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  • Incorrect string encodings

    - by James
    Note: I have read all of the related PHP, UTF-8, character encoding articles that are usually suggested, but my question relates to data inserted before I applied such techniques. I am wishing to retrospectively fix all character encoding problems. Now all connections are set as utf8 using PDO. PDO::MYSQL_ATTR_INIT_COMMAND => 'SET NAMES utf8' Unfortunately, a large amount of data was inserted that is of questionable encoding before I had implemented correct character encoding practices. As displayed by: $sql = "SELECT name FROM data LIMIT 3"; foreach ($pdo->query($sql) as $row) { $name = $row['name']; echo $name . "\n"; echo utf8_encode($name) . "\n"; echo utf8_decode($name) . "\n"; echo htmlspecialchars($name, ENT_QUOTES, 'UTF-8') . "\n"; echo htmlspecialchars(utf8_encode($name), ENT_QUOTES, 'UTF-8') . "\n"; echo htmlspecialchars(utf8_decode($name), ENT_QUOTES, 'UTF-8') . "\n"; echo '<hr/>'; } Which produces: Antonín Dvořák AntonÃÆÃ­n DvoÃâ¦Ãâ¢ÃÆÃ¡k Anton??­n Dvo??????¡k Antonín Dvořák AntonÃÆÃ­n DvoÃâ¦Ãâ¢ÃÆÃ¡k ---------- Ô±Ö€Õ¡Õ´ Ô½Õ¡Õ¹Õ¡Õ¿Ö€ÕµÕ¡Õ¶ ñÃâ¬Ã¡Ã´ ýáùáÿÃâ¬ÃµÃ¡Ã¶ ????? ?????????? Ô±Ö€Õ¡Õ´ Ô½Õ¡Õ¹Õ¡Õ¿Ö€ÕµÕ¡Õ¶ ñÃâ¬Ã¡Ã´ ýáùáÿÃâ¬ÃµÃ¡Ã¶ ---------- Tiësto Tiësto Tiësto Tiësto Tiësto Tiësto ---------- When removing 'SET NAMES utf8' with PDO it produces the data: Antonín DvoÅák Antonín DvoÃÂák Antonín Dvorák Antonín DvoÅák Antonín DvoÃÂák Antonín Dvorák ---------- ???? ????????? Ô±ÖÕ¡Õ´ Ô½Õ¡Õ¹Õ¡Õ¿ÖÕµÕ¡Õ¶ ???? ????????? ???? ????????? Ô±ÖÕ¡Õ´ Ô½Õ¡Õ¹Õ¡Õ¿ÖÕµÕ¡Õ¶ ???? ????????? ---------- Tiësto Tiësto Ti?sto Tiësto Tiësto ---------- And here is a dump of the database rows concerned: DROP TABLE IF EXISTS `data`; CREATE TABLE IF NOT EXISTS `data` ( `id` int(10) unsigned NOT NULL AUTO_INCREMENT, `name` varchar(80) NOT NULL, PRIMARY KEY (`id`), KEY `name` (`name`(10)), ) ENGINE=InnoDB DEFAULT CHARSET=utf8 AUTO_INCREMENT=0; INSERT INTO `data` (`id`, `name`) VALUES (0, 'Antonín Dvořák'), (1, 'Ô±Ö€Õ¡Õ´ Ô½Õ¡Õ¹Õ¡Õ¿Ö€ÕµÕ¡Õ¶'), (2, 'Tiësto'); The 3rd and 6th lines of the 3rd row "Tiësto" are then correctly echoed. I'm just unsure what is the best way to correct encodings/detect the encodings of bad strings and correct, etc.

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  • query excuting problem

    - by srini-r85
    hi, i tried to execute following query in php script. $db_selected = mysql_select_db("lumiinc1_sndemo1", $con); if ($db_selected) { echo "database connected"; } else { die ("Can\'t use db : " . mysql_error()); } $sql = "INSERT INTO `markers` ( `name`, `address`, `lat`, `lng`, `id` ) SELECT `name`, `street`, `latitude`, `longitude`, `lid` FROM `location` WHERE NOT EXISTS ( SELECT * FROM `markers` WHERE `location`.`lid` = `markers`.`id` )"; $result = mysql_query($sql); if ($result) { echo "Query executed OK"; } else { die("Invalid query: " . mysql_error()); } script does not show any error.also query executed.but i didn't get my expected result.at the same i try this query in phpmyAdmin i got my expected result. i dont know the cause of this problem. plz any one find the problem . thanks

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  • I can't insert data into my database

    - by Ken
    I don't know why, but my data doesn't go into my database 'users' with the table 'data'. <html> <body> <?php date_default_timezone_set("America/Los_Angeles"); include("mainmenu.php"); $con = mysql_connect("localhost", "root", "g00dfor@boy"); if(!$con){ die(mysql_error()); } $usrname = $_POST['usrname']; $fname = $_POST['fname']; $lname = $_POST['lname']; $password = $_POST['password']; $email = $_POST['email']; mysql_select_db(`users`, $con) or die(mysql_error()); mysql_query(INSERT INTO `users`.`data` (`id`, `usrname`, `fname`, `lname`, `email`, `password`) VALUES (NULL, '$usrname', '$fname', '$lname', '$email', 'password')) or die(mysql_error()); mysql_close($con) echo("Thank you for registering!"); ?> </body> </html> All i get is a blank page.

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  • Flexible forms and supporting database structure

    - by sunwukung
    I have been tasked with creating an application that allows administrators to alter the content of the user input form (i.e. add arbitrary fields) - the contents of which get stored in a database. Think Modx/Wordpress/Expression Engine template variables. The approach I've been looking at is implementing concrete tables where the specification is consistent (i.e. user profiles, user content etc) and some generic field data tables (i.e. text, boolean) to store non-specific values. Forms (and model fields) would be generated by first querying the tables and retrieving the relevant columns - although I've yet to think about how I would setup validation. I've taken a look at this problem, and it seems to be indicating an EAV type approach - which, from my brief research - looks like it could be a greater burden than the blessings it's flexibility would bring. I've read a couple of posts here, however, which suggest this is a dangerous route: How to design a generic database whose layout may change over time? Dynamic Database Schema I'd appreciate some advice on this matter if anyone has some to give regards SWK

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  • User has many computers, computers have many attributes in different tables, best way to JOIN?

    - by krismeld
    I have a table for users: USERS: ID | NAME | ---------------- 1 | JOHN | 2 | STEVE | a table for computers: COMPUTERS: ID | USER_ID | ------------------ 13 | 1 | 14 | 1 | a table for processors: PROCESSORS: ID | NAME | --------------------------- 27 | PROCESSOR TYPE 1 | 28 | PROCESSOR TYPE 2 | and a table for harddrives: HARDDRIVES: ID | NAME | ---------------------------| 35 | HARDDRIVE TYPE 25 | 36 | HARDDRIVE TYPE 90 | Each computer can have many attributes from the different attributes tables (processors, harddrives etc), so I have intersection tables like this, to link the attributes to the computers: COMPUTER_PROCESSORS: C_ID | P_ID | --------------| 13 | 27 | 13 | 28 | 14 | 27 | COMPUTER_HARDDRIVES: C_ID | H_ID | --------------| 13 | 35 | So user JOHN, with id 1 owns computer 13 and 14. Computer 13 has processor 27 and 28, and computer 13 has harddrive 35. Computer 14 has processor 27 and no harddrive. Given a user's id, I would like to retrieve a list of that user's computers with each computers attributes. I have figured out a query that gives me a somewhat of a result: SELECT computers.id, processors.id AS p_id, processors.name AS p_name, harddrives.id AS h_id, harddrives.name AS h_name, FROM computers JOIN computer_processors ON (computer_processors.c_id = computers.id) JOIN processors ON (processors.id = computer_processors.p_id) JOIN computer_harddrives ON (computer_harddrives.c_id = computers.id) JOIN harddrives ON (harddrives.id = computer_harddrives.h_id) WHERE computers.user_id = 1 Result: ID | P_ID | P_NAME | H_ID | H_NAME | ----------------------------------------------------------- 13 | 27 | PROCESSOR TYPE 1 | 35 | HARDDRIVE TYPE 25 | 13 | 28 | PROCESSOR TYPE 2 | 35 | HARDDRIVE TYPE 25 | But this has several problems... Computer 14 doesnt show up, because it has no harddrive. Can I somehow make an OUTER JOIN to make sure that all computers show up, even if there a some attributes they don't have? Computer 13 shows up twice, with the same harddrive listet for both. When more attributes are added to a computer (like 3 blocks of ram), the number of rows returned for that computer gets pretty big, and it makes it had to sort the result out in application code. Can I somehow make a query, that groups the two returned rows together? Or a query that returns NULL in the h_name column in the second row, so that all values returned are unique? EDIT: What I would like to return is something like this: ID | P_ID | P_NAME | H_ID | H_NAME | ----------------------------------------------------------- 13 | 27 | PROCESSOR TYPE 1 | 35 | HARDDRIVE TYPE 25 | 13 | 28 | PROCESSOR TYPE 2 | 35 | NULL | 14 | 27 | PROCESSOR TYPE 1 | NULL | NULL | Or whatever result that make it easy to turn it into an array like this [13] => [P_NAME] => [0] => PROCESSOR TYPE 1 [1] => PROCESSOR TYPE 2 [H_NAME] => [0] => HARDDRIVE TYPE 25 [14] => [P_NAME] => [0] => PROCESSOR TYPE 1

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  • Database contents setting themselves to 0

    - by Luis Armando
    I have a Database that contains 4 tables, however I'm using 1 of them which is separated from the others. In this table I have 4 fields which are varchar and the rest are ints (11 other fields), when the users fill up the DB everything gets saved correctly, however it has happened 3 times so far that the database values for the int's reset to 0 without any apparent reason. At first, I thought, it was because those fields (where the numbers should go) were varchars not ints. However since I changed it, it happened again. I've already double checked my code and I have nothing that even updates or inserts a 0 value. Also I'm using codeigniter and active records which protect against SQL injections AND have XSS filtering enabled, could anyone point out something I might be missing or a reason for this to be happening? Also, I'm pretty sure about the answer of this but, is there ANY way to recover some data?? Other than having to ask everyone to fill in everything again.. =/ ** EDIT ** The Storage Engine is MyISAM and Collation is latin1_swedish_ci, Pack Keys are default, for all intents and purposes it's a normal DB

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  • How do I replace NOT EXISTS with JOIN?

    - by YelizavetaYR
    I've got the following query: select distinct a.id, a.name from Employee a join Dependencies b on a.id = b.eid where not exists ( select * from Dependencies d where b.id = d.id and d.name = 'Apple' ) and exists ( select * from Dependencies c where b.id = c.id and c.name = 'Orange' ); I have two tables, relatively simple. The first Employee has an id column and a name column The second table Dependencies has 3 column, an id, an eid (employee id to link) and names (apple, orange etc). the data looks like this Employee table looks like this id | name ----------- 1 | Pat 2 | Tom 3 | Rob 4 | Sam Dependencies id | eid | Name -------------------- 1 | 1 | Orange 2 | 1 | Apple 3 | 2 | Strawberry 4 | 2 | Apple 5 | 3 | Orange 6 | 3 | Banana As you can see Pat has both Orange and Apple and he needs to be excluded and it has to be via joins and i can't seem to get it to work. Ultimately the data should only return Rob

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  • Error in computed Field of select Query

    - by Shehzad Bilal
    This Query is giving me an error of #1054 - Unknown column 'totalamount' in 'where clause' SELECT (amount1 + amount2) as totalamount FROM `Donation` WHERE totalamount > 1000 I know i can resolve this error by using group by clause and replace my where condition with having clause. But is there any other solution beside using having clause. If group by is the only solution then I want to know why I have to use group by clause even I havent use any aggregate function thanks.

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  • Hibernate deletion issue

    - by muffytyrone
    I'm trying to write a Java app that imports a data file. The process is as follows Create Transaction Delete all rows from datatable Load data file into datatable Commit OR Rollback if any errors were encountered. The data loaded in step 3 is mostly the same as the data deleted in step3. The deletion is performed using the following DetachedCriteria criteria = DetachedCriteria.forClass(myObject.class); List<myObject> myObjects = hibernateTemplate.findByCriteria(criteria); hibernateTemplate.deleteAll(myObjects); When I then load the datafile, i get the following exception nested exception is org.hibernate.NonUniqueObjectException: a different object with the same identifier value was already associated with the session: The whole process needs to take place in transaction. And I don't really want to have to compare the import file / data table and then perform an insert/update/delete to get them into sync. Any help would be appreciated.

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  • Complex query with two tables and multilpe data and price ranges

    - by TiuTalk
    Let's suppose that I have these tables: [ properties ] id (INT, PK) name (VARCHAR) [ properties_prices ] id (INT, PK) property_id (INT, FK) date_begin (DATE) date_end (DATE) price_per_day (DECIMAL) price_per_week (DECIMAL) price_per_month (DECIMAL) And my visitor runs a search like: List the first 10 (pagination) properties where the price per day (price_per_day field) is between 10 and 100 on the period for 1st may until 31 december I know thats a huge query, and I need to paginate the results, so I must do all the calculation and login in only one query... that's why i'm here! :)

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  • JOIN (SELECT DISTINCT [..] substitute

    - by FRKT
    Hello, I'd like to find a substitute for using SELECT DISTINCT in a derived table. Let's say I have three tables: CREATE TABLE `trades` ( `tradeID` int(11) unsigned NOT NULL AUTO_INCREMENT, `employeeID` int(11) unsigned NOT NULL, `corporationID` int(11) unsigned NOT NULL, `profit` int(11) NOT NULL, KEY `tradeID` (`tradeID`), KEY `employeeID` (`employeeID`), KEY `corporationID` (`corporationID`) ) ENGINE=MyISAM DEFAULT CHARSET=latin1 CREATE TABLE `corporations` ( `corporationID` int(11) unsigned NOT NULL AUTO_INCREMENT, `name` varchar(255) NOT NULL, PRIMARY KEY (`corporationID`) ) ENGINE=MyISAM DEFAULT CHARSET=latin1 CREATE TABLE `employees` ( `employeeID` int(11) unsigned NOT NULL AUTO_INCREMENT, `name` varchar(255) NOT NULL, PRIMARY KEY (`employeeID`) ) ENGINE=MyISAM DEFAULT CHARSET=latin1 Let's say I'd like to find out how much profit a specific employee has generated. Simple: SELECT SUM(profit) FROM trades JOIN employees ON trades.employeeID = employees.employeeID AND employees.employeeID = 1; It gets trickier if I'd like to query how much revenue a specific corporation has, however. I cannot simply replicate the aforementioned query, because two or more employees from the same company might be involved in the same trade. This query should do the trick: SELECT SUM(profit) FROM trades JOIN (SELECT DISTINCT tradeID FROM trades WHERE trades.corporationID = 1) ... unfortunately, DISTINCT JOINs seem crazy ineffective. Is there any alternative I can use to determine how much revenue a corporation has, taking into account that a corporation might be listed several times with the same tradeID?

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  • How to use SQL - INSERT...ON DUPLICATE KEY UPDATE?

    - by Probocop
    Hi, I have a script which captures tweets and puts them into a database. I will be running the script on a cronjob and then displaying the tweets on my site from the database to prevent hitting the limit on the twitter API. So I don't want to have duplicate tweets in my database, I understand I can use 'INSERT...ON DUPLICATE KEY UPDATE' to achieve this, but I don't quite understand how to use it. My database structure is as follows. Table - Hash id (auto_increment) tweet user user_url And currently my SQL to insert is as follows: $tweet = $clean_content[0]; $user_url = $clean_uri[0]; $user = $clean_name[0]; $query='INSERT INTO hash (tweet, user, user_url) VALUES ("'.$tweet.'", "'.$user.'", "'.$user_url.'")'; mysql_query($query); How would I correctly use 'INSERT...ON DUPLICATE KEY UPDATE' to insert only if it doesn't exist, and update if it does? Thanks

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  • Unknown Column?

    - by Kenny
    ok im trying to get mutual friends between these Two users, user1 and user92 This is the sql that is successful in displaying them SELECT IF(user_a = 1 OR user_a = 92, user_b, user_a) friend FROM friendship WHERE (user_a = 1 OR user_a = 92) OR (user_b = 1 OR user_b = 92) GROUP BY 1 HAVING COUNT(*) > 1 THis is how it looks friend 61 72 73 74 75 76 77 78 79 80 81 So now i want to select all users after the number 72, and i try to do it with this sql but its not working? It gives me the error, "unknown coulum name friend in where clause" SELECT IF(user_a = 1 OR user_a = 92, user_b, user_a) friend FROM friendship WHERE friend > 72 and (user_a = 1 OR user_a = 92) OR (user_b = 1 OR user_b = 92) GROUP BY 1 HAVING COUNT(*) > 1 what am i doing wrong? or what is the correct way?? thx

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  • Fetch image from folder via datatable does not work after placing image in subdirectory

    - by Arnold Bishkoff
    I am having trouble wrapping my head around the following I have code that fetches an image via smarty in a line img src="getsnap.php?picid={$data[$smarty.section.sec.index].picno|default:$nextpic}&typ=pic&width={$config.disp_snap_width}&height={$config.disp_snap_height}" class="smallpic" alt="" / this works if i pull the image from /temp/userimages/userid/imageNo.ext but because an OS can segfault if you store too many folders or images in a directory i have code that assigns the user image to a subdirectory based upon division of a subdir per 1000 userids. so in thise case i have user id 94 whos images get stored in /siteroot/temp/userimages/000000/94/pic_1.jpg (through 10) or tn_1 (through 10).jpg here is the code for getsnap.php <?php ob_start(); if ( !defined( 'SMARTY_DIR' ) ) { include_once( 'init.php' ); } include('core/snaps_functions.php'); if (isset($_REQUEST['username']) && $_REQUEST['username'] != '') { $userid = $osDB-getOne('select id from ! where username = ?',array(USER_TABLE, $_REQUEST['username']) ); } else { // include ( 'sessioninc.php' ); if( !isset($_GET['id']) || (isset($_GET['id'])&& (int)$_GET['id'] <= 0 ) ) { $userid = $_SESSION['UserId']; } else { $userid = $_GET['id']; } } if (!isset($_GET['picid']) ) { if ((isset($_REQUEST['type']) && $_REQUEST['type'] != 'gallery') || !isset($_REQUEST['type']) ) { $defpic = $osDB-getOne('select picno from ! where userid = ? and ( album_id is null or album_id = ?) and default_pic = ? and active = ? ',array(USER_SNAP_TABLE, $userid,'0','Y','Y' ) ); if ($defpic != '') { $picid = $defpic; } else { $picid = $osDB-getOne('select picno from ! where userid = ? and ( album_id is null or album_id = ?) and active=? order by rand()',array(USER_SNAP_TABLE, $userid,'0','Y' ) ); } unset( $defpic); } } else { $picid = $_GET['picid']; } $typ = isset( $_GET['typ'])?$_GET['typ']:'pic' ; $cond = ''; if ( ($config['snaps_require_approval'] == 'Y' || $config['snaps_require_approval'] == '1') && $userid != $_SESSION['UserId'] ) { $cond = " and active = 'Y' "; } $sql = 'select * from ! where userid = ? and picno = ? '.$cond; //Get the pic $row =& $osDB-getRow ( $sql, array( USER_SNAP_TABLE, $userid, $picid ) ); //Okay pic was found in the DB, Lets actually do something // $id = $userid; $dir = str_pad(($id - ($id % 1000))/100000,6,'0',STR_PAD_LEFT); $zimg = USER_IMAGES_DIR.$dir; $img = getPicture($zimg, $userid, $picid, $typ, $row); //$img = getPicture($userid, $picid, $typ, $row); //$img = getPicture($dir, $userid, $picid, $typ, $row); $ext = ($typ = 'tn')?$row['tnext']:$row['picext']; // Now pic is built as // something pic_x.ext ie pic_2.jpg if ( $img != '' && ( ( hasRight('seepictureprofile') && ( $config['snaps_require_approval'] == 'Y' && $row['active'] == 'Y' ) ||$config['snaps_require_approval'] == 'N' ) || $userid == $_SESSION['UserId'] ) ) { $img2 = $img; //$img2 = $dir.'/'.$img; } else { $gender = $osDB-getOne( 'select gender from ! where id = ?', array( USER_TABLE, $userid ) ) ; if ($gender == 'M') { $nopic = SKIN_IMAGES_DIR.'male.jpg'; } elseif ($gender == 'F') { $nopic = SKIN_IMAGES_DIR.'female.jpg'; } elseif ($gender == 'D') { $nopic = SKIN_IMAGES_DIR.'director.jpg'; } $img2 = imagecreatefromjpeg($nopic); $ext = 'jpg'; } ob_end_clean(); header("Pragma: public"); header("Content-Type: image/".$ext); header("Content-Transfer-Encoding: binary"); header("Cache-Control: must-revalidate"); $ExpStr = "Expires: " . gmdate("D, d M Y H:i:s", time() - 30) . " GMT"; header($ExpStr); $id = $userid; $dir = str_pad(($id - ($id % 1000))/100000,6,'0',STR_PAD_LEFT); $zimg = USER_IMAGES_DIR.$dir; //header("Content-Disposition: attachment; filename=profile_".$userid."_".$typ.".".$ext); //header("Content-Disposition: attachment; filename=$dir.'/'.profile_".$userid."".$typ.".".$ext); //header("Content-Disposition: attachment; filename=profile"$dir".'/'.".$userid."_".$typ.".".$ext); header("Content-Disposition: attachment; filename=profile_".$userid."_".$typ.".".$ext); /* if ($_SESSION['browser'] != 'MSIE') { header("Content-Disposition: inline" ); } */ if ($ext == 'jpg') { imagejpeg($img2); } elseif ($ext == 'gif') { imagegif($img2); } elseif ($ext == 'png') { imagepng($img2); } elseif ($ext == 'bmp') { imagewbmp($img2); } imagedestroy($img2); ?

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  • passing a scalar query result to coalesce

    - by Fakrudeen
    How can I pass the result from a scalar [single row, single value] query to coalesce? I am trying to pick the priority as (the biggest priority so far in the table) + 1. [0 if it is the first row.] create trigger priority_SuperRuleSamples before insert on SuperRuleSamples FOR EACH ROW SET NEW.Priority=coalesce(NEW.Priority, coalesce( select Priority from SuperRuleSamples order by Priority desc limit 1, -1 )+1 )

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  • How to exclude rows where matching join is in an SQL tree

    - by Greg K
    Sorry for the poor title, I couldn't think how to concisely describe this problem. I have a set of items that should have a 1-to-1 relationship with an attribute. I have a query to return those rows where the data is wrong and this relationship has been broken (1-to-many). I'm gathering these rows to fix them and restore this 1-to-1 relationship. This is a theoretical simplification of my actual problem but I'll post example table schema here as it was requested. item table: +------------+------------+-----------+ | item_id | name | attr_id | +------------+------------+-----------+ | 1 | BMW 320d | 20 | | 1 | BMW 320d | 21 | | 2 | BMW 335i | 23 | | 2 | BMW 335i | 34 | +------------+------------+-----------+ attribute table: +---------+-----------------+------------+ | attr_id | value | parent_id | +---------+-----------------+------------+ | 20 | SE | 21 | | 21 | M Sport | 0 | | 23 | AC | 24 | | 24 | Climate control | 0 | .... | 34 | Leather seats | 0 | +---------+-----------------+------------+ A simple query to return items with more than one attribute. SELECT item_id, COUNT(DISTINCT(attr_id)) AS attributes FROM item GROUP BY item_id HAVING attributes > 1 This gets me a result set like so: +-----------+------------+ | item_id | attributes | +-----------+------------+ | 1 | 2 | | 2 | 2 | | 3 | 2 | -- etc. -- However, there's an exception. The attribute table can hold a tree structure, via parent links in the table. For certain rows, parent_id can hold the ID of another attribute. There's only one level to this tree. Example: +---------+-----------------+------------+ | attr_id | value | parent_id | +---------+-----------------+------------+ | 20 | SE | 21 | | 21 | M Sport | 0 | .... I do not want to retrieve items in my original query where, for a pair of associated attributes, they related like attributes 20 & 21. I do want to retrieve items where: the attributes have no parent for two or more attributes they are not related (e.g. attributes 23 & 34) Example result desired, just the item ID: +------------+ | item_id | +------------+ | 2 | +------------+ How can I join against attributes from items and exclude these rows? Do I use a temporary table or can I achieve this from a single query? Thanks.

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  • problem with joomla, php and json

    - by sebastian
    hi, i have a problem with a joomla component. i'm, unsing php and json for some dynamic drop down boxes. here is the code:` jQuery( function () { //jQuery.ajaxSetup({error : function (a,b) {console.dir(a); console.dir(b);}}); jQuery("#util, #loc").change( function() { var locatie = jQuery("#loc").val(); var utilitate = jQuery("#util").val(); if ( (locatie!= '---') && (utilitate!='---') ) jQuery.getJSON( "index.php?option=com_calculator&opt=json_contor&format=raw", { locatie: locatie, utilitate: utilitate }, function (data) { var html = ""; if ( data.success == 'ok' ) for (var i in data.val) html += "<option name=den_contor value ='"+ i+"' >" + data.val[i]+ " </option>"; jQuery("#den_contor").html( html ) } ) }) }); the query works, but only on one PC. we have exactly the same xampp server, exactly the same files. on one pc it works, and on a online server and on my pc it doesn't. EDIT: i have three drop down boxes, the first is populated directly from the database, the second has 4 predefined values. and the third is populated depending on combination of the first two. i have a test site online. http://contor.redxart.com must be logged in to use Calculator in the menu. you can make an new account :) "Adaugare Index" is the part that isn't working any ideas? thanks, sebastian

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  • Correct Sql Script for Formula

    - by Madan Madan
    Can anyone help me write SQL script for the following formula? If DEP = 1 If DROP 1 PLV = 334.86 * exp(0.3541 * ACTIVE_DAYS) + 0.25 * DROP + 20 * DEP Else If DROP < 0 PLV = DROP + 70 * ACTIVE_DAYS Else PLV = 0.25 * DROP + 70 * ACTIVE_DAYS The SQL script which I have is the following SELECT IF(dep=1, if(dep=1, (334.86 * exp(0.3541 * act_days)) + (0.25 * 'drop') + (20 * dep), if('drop'<0, 'drop' + (70 * act_days), (0.25 * 'drop') + (70 * act_days))),'0') as PLV But the above query is not right as something is missing where the formula says Else PLV = 0.26 * DROP Thanks,

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