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  • How to add SQL elements to an array in PHP

    - by DanLeaningphp
    So this question is probably pretty basic. I am wanting to create an array from selected elements from a SQL table. I am currently using: $rcount = mysql_num_rows($result); for ($j = 0; $j <= $rcount; $j++) { $row = mysql_fetch_row($result); $patients = array($row[0] => $row[2]); } I would like this to return an array like this: $patients = (bob=>1, sam=>2, john=>3, etc...) Unfortunately, in its current form, this code is either copying nothing to the array or only copying the last element.

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  • how to link table to table

    - by Niño Seymour L. Rodriguez
    I am a comsci student and I'm taking up database now. I got a problem in or should I say I dont know how to link table to table. It is not like you'll just use a foreign key and connect it to the primary key. The outcome should be like this: In the table Course there are three fields namely "course_id", "Description" and "subjects". When you click the name field Subject, a table named Subject should appear. Can you help me with this? hope you understnd my grammar, hehe..im not good in english......it will be a big help if you can answer it.........thank you po..............

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  • update myqsl table

    - by Simon
    how can i write the query, to update the table videos, and set the value of field name to 'something' where the average is max(), or UPDATE the table, where average has the second value by size!!! i think the query must look like this!!! UPDATE videos SET name = 'something' WHERE average IN (SELECT `average` FROM `videos` ORDER BY `average` DESC LIMIT 1) but it doesn't work!!!

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  • What's wrong with this SQL query?

    - by ThinkingInBits
    I have two tables: photographs, and photograph_tags. Photograph_tags contains a column called photograph_id (id in photographs). You can have many tags for one photograph. I have a photograph row related to three tags: boy, stream, and water. However, running the following query returns 0 rows SELECT p.* FROM photographs p, photograph_tags c WHERE c.photograph_id = p.id AND (c.value IN ('dog', 'water', 'stream')) GROUP BY p.id HAVING COUNT( p.id )=3 Is something wrong with this query?

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  • Editing/Updating one of the results in a search query.

    - by eggman20
    Hi guys. I'm creating a page that searches for an item and then be able to edit/update it. I was able to do it when it returns just one result but when it gives me multiple results I could only edit the very last item. Below is my code: ....... $dj =$_POST[djnum]; $sql= "SELECT * From dj WHERE datajack LIKE '$dj%'"; $result = mysql_query($sql); //more code in here// while ($info =mysql_fetch_array($result)) { // display the result echo "<form action=\"dj_update.php\" method=\"POST\"><input type=\"hidden\" name=\"djnumber\" value=\"".$info['datajack']."\">"; echo "<tr><td>DJ ".$info['datajack']."</td>"; echo "<td>".$info['building']."&nbsp;</td>"; echo "<td>Rm ".$info['room']."&nbsp;</td>"; echo "<td>".$info['switch']."&nbsp;</td>"; echo "<td>".$info['port']."&nbsp;</td>"; echo "<td>".$info['notes']."&nbsp;</td>"; echo "<td style=\"text-align:center;\"><input type=\"Submit\" value=\"Edit\" ></td></tr>"; } // more code here // Then this is the screen shot of the result: The idea is the user should be able to click on "Edit" and be able to edit/update that particular item. But when I click any of the Edit button I could only edit the last item. What am I missing here? Is there an easier way to do this? Thanks guys and Happy new year!

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  • Recursive Function To Create Array

    - by mTuran
    Hi, i use kohana framework and i am trying to code recursive function to create category tree. My Categories Table id int(11) NO PRI NULL auto_increment name varchar(50) NO NULL parent_id int(11) NO NULL projects_count int(11) NO NULL My Example Which Is Not Work public static function category_list($parent_id = 0) { $result = Database::instance()->query(' SELECT name, projects_count FROM project_categories WHERE parent_id = ?', array($parent_id) ); $project_categories = array(); foreach($result as $row) { $project_categories[] = $row; Project_Categories_Model::factory()->category_list($parent_id + 1); } return $project_categories; }

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  • Selecting all but one field?

    - by gsquare567
    instead of SELECT * FROM mytable, i would like to select all fields EXCEPT one (namely, the 'serialized' field, which stores a serialized object). this is because i think that losing that field will speed up my query by a lot. however, i have so many fields and am quite the lazy guy. is there a way to say... `SELECT ALL_ROWS_EXCEPT(serialized) FROM mytable` ? thanks!

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  • How to add space between the images being fetched using database through php

    - by ParveenArora
    I am using following code to fetch images using database with php. while($row = mysql_fetch_array($result)) //To excute result query { echo "<a href='http://".$row['website']."' target='_blank'><img src=\"" . $PathImage . $row['logo'] . "\" height = $FooterWidth /></a>XX; } Here I am using $row[logo] is fetching the path of images stored on the server and XX to put the spaced between the images having the same color of text XX as background, and but I want to use the proper method I know this can be done using table but I want to do it without using table. Any Suggestions?

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  • Correlate GROUP BY and LEFT JOIN on multiple criteria to show latest record?

    - by Sunbird
    In a simple stock management database, quantity of new stock is added and shipped until quantity reaches zero. Each stock movement is assigned a reference, only the latest reference is used. In the example provided, the latest references are never shown, the stock ID's 1,4 should have references charlie, foxtrot respectively, but instead show alpha, delta. How can a GROUP BY and LEFT JOIN on multiple criteria be correlated to show the latest record? http://sqlfiddle.com/#!2/6bf37/107 CREATE TABLE stock ( id tinyint PRIMARY KEY, quantity int, parent_id tinyint ); CREATE TABLE stock_reference ( id tinyint PRIMARY KEY, stock_id tinyint, stock_reference_type_id tinyint, reference varchar(50) ); CREATE TABLE stock_reference_type ( id tinyint PRIMARY KEY, name varchar(50) ); INSERT INTO stock VALUES (1, 10, 1), (2, -5, 1), (3, -5, 1), (4, 20, 4), (5, -10, 4), (6, -5, 4); INSERT INTO stock_reference VALUES (1, 1, 1, 'Alpha'), (2, 2, 1, 'Beta'), (3, 3, 1, 'Charlie'), (4, 4, 1, 'Delta'), (5, 5, 1, 'Echo'), (6, 6, 1, 'Foxtrot'); INSERT INTO stock_reference_type VALUES (1, 'Customer Reference'); SELECT stock.id, SUM(stock.quantity) as quantity, customer.reference FROM stock LEFT JOIN stock_reference AS customer ON stock.id = customer.stock_id AND stock_reference_type_id = 1 GROUP BY stock.parent_id

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  • how to have defined connection within function for pdo communication with DB

    - by Scarface
    hey guys I just started trying to convert my query structure to PDO and I have come across a weird problem. When I call a pdo query connection within a function and the connection is included outside the function, the connection becomes undefined. Anyone know what I am doing wrong here? I was just playing with it, my example is below. include("includes/connection.php"); function query(){ $user='user'; $id='100'; $sql = 'SELECT * FROM users'; $stmt = $conn->prepare($sql); $result=$stmt->execute(array($user, $id)); // now iterate over the result as if we obtained // the $stmt in a call to PDO::query() while($r = $stmt->fetch(PDO::FETCH_ASSOC)) { echo "$r[username] $r[id] \n"; } } query();

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  • SQL Query Math Gymnastics

    - by keruilin
    I have two tables of concern here: users and race_weeks. User has many race_weeks, and race_week belongs to User. Therefore, user_id is a fk in the race_weeks table. I need to perform some challenging math on fields in the race_weeks table in order to return users with the most all-time points. Here are the fields that we need to manipulate in the race_weeks table. races_won (int) races_lost (int) races_tied (int) points_won (int, pos or neg) recordable_type(varchar, Robots can race, but we're only concerned about type 'User') Just so that you fully understand the business logic at work here, over the course of a week a user can participate in many races. The race_week record represents the summary results of the user's races for that week. A user is considered active for the week if races_won, races_lost, or races_tied is greater than 0. Otherwise the user is inactive. So here's what we need to do in our query in order to return users with the most points won (actually net_points_won): Calculate each user's net_points_won (not a field in the DB). To calculate net_points, you take (1000 * count_of_active_weeks) - sum(points__won). (Why 1000? Just imagine that every week the user is spotted a 1000 points to compete and enter races. We want to factor-out what we spot the user because the user could enter only one race for the week for 100 points, and be sitting on 900, which we would skew who actually EARNED the most points.) This one is a little convoluted, so let me know if I can clarify further.

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  • uploading image & getting back from database

    - by Anup Prakash
    Putting a set of code which is pushing image to database and fetching back from database: <!-- <?php error_reporting(0); // Connect to database $errmsg = ""; if (! @mysql_connect("localhost","root","")) { $errmsg = "Cannot connect to database"; } @mysql_select_db("test"); $q = <<<CREATE create table image ( pid int primary key not null auto_increment, title text, imgdata longblob, friend text) CREATE; @mysql_query($q); // Insert any new image into database if (isset($_POST['submit'])) { move_uploaded_file($_FILES['imagefile']['tmp_name'],"latest.img"); $instr = fopen("latest.img","rb"); $image = addslashes(fread($instr,filesize("latest.img"))); if (strlen($instr) < 149000) { $image_query="insert into image (title, imgdata,friend) values (\"". $_REQUEST['title']. "\", \"". $image. "\",'".$_REQUEST['friend']."')"; mysql_query ($image_query) or die("query error"); } else { $errmsg = "Too large!"; } $resultbytes=''; // Find out about latest image $query = "select * from image where pid=1"; $result = @mysql_query("$query"); $resultrow = @mysql_fetch_assoc($result); $gotten = @mysql_query("select * from image order by pid desc limit 1"); if ($row = @mysql_fetch_assoc($gotten)) { $title = htmlspecialchars($row[title]); $bytes = $row[imgdata]; $resultbytes = $row[imgdata]; $friend=$row[friend]; } else { $errmsg = "There is no image in the database yet"; $title = "no database image available"; // Put up a picture of our training centre $instr = fopen("../wellimg/ctco.jpg","rb"); $bytes = fread($instr,filesize("../wellimg/ctco.jpg")); } if ($resultbytes!='') { echo $resultbytes; } } ?> <html> <head> <title>Upload an image to a database</title> </head> <body bgcolor="#FFFF66"> <form enctype="multipart/form-data" name="file_upload" method="post"> <center> <div id="image" align="center"> <h2>Heres the latest picture</h2> <font color=red><?php echo $errmsg; ?></font> <b><?php echo $title ?></center> </div> <hr> <h2>Please upload a new picture and title</h2> <table align="center"> <tr> <td>Select image to upload: </td> <td><input type="file" name="imagefile"></td> </tr> <tr> <td>Enter the title for picture: </td> <td><input type="text" name="title"></td> </tr> <tr> <td>Enter your friend's name:</td> <td><input type="text" name="friend"></td> </tr> <tr> <td><input type="submit" name="submit" value="submit"></td> <td></td> </tr> </table> </form> </body> </html> --> Above set of code has one problem. The problem is whenever i pressing the "submit" button. It is just displaying the image on a page. But it is leaving all the html codes. even any new line message after the // Printing image on browser echo $resultbytes; //************************// So, for this i put this set of code in html tag: This is other sample code: <!-- <?php error_reporting(0); // Connect to database $errmsg = ""; if (! @mysql_connect("localhost","root","")) { $errmsg = "Cannot connect to database"; } @mysql_select_db("test"); $q = <<<CREATE create table image ( pid int primary key not null auto_increment, title text, imgdata longblob, friend text) CREATE; @mysql_query($q); // Insert any new image into database if (isset($_POST['submit'])) { move_uploaded_file($_FILES['imagefile']['tmp_name'],"latest.img"); $instr = fopen("latest.img","rb"); $image = addslashes(fread($instr,filesize("latest.img"))); if (strlen($instr) < 149000) { $image_query="insert into image (title, imgdata,friend) values (\"". $_REQUEST['title']. "\", \"". $image. "\",'".$_REQUEST['friend']."')"; mysql_query ($image_query) or die("query error"); } else { $errmsg = "Too large!"; } $resultbytes=''; // Find out about latest image $query = "select * from image where pid=1"; $result = @mysql_query("$query"); $resultrow = @mysql_fetch_assoc($result); $gotten = @mysql_query("select * from image order by pid desc limit 1"); if ($row = @mysql_fetch_assoc($gotten)) { $title = htmlspecialchars($row[title]); $bytes = $row[imgdata]; $resultbytes = $row[imgdata]; $friend=$row[friend]; } else { $errmsg = "There is no image in the database yet"; $title = "no database image available"; // Put up a picture of our training centre $instr = fopen("../wellimg/ctco.jpg","rb"); $bytes = fread($instr,filesize("../wellimg/ctco.jpg")); } } ?> <html> <head> <title>Upload an image to a database</title> </head> <body bgcolor="#FFFF66"> <form enctype="multipart/form-data" name="file_upload" method="post"> <center> <div id="image" align="center"> <h2>Heres the latest picture</h2> <?php if ($resultbytes!='') { // Printing image on browser echo $resultbytes; } ?> <font color=red><?php echo $errmsg; ?></font> <b><?php echo $title ?></center> </div> <hr> <h2>Please upload a new picture and title</h2> <table align="center"> <tr> <td>Select image to upload: </td> <td><input type="file" name="imagefile"></td> </tr> <tr> <td>Enter the title for picture: </td> <td><input type="text" name="title"></td> </tr> <tr> <td>Enter your friend's name:</td> <td><input type="text" name="friend"></td> </tr> <tr> <td><input type="submit" name="submit" value="submit"></td> <td></td> </tr> </table> </form> </body> </html> --> ** But in this It is showing the image in format of special charaters and digits. 1) So, Please help me to print the image with some HTML code. So that i can print it in my form to display the image. 2) Is there any way to convert the database image into real image, so that i can store it into my hard-disk and call it from tag? Please help me.

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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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  • URL Rewrite query database?

    - by Liam
    Im trying to understand how URL rewriting works. I have the following link... mysite.com/profile.php?id=23 I want to rewrite the above url with the Users first and last name... mysite.com/directory/liam-gallagher From what Ive read however you specify the rule for what the url should be output as, But how do i query my table to get each users name? Sorry if this is hard to understand, ive confused myself!

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  • PHP While() Stop Looping

    - by Axel
    Hi, i have a php loop which displays only one record even if there is hundreds. here is the code: <?php $result1 = mysql_query("SELECT * FROM posts") or die(mysql_error()); $numexem = mysql_num_rows($result1); $s="0"; while($s<$numexem){ $postid=mysql_result($result1,$s,"id"); echo "Post id:".$postid; $result2 = mysql_query("SELECT * FROM pics WHERE postid='$postid'") or die(mysql_error()); $rows = mysql_fetch_array($result2) or die(mysql_error()); $pnum = mysql_num_rows($result2); echo " There is ".$pnum." Attached Pictures"; $s++; } ?> I'm wondering if the loop stop because there is other SQL query inside it or what? and i don't think so. Thanks

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  • how to set a status

    - by ejah85
    hello guys..here i've a problem where i want to set the status whether it is approved or reject.. the condition are if admin select the registration number and driver name, that means the status is approve otherwise, if admin fill up the reason, that means the request is reject.. here is the code to set status if ($reason =='null'){ $query2 = "UPDATE usage SET status ='APPROVED' WHERE '$bookingno'=bookingno"; $result2 = @mysql_query($query2); } elseif (($regno =='null')&&($d_name =='null')) { $query3 = "UPDATE usage SET status ='REJECT' WHERE '$bookingno'=bookingno"; $result3 = @mysql_query($query3); } when i save the data, the status field are not updates..

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  • Can I join two tables whereby the joined table is sorted by a certain column?

    - by Ferdy
    I'm not much of a database guru so I need some help on a query I'm working on. In my photo community project I want to richly visualize tags by not only showing the tag name and counter (# of images inside them), I also want to show a thumb of the most popular image inside the tag (most karma). The table setup is as follow: Image table holds basic image metadata, important is the karma field Imagefile table holds multiple entries per image, one for each format Tag table holds tag definitions Tag_map table maps tags to images In my usual trial and error query authoring I have come this far: SELECT * FROM (SELECT tag.name, tag.id, COUNT(tag_map.tag_id) as cnt FROM tag INNER JOIN tag_map ON (tag.id = tag_map.tag_id) INNER JOIN image ON tag_map.image_id = image.id INNER JOIN imagefile on image.id = imagefile.image_id WHERE imagefile.type = 'smallthumb' GROUP BY tag.name ORDER BY cnt DESC) as T1 WHERE cnt > 0 ORDER BY cnt DESC [column clause of inner query snipped for the sake of simplicity] This query gives me somewhat what I need. The outer query makes sure that only tags are returned for which there is at least 1 image. The inner query returns the tag details, such as its name, count (# of images) and the thumb. In addition, I can sort the inner query as I want (by most images, alphabetically, most recent, etc) So far so good. The problem however is that this query does not match the most popular image (most karma) of the tag, it seems to always take the most recent one in the tag. How can I make sure that the most popular image is matched with the tag?

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  • Unnecessary Error Message Being Displayed

    - by ThatMacLad
    I've set up a form to update my blog and it was working fine up until about this morning. It keeps on turning up with an Invalid Entry ID error on the edit post page when I click the update button despite the fact that it updates the homepage. All help is seriously appreciated. <html> <head> <title>Ultan's Blog | New Post</title> <link rel="stylesheet" href="css/editpost.css" type="text/css" /> </head> <body> <div class="new-form"> <div class="header"> </div> <div class="form-bg"> <?php mysql_connect ('localhost', 'root', 'root') ; mysql_select_db ('tmlblog'); if (isset($_POST['update'])) { $id = htmlspecialchars(strip_tags($_POST['id'])); $month = htmlspecialchars(strip_tags($_POST['month'])); $date = htmlspecialchars(strip_tags($_POST['date'])); $year = htmlspecialchars(strip_tags($_POST['year'])); $time = htmlspecialchars(strip_tags($_POST['time'])); $entry = $_POST['entry']; $title = htmlspecialchars(strip_tags($_POST['title'])); if (isset($_POST['password'])) $password = htmlspecialchars(strip_tags($_POST['password'])); else $password = ""; $entry = nl2br($entry); if (!get_magic_quotes_gpc()) { $title = addslashes($title); $entry = addslashes($entry); } $timestamp = strtotime ($month . " " . $date . " " . $year . " " . $time); $result = mysql_query("UPDATE php_blog SET timestamp='$timestamp', title='$title', entry='$entry', password='$password' WHERE id='$id' LIMIT 1") or print ("Can't update entry.<br />" . mysql_error()); header("Location: post.php?id=" . $id); } if (isset($_POST['delete'])) { $id = (int)$_POST['id']; $result = mysql_query("DELETE FROM php_blog WHERE id='$id'") or print ("Can't delete entry.<br />" . mysql_error()); if ($result != false) { print "The entry has been successfully deleted from the database."; exit; } } if (!isset($_GET['id']) || empty($_GET['id']) || !is_numeric($_GET['id'])) { die("Invalid entry ID."); } else { $id = (int)$_GET['id']; } $result = mysql_query ("SELECT * FROM php_blog WHERE id='$id'") or print ("Can't select entry.<br />" . $sql . "<br />" . mysql_error()); while ($row = mysql_fetch_array($result)) { $old_timestamp = $row['timestamp']; $old_title = stripslashes($row['title']); $old_entry = stripslashes($row['entry']); $old_password = $row['password']; $old_title = str_replace('"','\'',$old_title); $old_entry = str_replace('<br />', '', $old_entry); $old_month = date("F",$old_timestamp); $old_date = date("d",$old_timestamp); $old_year = date("Y",$old_timestamp); $old_time = date("H:i",$old_timestamp); } ?> <form method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>"> <p><input type="hidden" name="id" value="<?php echo $id; ?>" /> <strong><label for="month">Date (month, day, year):</label></strong> <select name="month" id="month"> <option value="<?php echo $old_month; ?>"><?php echo $old_month; ?></option> <option value="January">January</option> <option value="February">February</option> <option value="March">March</option> <option value="April">April</option> <option value="May">May</option> <option value="June">June</option> <option value="July">July</option> <option value="August">August</option> <option value="September">September</option> <option value="October">October</option> <option value="November">November</option> <option value="December">December</option> </select> <input type="text" name="date" id="date" size="2" value="<?php echo $old_date; ?>" /> <select name="year" id="year"> <option value="<?php echo $old_year; ?>"><?php echo $old_year; ?></option> <option value="2004">2004</option> <option value="2005">2005</option> <option value="2006">2006</option> <option value="2007">2007</option> <option value="2008">2008</option> <option value="2009">2009</option> <option value="2010">2010</option> </select> <strong><label for="time">Time:</label></strong> <input type="text" name="time" id="time" size="5" value="<?php echo $old_time; ?>" /></p> <p><strong><label for="title">Title:</label></strong> <input type="text" name="title" id="title" value="<?php echo $old_title; ?>" size="40" /> </p> <p><strong><label for="password">Password protect?</label></strong> <input type="checkbox" name="password" id="password" value="1"<?php if($old_password == 1) echo " checked=\"checked\""; ?> /></p> <p><textarea cols="80" rows="20" name="entry" id="entry"><?php echo $old_entry; ?></textarea></p> <p><input type="submit" name="update" id="update" value="Update"></p> </form> <p><strong>Be absolutely sure that this is the post that you wish to remove from the blog!</strong><br /> </p> <form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post"> <input type="hidden" name="id" id="id" value="<?php echo $id; ?>" /> <input type="submit" name="delete" id="delete" value="Delete" /> </form> </div> </div> </div> <div class="bottom"></div> </body> </html>

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  • Dynamically removing records when certain columns = 0; data cleansing

    - by cdburgess
    I have a simple card table: CREATE TABLE `users_individual_cards` ( `id` int(11) NOT NULL AUTO_INCREMENT, `user_id` char(36) NOT NULL, `individual_card_id` int(11) NOT NULL, `own` int(10) unsigned NOT NULL, `want` int(10) unsigned NOT NULL, `trade` int(10) unsigned NOT NULL, PRIMARY KEY (`id`), UNIQUE KEY `user_id` (`user_id`,`individual_card_id`), KEY `user_id_2` (`user_id`), KEY `individual_card_id` (`individual_card_id`) ) ENGINE=MyISAM DEFAULT CHARSET=latin1 AUTO_INCREMENT=1; I have ajax to add and remove the records based on OWN, WANT, and TRADE. However, if the user removes all of the OWN, WANT, and TRADE cards, they go to zero but it will leave the record in the database. I would prefer to have the record removed. Is checking after each "update" to see if all the columns = 0 the only way to do this? Or can I set a conditional trigger with something like: //psuedo sql AFTER update IF (OWN = 0, WANT = 0, TRADE = 0) DELETE What is the best way to do this? Can you help with the syntax?

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  • One to two relationship in Doctrine with YAML

    - by Jeremy DeGroot
    I'm working on my first Symfony project with Doctrine, and I've run into a hitch. I'm trying to express a game with two players. The relationship I want to have is PlayerOne and PlayerTwo each being keyed to an ID in the Users table. This is part of what I've got so far: Game: actAs: { Timestampable:- } columns: id: { type: integer, notnull: true, unique: true } startDate: { type: timestamp, notnull: true } playerOne: { type: integer, notnull: true } playerTwo: { type: integer, notnull: true } winner: { type: integer, notnull:true, default:0 } relations: User: { onUpdate: cascade, local: playerOne, foreign: id} User: { onUpdate: cascade, local: playerTwo, foreign: id} That doesn't work. It builds fine, but the SQL it generates only includes a constraint for playerTwo. I've tried a few other things: User: { onUpdate: cascade, local: [playerOne, playerTwo], foreign: id} Also: User: [{ onUpdate: cascade, local: playerOne, foreign: id}, { onUpdate: cascade, local: playerTwo, foreign: id}] Those last two throw errors when I try to build. Is there anyone out there who understands what I'm trying to do and can help me achieve it?

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  • Favouriting things in a database - most efficient method of keeping track?

    - by a2h
    I'm working on a forum-like webapp where I'd like to allow users to favourite an item so that they can keep track of it, and also so that others can see how many times an item's been favourited. The problem is, I'm unsure on the best practices for databases, which includes this situation. I have two ideas in my head on how to do this: Add an extra column to the user table and store things like so: "|2|5|73|" Add an extra table with at least two columns, one for referencing an item, the other for referencing a user. I feel uncomfortable about going for the second method as it involves an extra table, and potentially more queries would be required. Perhaps these beliefs aren't an issue, as I have little understanding of databases beyond simply working with table layouts and basic queries.

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