Search Results

Search found 36111 results on 1445 pages for 'mysql update'.

Page 449/1445 | < Previous Page | 445 446 447 448 449 450 451 452 453 454 455 456  | Next Page >

  • 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?

    Read the article

  • 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..............

    Read the article

  • How do I use Django to insert a Geometry Field into the database?

    - by alex
    class LocationLog(models.Model): user = models.ForeignKey(User) utm = models.GeometryField(spatial_index=True) This is my database model. I would like to insert a row. I want to insert a circle at point -55, 333. With a radius of 10. How can I put this circle into the geometry field? Of course, then I would want to check which circles overlap a given circle. (my select statement)

    Read the article

  • Unknown column even thoug it exits

    - by george
    I have SELECT servisler.geo_location, servisler.ADRES_MERKEZ, servisler.ADRES_ILCE, servisler.ADRES_IL, servisler.FIRMA_UNVANI, servisler.ADRES_ISTEL, servisler.YETKILI_ADISOYADI, urun_gruplari.GRUP_ADI FROM servisler INNER JOIN urun_gruplari ON kullanici_cihaz.URUN_GRUP_NO= urun_gruplari.RECNO INNER JOIN kullanici ON kullanici.SERVIS_RECNO = servisler.RECNO INNER JOIN kullanici_cihaz ON kullanici.RECNO = kullanici_cihaz.KUL_RECNO AND kullanici_cihaz.URUN_GRUP_NO = urun_gruplari.RECNO where kullanici.kullanici = 'MAR.EDI.003' but it says [Err] 1054 - Unknown column 'kullanici_cihaz.URUN_GRUP_NO' in 'on clause' enen though the column exits. What is its problem? schema Server version: 5.1.33-community-log

    Read the article

  • NSPredicate cause update editing to return NSFetchedResultsChangeDelete not NSFetchedResultsChangeUp

    - by Matthew Weiss
    I have predicate inside of - (NSFetchedResultsController *)fetchedResultsController in a standard way starting from the CoreDataBook example. NSPredicate *predicate = [NSPredicate predicateWithFormat:@"state=%@ && date = %@ && date < %@", @"1",fromDate,toDate]; [fetchRequest setPredicate:predicate]; This works fine however when editing an item, it returns with NSFetchedResultsChangeDelete not Update. When the main view returns, it is missing the item. If I restart the simulator the delete was not saved and the correct editing result is shown the the predicate working correctly. case NSFetchedResultsChangeDelete: [tableView deleteRowsAtIndexPaths:[NSArray arrayWithObject:indexPath] withRowAnimation:UITableViewRowAnimationFade]; break; I can confirm the behavior by commenting out the two predicate lines ONLY and then all works as it should correctly returning with the full set after editing and calling NSFetchedResultsChangeUpdate instead of NSFetchedResultsChangeDelete. I have read http://matteocaldari.it/2009/11/multiple-contexts-controllers-delegates-and-coredata-bug who reports similar behavior but I have not found a work around to my problem. I can

    Read the article

  • Database time data retrieval, time based queries

    - by Raphael Pineda
    I am new to time manipulation or time arithmetic operations and am currently developing a navigation system with Web server based information and currently I have this Database that contains a table peek hours whose columns are id, start_time, end_time , edge_id, day_of_the_week, edge_weight ------------------------------------------------------------------------ | Peek Hours | ------------------------------------------------------------------------ | | | | | | | | id | start_time | end_time | edge_id | day_of_the_week | edge_weight | | | | | | | | ------------------------------------------------------------------------ I am using PHP as a webservice and so based on the current time i want to get all the records that would fit this equation start_time< current_time < end_time

    Read the article

  • SQL: GROUP BY after JOIN without overriding rows?

    - by krismeld
    I have a table of basketball leagues, a table af teams and a table of players like this: LEAGUES ID | NAME | ------------------ 1 | NBA | 2 | ABA | TEAMS: ID | NAME | LEAGUE_ID ------------------------------ 20 | BULLS | 1 21 | KNICKS | 2 PLAYERS: ID | TEAM_ID | FIRST_NAME | LAST_NAME | --------------------------------------------- 1 | 21 | John | Starks | 2 | 21 | Patrick | Ewing | Given a League ID, I would like to retrieve all the players' names and their team ID from all the teams in that league, so I do this: SELECT t.id AS team_id, p.id AS player_id, p.first_name, p.last_name FROM teams AS t JOIN players AS p ON p.team_id = t.id WHERE t.league_id = 1 which returns: [0] => stdClass Object ( [team_id] => 21 [player_id] => 1 [first_name] => John [last_name] => Starks ) [1] => stdClass Object ( [team_id] => 21 [player_id] => 2 [first_name] => Patrick [last_name] => Ewing ) + around 500 more objects... Since I will use this result to populate a dropdown menu for each team containing each team's list of players, I would like to group my result by team ID, so the loop to create these dropdowns will only have to cycle through each team ID instead of all 500+ players each time. But when I use the GROUP BY like this: SELECT t.id AS team_id, p.id AS player_id, p.first_name, p.last_name FROM teams AS t JOIN players AS p ON p.team_id = t.id WHERE t.league_id = 1 GROUP BY t.id it only returns one player from each team like this, overriding all the other players on the same team because of the use of the same column names. [0] => stdClass Object ( [team_id] => 21 [player_id] => 2 [first_name] => Patrick [last_name] => Ewing ) [1] => stdClass Object ( [team_id] => 22 [player_id] => 31 [first_name] => Shawn [last_name] => Kemp ) etc... I would like to return something like this: [0] => stdClass Object ( [team_id] => 2 [player_id1] => 1 [first_name1] => John [last_name1] => Starks [player_id2] => 2 [first_name2] => Patrick [last_name2] => Ewing +10 more players from this team... ) +25 more teams... Is it possible somehow?

    Read the article

  • can you make an sql query for this situation?

    - by saurav
    i have a table as below. name and 10 cities in which he lived during his lifetime. name , city1 , city2 , city3 ,city4 , city5 ,city6 , city7 , city8 , city9 city10 suppose for a particular name i want to fetch other names in table matching with maximum number of cities. for example if i want to fetch other people who have lived in three or more cities lived by this person.

    Read the article

  • 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; }

    Read the article

  • MYSQ request | GROUP BY DAY

    - by user889349
    I have a table: orders, and need to make a request and get other table. My DB table: id close 1 2012-05-29 03:11:15 2 2012-05-30 03:11:40 3 2012-05-31 03:12:10 4 2012-05-31 03:14:13 5 2012-05-31 03:16:50 6 2012-05-31 03:40:07 7 2012-05-31 05:22:18 8 2012-05-31 05:22:22 9 2012-05-31 05:22:50 ... I need to make a request and get this table (GROUP BY DAY(close)): 1 2012-05-29 03:11:15 2 2012-05-30 03:11:40 9 2012-05-31 05:22:50 /*This is a last record on this day (05-31)*/ Thanks!

    Read the article

  • PHP coding a price comparaison tool

    - by Tristan
    Hello, it's the first time I developp such tool you all know (the possibility to compare articles according to price and/or options) Since I never did that i want to tell me what do you think of the way i see that : On the database we would have : offer / price / option 1 / option 2 / option 3 / IDseller / IDoffer best buy / 15$ / full FTP / web hosting / php.ini / 10 / 1 .../..../.... And the request made by the client : "SELECT * FROM offers WHERE price <= 20 AND option1 = fullFTP"; I don't know if it seems OK to you. Plus i was wondering, how to avoid multiples entries for the same seller. Imagine you have multiple offers with a price <= 20 with the option FullFTP for the same seller, i don't want him to be shown 5 times on the comparator. If you have any advices ;) Thanks

    Read the article

  • declaring constraint to consider prog logic

    - by shantanuo
    I can open a trip only once but can close it multiple times. I can not declare the Trip_no + status as primary key since there can be multiple entries while closing the trip. Is there any way that will assure me that a trip number is opened only once? For e.g. there should not be the second row with "Open" status for trip No. 3 since it is already there in the following table. Trip No | Status 1 Open 1 Close 1 Close 2 Open 2 Close 3 Open 3 Close 3 Close 3 Close 3 Close

    Read the article

  • 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();

    Read the article

  • 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?

    Read the article

  • LINQ to Sql: Insert instead of Update

    - by Christina Mayers
    I am stuck with this problems for a long time now. Everything I try to do is insert a row in my DB if it's new information - if not update the existing one. I've updated many entities in my life before - but what's wrong with this code is beyond me (probably something pretty basic) I guess I can't see the wood for the trees... private Models.databaseDataContext db = new Models.databaseDataContext(); internal void StoreInformations(IEnumerable<EntityType> iEnumerable) { foreach (EntityType item in iEnumerable) { EntityType type = db.EntityType.Where(t => t.Room == item.Room).FirstOrDefault(); if (type == null) { db.EntityType.InsertOnSubmit(item); } else { type.Date = item.Date; type.LastUpdate = DateTime.Now(); type.End = item.End; } } } internal void Save() { db.SubmitChanges(); } Edit: just checked the ChangeSet, there are no updates only inserts. For now I've settled with foreach (EntityType item in iEnumerable) { EntityType type = db.EntityType.Where(t => t.Room == item.Room).FirstOrDefault(); if (type != null) { db.Exams.DeleteOnSubmit(type); } db.EntityType.InsertOnSubmit(item); } but I'd love to do updates and lose these unnecessary delete statements.

    Read the article

  • Zend Framework Error:Invalid parameter number: no parameters were bound'

    - by roast_soul
    I'm using the Zend Frameworker 1.12. According to the help file, I used the Zend_Db_Statement to execute my sql. Below is my php code: $sql = "delete from options where id=?"; $stmt = new Zend_Db_Statement_Mysqli($this->getAdapter(), $sql); return $stmt->execute(array('1')); But the error is exception 'PDOException' with message 'SQLSTATE[HY093]: Invalid parameter number: no parameters were bound' in D:\Zend\workspaces\DefaultWorkspace.metadata.plugins\org.zend.php.framework.resource\resources\ZendFramework-1\library\Zend\Db\Statement\Mysqli.php:209 Stack trace: ......... ......... I googled for days, but nothing works. Any one know how to fix it?

    Read the article

  • what is the question for the query?

    - by Kevinniceguy
    Sorry...I mean what question will be for this query? SELECT SUM(price) FROM Room r, Hotel h WHERE r.hotelNo = h.hotelNo and hotelName = 'Paris Hilton' and roomNo NOT IN (SELECT roomNo FROM Booking b, Hotel h WHERE (dateFrom <= CURRENT_DATE AND dateTo >= CURRENT_DATE) AND b.hotelNo = h.hotelNo AND hotelName = 'Paris Hilton');

    Read the article

  • sql query question / count

    - by scheibenkleister
    Hi, I have houses that belongs to streets. A user can buy several houses. How do I find out, if the user owns an entire street? street table with columns (id/name) house table with columns (id/street_id [foreign key] owner table with columns (id/house_id/user_id) [join table with foreign keys] So far, I'm using count which returns the result: select count(*), street_id from owner left join house on owner.house_id = house.id group by street_id where user_id = 1 count(*) | street_id 3 | 1 2 | 2 A more general count: select count(*) from house group by street_id returns: count(*) | street_id 3 | 1 3 | 2 How can I find out, that user 1 owns the entire street 1 but not street 2? Thanks.

    Read the article

  • 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!

    Read the article

  • 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?

    Read the article

  • PHP Login, Store Session Variables.

    - by Andreas Carlbom
    Yo. I'm trying to make a simple login system in PHP and my problem is this: I don't really understand sessions. Now, when I log a user in, I run session_register("user"); but I don't really understand what I'm up to. Does that session variable contain any identifiable information, so that I for example can get it out via $_SESSION["user"] or will I have to store the username in a separate variable? Thanks.

    Read the article

  • Need an alternative to two left joins.

    - by Scarface
    Hey guys quick question, I always use left join, but when I left join twice I always get funny results, usually duplicates. I am currently working on a query that Left Joins twice to retrieve the necessary information needed but I was wondering if it were possible to build another select statement in so then I do not need two left joins or two queries or if there were a better way. For example, if I could select the topic.creator in table.topic first AS something, then I could select that variable in users and left join table.scrusersonline. Thanks in advance for any advice. SELECT * FROM scrusersonline LEFT JOIN users ON users.id = scrusersonline.id LEFT JOIN topic ON users.username = topic.creator WHERE scrusersonline.topic_id = '$topic_id' The whole point of this query is to check if the topic.creator is online by retrieving his name from table.topic and matching his id in table.users, then checking if he is in table.scrusersonline. It produces duplicate entries unfortunately and is thus inaccurate in my mind.

    Read the article

  • Create Downloadable CSV File from PHP Script

    - by Aphex22
    How would I create a formatted version of the following PHP script as a downloadable CSV file from the code below (1.0) At the moment the fputcsv function is currently dumping the unparsed PHP/HTML code into a CSV file. This is incorrect. The downloaded CSV file should contain the columns and rows generated from the code at (1.0) as shown in the image link below. I've tried using the following code at the top of the PHP file: // output headers so that the file is downloaded rather than displayed header('Content-Type: text/csv; charset=utf-8'); header('Content-Disposition: attachment; filename=amazon.csv'); // create a file pointer connected to the output stream $output = fopen('php://output', 'w'); $mysql_hostname = ""; $mysql_user = ""; $mysql_password = ""; $mysql_database = ""; $bd = mysql_connect($mysql_hostname, $mysql_user, $mysql_password) or die("Could not connect database"); mysql_select_db($mysql_database, $bd) or die("Could not select database"); $sql = "select * from product WHERE on_amazon = 'on' AND active = 'on'"; $result = mysql_query($sql) or die ( mysql_error() ); // loop over the rows, outputting them while ($sql_result = mysql_fetch_assoc($sql)) fputcsv($output, $sql_result); 1.0 The start of the code outputs the column headings for the CSV file: // set headers echo " item_sku, external_product_id, external_product_id_type, item_name, brand_name, manufacturer, product_description, feed_product_type, update_delete, part_number, model, standard_price, list_price, currency, quantity, product_tax_code, product_site_launch_date, merchant_release_date, restock_date ... <br>"; And then follows PHP script for the column values // load all stock while ($line = mysql_fetch_assoc($result) ) { ?> <?php $size_suffix = array ("",'_chain','_con_b','_con_c'); $arrayLength = count ($size_suffix); for($y=0;$y<$arrayLength;$y++) { //Possible size array to loop through when checking quantity $con_size = array (36,365,37,375,38,385,39,395,40,405,41,415,42,425,43,435,44,445,45,455,46,465,47,475,48,485); $arrlength=count($con_size); for($x=0;$x<$arrlength;$x++) { // check if size is available if($line['quantity_c_size_'.$con_size[$x].$size_suffix[$y]] > 0 ) { ?> <!-- item sku --> <?=$line['product_id']?>, <!-- external product id --> <?=$line['code_size_'.$con_size[$x].'']?>, <? // external product id type $barcode = $line['code_size_'.$con_size[$x]]; $trim_barcode = trim($barcode); $count = strlen($trim_barcode); if ($count == 12) { echo "UPC"; } if ($count == 13) { echo "EAN"; } elseif ($count < 12) { echo " "; } ?>, <!-- item name --> <?=$line['title']?>, <? // brand_name $brand = $line['jys_brand']; echo ucfirst($brand); ?>, <? // manufacturer $brand = $line['jys_brand']; echo ucfirst($brand); ?>, <!-- product description --> <?=preg_replace('/[^\da-z]/i', ' ', $line['amazon_desc']) ?>, <!-- feed product type --> Shoes, , , , <!-- standard price --> <?=$line['price']?>, , <!-- currency --> GBP, <!-- quantity --> <?=$line['quantity_size_'.$con_size[$x].$size_suffix[$y]]?>, , <!-- product site launch date --> <?=$line['added_y']?>-<?=$line['added_m']?>-<?=$line['added_d']?>, <!-- merchat release date --> <?=$line['added_y']?>-<?=$line['added_m']?>-<?=$line['added_d']?>, , , , , <!-- item package quantity --> 1, , , , , <!-- fulfillment latency --> 2, <!-- max aggregate ship quantity --> 1, , , , , , , , , , , , , , , , , , , , , , , , , , , , , , <!-- main image url, url1, url2, url3 --> http://www.getashoe.co.uk/full/<?=$line['product_id']?>_1.jpg, http://www.getashoe.co.uk/full/<?=$line['product_id']?>_2.jpg, http://www.getashoe.co.uk/full/<?=$line['product_id']?>_3.jpg, http://www.getashoe.co.uk/full/<?=$line['product_id']?>_4.jpg, , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , <!-- heel height --> <?=$line['heel']?>, , , , , , , , , , , <!-- colour name --> <?=$line['colour']?>, <!-- colour map --> <? $colour = preg_replace('/[()]/i', ' ', $line['colour']); if (preg_match( '/[\/].*/i', $colour)) { echo 'Multicolour'; } if (preg_match( '/off.*/i', $colour)) { echo 'Off-White'; } elseif( preg_match( '/white.*/i', $colour)) { echo 'White'; } elseif( preg_match( '/moro.*/i', $colour)) { echo 'Brown'; } elseif( preg_match( '/morado.*/i', $colour)) { echo 'Purple'; } elseif( preg_match( '/cream.*/i', $colour)) { echo 'Off-White'; } elseif( preg_match( '/pewter.*/i', $colour)) { echo 'Silver'; } elseif( preg_match( '/yellow.*/i', $colour)) { echo 'Yellow'; } elseif( preg_match( '/camel.*/i', $colour)) { echo 'Beige'; } elseif( preg_match( '/navy.*/i', $colour)) { echo 'Blue'; } elseif( preg_match( '/tan.*/i', $colour)) { echo 'Brown'; } elseif( preg_match( '/rainbow.*/i', $colour)) { echo 'Multicolour'; } elseif( preg_match( '/orange.*/i', $colour)) { echo 'Orange'; } elseif( preg_match( '/leopard.*/i', $colour)) { echo 'Multicolour'; } elseif( preg_match( '/red.*/i', $colour)) { echo 'Red'; } elseif( preg_match( '/pink.*/i', $colour)) { echo 'Pink'; } elseif( preg_match( '/purple.*/i', $colour)) { echo 'Purple'; } elseif( preg_match( '/blue.*/i', $colour)) { echo 'Blue'; } elseif( preg_match( '/green.*/i', $colour)) { echo 'Green'; } elseif( preg_match( '/brown.*/i', $colour)) { echo 'Brown'; } elseif( preg_match( '/grey.*/i', $colour)) { echo 'Grey'; } elseif( preg_match( '/black.*/i', $colour)) { echo 'Black'; } elseif( preg_match( '/gold.*/i', $colour)) { echo 'Gold'; } elseif( preg_match( '/silver.*/i', $colour)) { echo 'Silver'; } elseif( preg_match( '/multi.*/i', $colour)) { echo 'Multicolour'; } elseif( preg_match( '/beige.*/i', $colour)) { echo 'Beige'; } elseif( preg_match( '/nude.*/i', $colour)) { echo 'Beige'; } ?>, <!-- size name --> <? echo $con_size[$x];?>, <!-- size map --> <? if ($con_size[$x] == 36) { echo "3 UK"; } elseif ($con_size[$x] == 37 ) { echo "4 UK"; } elseif ($con_size[$x] == 38) { echo "5 UK"; } elseif ($con_size[$x] == 39 ) { echo "6 UK"; } elseif ($con_size[$x] == 40 ) { echo "7 UK"; } elseif ($con_size[$x] == 41) { echo "8 UK"; } elseif ($con_size[$x] == 42) { echo "9 UK"; } elseif ($con_size[$x] == 43) { echo "10 UK"; } elseif ($con_size[$x] == 44 ) { echo "11 UK"; } elseif ($con_size[$x] == 45 ) { echo "12 UK"; } elseif ($con_size[$x] == 46 ) { echo "13 UK"; } elseif ($con_size[$x] == 47 ) { echo "14 UK"; } elseif ($con_size[$x] == 48 ) { echo "15 UK"; } elseif ($con_size[$x] == 365) { echo "3.5 UK"; } elseif ($con_size[$x] == 375 ) { echo "4.5 UK"; } elseif ($con_size[$x] == 385) { echo "5.5 UK"; } elseif ($con_size[$x] == 395 ) { echo "6.5 UK"; } elseif ($con_size[$x] == 405 ) { echo "7.5 UK"; } elseif ($con_size[$x] == 415) { echo "8.5 UK"; } elseif ($con_size[$x] == 425) { echo "9.5 UK"; } elseif ($con_size[$x] == 435) { echo "10.5 UK"; } elseif ($con_size[$x] == 445 ) { echo "11.5 UK"; } elseif ($con_size[$x] == 455 ) { echo "12.5 UK"; } elseif ($con_size[$x] == 465 ) { echo "13.5 UK"; } elseif ($con_size[$x] == 475 ) { echo "14.5 UK"; } elseif ($con_size[$x] == 485 ) { echo "15.5 UK"; } ?>, <br> <? // finish checking if size is available } } } ?> I've included an image of how the CSV file should appear. https://i.imgur.com/ZU3IFer.png Any help would be great.

    Read the article

< Previous Page | 445 446 447 448 449 450 451 452 453 454 455 456  | Next Page >