("Y-m-d-H-i-s") . '.gz'; $command = "mysqldump --opt -h $dbhost -u $dbuser -p $dbpass $dbname | gzip > $backupFile"; system($command); Also why the hell can you not reply yo your own post with answering it? :( :::::::: EDIT :::::::: Ok Im having trouble finding out how to select a full database for backup as an *.sql file rather than only an individual table. On the localhost I have several databases with one named "foo" and it is that which I want to backup and not any of the individual tables inside the database "foo". The code to connect to the database; //Database Information $dbhost = "localhost"; $dbname = "foo"; $dbuser = "bar"; $dbpass = "rulz"; //Connect to database mysql_connect ($dbhost, $dbuser, $dbpass) or die("Could not connect: ".mysql_error()); mysql_select_db($dbname) or die(mysql_error()); The code to backup the database; // Grab the time to know when this post was submitted $time = date('Y-m-d-H-i-s'); $tableName = 'foo'; $backupFile = '/sql/backup/'. $time .'.sql'; $query = "SELECT * INTO OUTFILE '". $backupFile ."' FROM ". $tableName .""; $result = mysql_query($query)or die("Database query died: " . mysql_error()); My brain is hurting near to the end of the day so no doubts i've missed something out very obvious. Thanks in advance to anyone helping me out.

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  • LINQ expression precedence with Skip(), Take() and OrderBy()

    - by Robert Koritnik
    I'm using LINQ to Entities and display paged results. But I'm having issues with the combination of Skip(), Take() and OrderBy() calls. Everything works fine, except that OrderBy() is assigned too late. It's executed after result set has been cut down by Skip() and Take(). So each page of results has items in order. But ordering is done on a page handful of data instead of ordering of the whole set and then limiting those records with Skip() and Take(). How do I set precedence with these statements? My example (simplified) var query = ctx.EntitySet.Where(/* filter */).OrderBy(/* expression */); int total = query.Count(); var result = query.Skip(n).Take(x).ToList();

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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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  • Rails: Generated tokens missing occasionally

    - by Vincent Chan
    We generate an unique token for each user and store it on database. Everything is working fine in the local environment. However, after we upload the codes to the production server on Engine Yard, things become weird. We tried to register an account right after the deploy. It is working fine and we can see the token in the db. But after that, when we register new accounts, we cannot see any tokens. We only have NULL in the db. Not sure what caused this problem because we can't re-produce this in the local machine. Thanks for your help.

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  • [WEB] Local/Dev/Live deployment - best workflow

    - by Adam Kiss
    Hello, situation We our little company with 3 people, each has a localhost webserver and most projects (previous and current) are on one PC network shared disk. We have virtual server, where some of our clients' sites and our site. Our standard workflow is: Coder PC ? Programmer localhost ? dev domain (client.company.com) ? live version (client.com) It often happens, that there are two or three guys working on same projects at the same time - one is on dev version, two are on localhost. When finished, we try to synchronize the files on dev version and ideally not to mess (thanks ILMV:]) up any files, which *knock knock * doesn't happen often. And then one of us deploys dev version on live webserver. question we are looking for a way to simplify this workflow while updating websites - ideally some sort of diff uploader or VCS probably (Git/SVN/VCS/...), but we are not completely sure where to begin or what way would be ideal, therefore I ask you, fellow stackoverflowers for your experience with website / application deployment and recommended workflow. We probably will also need to use Mac in process, so if it won't be a problem, that would be even better. Thank you

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  • please help me in this query

    - by testkhan
    I have three tables (user, friends, posts) and two users (user1 and user2). When user1 adds user2 as friend then user1 can see the posts of user2 just like on Facebook. But only the posts after the date when user1 added user2 as friend. My query is like this: $sql = mysql_query("SELECT * FROM posts p JOIN friends f ON p.currentuserid = f.friendid AND p.time >= f.friend_since OR p.currentuserid='user1id' WHERE f.myid='user1id' ORDER BY p.postid DESC LIMIT 20"); it is working all the way fine but with a little problem.....!! it displays user2, user3 (all the users as friends of user1) posts for single time but shows user1 posts multiple.......i.e user2. hi user1. userssfsfsfsfsdf user1. userssfsfsfsfsdf user3. dddddddd user1. sdfsdsdfsdsfsf user1. sdfsdsdfsdsfsf but i in database it is single entry/post why it is happening........!! How can I fix it?

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  • PHP: table structure

    - by A3efan
    I'm developing a website that has some audio courses, each course can have multiple lessons. I want to display each course in its own table with its different lessons. This is my sql statement: Table: courses id, title Table: lessons id, cid (course id), title, date, file $sql = "SELECT lessons.*, courses.title AS course FROM lessons INNER JOIN courses ON courses.id = lessons.cid GROUP BY lessons.id ORDER BY lessons.id" ; Can someone help me with the PHP code? This is the I code I have written: mysql_select_db($database_config, $config); mysql_query("set names utf8"); $sql = "SELECT lessons.*, courses.title AS course FROM lessons INNER JOIN courses ON courses.id = lessons.cid GROUP BY lessons.id ORDER BY lessons.id" ; $result = mysql_query($sql) or die(mysql_error()); while ($row = mysql_fetch_assoc($result)) { echo "<p><span class='heading1'>" . $row['course'] . "</span> </p> "; echo "<p class='datum'>Posted onder <a href='*'>*</a>, latest update on " . strftime("%A %d %B %Y %H:%M", strtotime($row['date'])); } echo "</p>"; echo "<class id='text'>"; echo "<p>...</p>"; echo "<table border: none cellpadding='1' cellspacing='1'>"; echo "<tr>"; echo "<th>Nr.</th>"; echo "<th width='450'>Lesso</th>"; echo "<th>Date</th>"; echo "<th>Download</th>"; echo "</tr>"; echo "<tr>"; echo "<td>" . $row['nr'] . "</td>"; echo "<td>" . $row['title'] . "</td>"; echo "<td>" . strftime("%d/%m/%Y", strtotime($row['date'])) . "</td>"; echo "<td><a href='audio/" . rawurlencode($row['file']) . "'>MP3</a></td>"; echo "</tr>"; echo "</table>"; echo "<br>"; } ?>

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  • SQL join produces one result only

    - by Rami
    Can anyone please tell me why this result is generation only one results? taking in mind that everything is set right and the three tables are populated correctly, i took out the group_concat and it worked but of course with a php undefined index error! SELECT `songs`.`song_name`, `songs`.`add_date`, `songs`.`song_id`, `songs`.`song_picture`, group_concat(DISTINCT artists.artist_name) as artist_name FROM (`songs`) JOIN `mtm_songs_artists` ON `songs`.`song_id` = `mtm_songs_artists`.`song_id` JOIN `artists` ON `artists`.`artist_id` = `mtm_songs_artists`.`artist_id` ORDER BY `songs`.`song_id` DESC LIMIT 10 so i'm guessing it's something related to group_concat. best regards, Rami

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  • PHP error problem.

    - by TaG
    I get the following error on line 8: Undefined index: real_name which is $privacy_policy = mysqli_real_escape_string($mysqli, $_POST['privacy_policy']); I was wondering how can I fix this problem? Here is the PHP. if (isset($_POST['submitted'])) { $mysqli = mysqli_connect("localhost", "root", "", "sitename"); $dbc = mysqli_query($mysqli,"SELECT users.* FROM users WHERE user_id=3"); $privacy_policy = mysqli_real_escape_string($mysqli, $_POST['privacy_policy']); if (mysqli_num_rows($dbc) == 0) { $mysqli = mysqli_connect("localhost", "root", "", "sitename"); $dbc = mysqli_query($mysqli,"INSERT INTO users (user_id, privacy_policy) VALUES ('$user_id', '$privacy_policy')"); } if ($dbc == TRUE) { $dbc = mysqli_query($mysqli,"UPDATE users SET privacy_policy = '$privacy_policy' WHERE user_id = '$user_id'"); echo '<p class="changes-saved">Your changes have been saved!</p>'; } if (!$dbc) { print mysqli_error($mysqli); return; } } Here is the HTML. <form method="post" action="index.php"> <fieldset> <ul> <li><input type="checkbox" name="privacy_policy" id="privacy_policy" value="yes" <?php if (isset($_POST['privacy_policy'])) { echo 'checked="checked"'; } else if($privacy_policy == "yes") { echo 'checked="checked"'; } ?> /></li> <li><input type="submit" name="submit" value="Save Changes" class="save-button" /> <input type="hidden" name="submitted" value="true" /> <input type="submit" name="submit" value="Preview Changes" class="preview-changes-button" /></li> </ul> </fieldset> </form>

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  • Trouble making login page?

    - by Ken
    Okay, so I want to make a simple login page. I've created a register page successfully, but i can't get the login thing down. login.php: <?php session_start(); include("mainmenu.php"); $usrname = mysql_real_escape_string($_POST['usrname']); $password = md5($_POST['password']); $con = mysql_connect("localhost", "root", "g00dfor@boy"); if(!$con){ die(mysql_error()); } mysql_select_db("users", $con) or die(mysql_error()); $login = "SELECT * FROM `users` WHERE (usrname = '$usrname' AND password = '$password')"; $result = mysql_query($login); if(mysql_num_rows($result) == 1 { $_SESSION = true; header('Location: indexlogin.php'); } else { echo = "Wrong username or password." ; } ?> indexlogin.php just echoes "Login successful." What am I doing wrong? Oh, and just FYI- my database is "users" and my table is "data".

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  • ob_start() is partially capturing data

    - by AAA
    I am using the following code: PHP: // Generate Guid function NewGuid() { $s = strtoupper(uniqid(rand(),true)); $guidText = substr($s,0,8) . '-' . substr($s,8,4) . '-' . substr($s,12,4). '-' . substr($s,16,4). '-' . substr($s,20); return $guidText; } // End Generate Guid $Guid = NewGuid(); $alphabet = '123456789abcdefghijkmnopqrstuvwxyzABCDEFGHJKLMNPQRSTUVWXYZ'; function base_encode($num, $alphabet) { $base_count = strlen($alphabet); $encoded = ''; while ($num >= $base_count) { $div = $num/$base_count; $mod = ($num-($base_count*intval($div))); $encoded = $alphabet[$mod] . $encoded; $num = intval($div); } if ($num) $encoded = $alphabet[$num] . $encoded; return $encoded; } function base_decode($num, $alphabet) { $decoded = 0; $multi = 1; while (strlen($num) > 0) { $digit = $num[strlen($num)-1]; $decoded += $multi * strpos($alphabet, $digit); $multi = $multi * strlen($alphabet); $num = substr($num, 0, -1); } return $decoded; } ob_start(); echo base_encode($Guid, $alphabet); //should output: bUKpk $theid = ob_get_contents(); ob_get_clean(); The problem: When i echo $theid, it shows the complete entry, but as it is being inserted into the database, only the first entry in the sequence gets inserted, for example for the entry buKPK, only 'b' is being inserted not the rest.

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  • Can you automatically create a mysqldump file that doesn't enforce foreign key constraints?

    - by Tai Squared
    When I run a mysqldump command on my database and then try to import it, it fails as it attempts to create the tables alphabetically, even though they may have a foreign key that references a table later in the file. There doesn't appear to be anything in the documentation and I've found answers like this that say to update the file after it's created to include: set FOREIGN_KEY_CHECKS = 0; ...original mysqldump file contents... set FOREIGN_KEY_CHECKS = 1; Is there no way to automatically set those lines or export the tables in the necessary order (without having to manually specify all table names as that can be tedious and error prone)? I could wrap those lines in a script, but was wondering if there is an easy way to ensure I can dump a file and then import it without manually updating it.

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  • In SQL, in what situation do we want to Index a field in a table, or 2 fields in a table at the same

    - by Jian Lin
    In SQL, it is obvious that whenever we want to do a search on millions of record, say CustomerID in a Transactios table, then we want to add an index for CustomerID. Is another situation we want to add an index to a field when we need to do inner join or outer join using that field as a criteria? Such as Inner join on t1.custumerID = t2.customerID. Then if we don't have an index on customerID on both tables, we are looking at O(n^2) because we need to loop through the 2 tables sequentially. If we have index on customerID on both tables, then it becomes O( (log n) ^ 2 ) and it is much faster. Any other situation where we want to add an index to a field in a table? What about adding index for 2 fields combined in a table. That is, one index, for 2 fields together?

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  • Need help in Select query..

    - by Parth
    If I have four rows for a field with different values with other fields similar and then other four rows with same condition, as given below here u can see there different rows for insert with only difference in the "newvalue" and "field" excluding "id" for the table_name=jos_menu, operation=INSERT and live=0 now here what select query should be used to get only single row from the table on every change of table_name...??

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  • Is it possible to LIMIT results from a JOIN query?

    - by Arms
    I've got a query that currently queries a Post table while LEFT JOINing a Comment table. It fetches all Posts and their respective Comments. However, I want to limit the number of Comments returned. I tried adding a sub-select, but ran into errors if I didn't LIMIT the results to 1. I'm really not sure how to go about this while still using only one query. Is this possible?

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  • PHP errors -> Warning: mysqli_stmt::execute(): Couldn't fetch mysqli_stmt | Warning: mysqli_stmt::c

    - by Tunji Gbadamosi
    I keep getting this error while trying to modify some tables. Here's my code: /** <- line 320 * * @param array $guests_array * @param array $tickets_array * @param integer $seat_count * @param integer $order_count * @param integer $guest_count */ private function book_guests($guests_array, $tickets_array, &$seat_count, &$order_count, &$guest_count){ /* @var $guests_array ArrayObject */ $sucess = false; if(sizeof($guests_array) >= 1){ //$this->mysqli->autocommit(FALSE); //insert the guests into guest, person, order, seat $menu_stmt = $this->mysqli->prepare("SELECT id FROM menu WHERE name=?"); $menu_stmt->bind_param('s',$menu); //$menu_stmt->bind_result($menu_id); $table_stmt = $this->mysqli->prepare("SELECT id FROM tables WHERE name=?"); $table_stmt->bind_param('s',$table); //$table_stmt->bind_result($table_id); $seat_stmt = $this->mysqli->prepare("SELECT id FROM seat WHERE name=? AND table_id=?"); $seat_stmt->bind_param('ss',$seat, $table_id); //$seat_stmt->bind_result($seat_id); for($i=0;$i<sizeof($guests_array);$i++){ $menu = $guests_array[$i]['menu']; $table = $guests_array[$i]['table']; $seat = $guests_array[$i]['seat']; //get menu id if($menu_stmt->execute()){ $menu_stmt->bind_result($menu_id); while($menu_stmt->fetch()) ; } $menu_stmt->close(); //get table id if($table_stmt->execute()){ $table_stmt->bind_result($table_id); while($table_stmt->fetch()) ; } $table_stmt->close(); //get seat id if($seat_stmt->execute()){ $seat_stmt->bind_result($seat_id); while($seat_stmt->fetch()) ; } $seat_stmt->close(); $dob = $this->create_date($guests_array[$i]['dob_day'], $guests_array[$i]['dob_month'], $guests_array[$i]['dob_year']); $id = $this->add_person($guests_array[$i]['first_name'], $guests_array[$i]['surname'], $dob, $guests_array[$i]['sex']); if(is_string($id)){ $seat = $this->add_seat($table_id, $seat_id, $id); /* @var $tickets_array ArrayObject */ $guest = $this->add_guest($id,$tickets_array[$i+1],$menu_id, $this->volunteer_id); /* @var $order integer */ $order = $this->add_order($this->volunteer_id, $table_id, $seat_id, $id); if($guest == 1 && $seat == 1 && $order == 1){ $seat_count += $seat; $guest_count += $guest; $order_count += $order; $success = true; } } } } return $success; } <- line 406 Here are the warnings: The person PRSN10500000LZPH has been added to the guest tablePRSN10500000LZPH added to table (1), seat (1)The order for person(PRSN10500000LZPH) is registered with volunteer (PRSN10500000LZPH) at table (1) and seat (1)PRSN10600000LZPH added to table (1), seat (13)The person PRSN10600000LZPH has been added to the guest tableThe order for person(PRSN10600000LZPH) is registered with volunteer (PRSN10500000LZPH) at table (1) and seat (13) Warning: mysqli_stmt::execute(): Couldn't fetch mysqli_stmt in /Users/olatunjigbadamosi/Sites/ST_Ambulance/FormDB.php on line 358 Warning: mysqli_stmt::close(): Couldn't fetch mysqli_stmt in /Users/olatunjigbadamosi/Sites/ST_Ambulance/FormDB.php on line 363 Warning: mysqli_stmt::execute(): Couldn't fetch mysqli_stmt in /Users/olatunjigbadamosi/Sites/ST_Ambulance/FormDB.php on line 366 Warning: mysqli_stmt::close(): Couldn't fetch mysqli_stmt in /Users/olatunjigbadamosi/Sites/ST_Ambulance/FormDB.php on line 371 Warning: mysqli_stmt::execute(): Couldn't fetch mysqli_stmt in /Users/olatunjigbadamosi/Sites/ST_Ambulance/FormDB.php on line 374 Warning: mysqli_stmt::close(): Couldn't fetch mysqli_stmt in /Users/olatunjigbadamosi/Sites/ST_Ambulance/FormDB.php on line 379 PRSN10700000LZPH added to table (1), seat (13)The person PRSN10700000LZPH has been added to the guest tableThe order for person(PRSN10700000LZPH) is registered with volunteer (PRSN10500000LZPH) at table (1) and seat (13) Warning: mysqli_stmt::execute(): Couldn't fetch mysqli_stmt in /Users/olatunjigbadamosi/Sites/ST_Ambulance/FormDB.php on line 358 Warning: mysqli_stmt::close(): Couldn't fetch mysqli_stmt in /Users/olatunjigbadamosi/Sites/ST_Ambulance/FormDB.php on line 363 Warning: mysqli_stmt::execute(): Couldn't fetch mysqli_stmt in /Users/olatunjigbadamosi/Sites/ST_Ambulance/FormDB.php on line 366 Warning: mysqli_stmt::close(): Couldn't fetch mysqli_stmt in /Users/olatunjigbadamosi/Sites/ST_Ambulance/FormDB.php on line 371 Warning: mysqli_stmt::execute(): Couldn't fetch mysqli_stmt in /Users/olatunjigbadamosi/Sites/ST_Ambulance/FormDB.php on line 374 Warning: mysqli_stmt::close(): Couldn't fetch mysqli_stmt in /Users/olatunjigbadamosi/Sites/ST_Ambulance/FormDB.php on line 379 PRSN10800000LZPH added to table (1), seat (13)The person PRSN10800000LZPH has been added to the guest tableThe order for person(PRSN10800000LZPH) is registered with volunteer (PRSN10500000LZPH) at table (1) and seat (13)

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  • Complex Join - involving date ranges and sum...

    - by calumbrodie
    I have two tables that I need to join... I want to join table1 and table2 on 'id' - however in table two id is not unique. I only want one value returned for table two, and this value represents the sum of a column called 'total_sold' - within a specified date range (say one month).. SELECT ta.id, tb.total_sold as total_sold_this_week FROM table_a as ta LEFT JOIN table_b as tb ON ta.id=tb.id AND tb.date_sold BETWEEN ADDDATE(NOW(),INTERVAL -3 WEEK) AND NOW() this works but does not SUM the rows - only returning one row for each id... how do I get the sum from table b instead of only one row??? Please criticise if format of question could use more work - I can rewrite and provide sample data if required - this is a trivialised version of a much larger problem. -Thanks

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