Search Results

Search found 63875 results on 2555 pages for 'mysql error 1045'.

Page 192/2555 | < Previous Page | 188 189 190 191 192 193 194 195 196 197 198 199  | Next Page >

  • MYSQL Select and group by date

    - by Gil
    I'm not sure how create a right query to get the result I'm looking for. What I have is 2 tables. first has ID, Name columns and second has date and adminID, which is referenced from table 1 column ID. Now, what I want to get is basically number of times each admin loged in per day during the month. From structure like this one I want to get per day and month data so result would be similar to 1, 2, 2 march total 5 for admin 4. ID | Date ------------------ 4 | 2010/03/01 4 | 2010/03/04 4 | 2010/03/04 4 | 2010/03/05 4 | 2010/03/05

    Read the article

  • MySQL DATETIME format comparison - is strtotime needed?

    - by Steffan
    I've been doing something along the lines of.. $dt1 = '1000-01-01 00:00:00'; //really some val from db $dt2 = '1000-01-01 00:00:10'; //another val maybe db maybe formatted if(strtotime($dt1) > strtotime($dt2){ //do something } Is the strtotime needed? can i do a more direct comparison on the datetime formatted strings? i.e. if($dt1 > $dt2){ //do something } Will that always work?

    Read the article

  • MYSQL CASE WHEN PROBLEM

    - by user305270
    SELECT `profiles`.* FROM `profiles` INNER JOIN `friendships` ON `profiles`.id = `friendships`.(CASE WHEN friendships.profile_id = 1 THEN`friend_id` ELSE `profile_id` END) How can i make the inner join like profile.id = friendships.(here will select the one key that is needed) but it doesnt work. please help :P it cant be: `profiles`.id = (CASE WHEN friendships.profile_id = 1 THEN `friendships`.`friend_id` ELSE `friendships`.`profile_id` END)

    Read the article

  • MySQL & PHP - Creating Multiple Parent Child Relations

    - by Ashok
    Hi, I'm trying to build a navigation system using categories table with hierarchies. Normally, the table would be defined as follows: id (int) - Primary key name (varchar) - Name of the Category parentid (int) - Parent ID of this Category referenced to same table (Self Join) But the catch is that I require that a category can be child to multiple parent categories.. Just like a Has and Belongs to Many (HABTM) relation. I know that if there are two tables, categories & items, we use a join table categories_items to list the HABTM relations. But here i'm not having two tables but only table but should somehow show HABTM relations to itself. Is this be possible using a single table? If yes, How? If not possible, what rules (table naming, fields) should I follow while creating the additional join table? I'm trying to achieve this using CakePHP, If someone can provide CakePHP solution for this problem, that would be awesome. Even if that's not possible, any solution about creating join table is appreciated. Thanks for your time.

    Read the article

  • Special characters in PHP / MySQL

    - by Jonathan
    Hi, I have in the database words that include special character (in Spanish mostly, like tildes). In the database everything is saved and shown correctly with PHPmyAdmin, but when I get the data (using PHP) and display it in a browser, I get a weird character, like a "?" with a square... I need a general fix so I don't need to escape each character every time, and also I would be able to insert special Spanish characters from a PHP form into the database... The HTML is correct: <meta http-equiv="content-type" content="text/html; charset=utf-8" /> All tables and databas are set to utf8_spanish The character I get: ? Any suggestions??? Thanks!!!

    Read the article

  • MySQL top count({column}) with a limit

    - by Josh K
    I have a table with an ip address column. I would like to find the top five addresses which are listed. Right now I'm planning it out the following: Select all distinct ip addresses Loop through them all saying count(id) where IP='{ip}' and storing the count List the top five counts. Downsides include what if I have 500 ip addresses. That's 500 queries I have to run to figure out what are the top five. I'd like to build a query like so select ip from table where 1 order by count({distinct ip}) asc limit 5

    Read the article

  • how to atomically claim a row or resource using UPDATE in mysql

    - by Igor
    i have a table of resources (lets say cars) which i want to claim atomically. if there's a limit of one resource per one user, i can do the following trick: UPDATE cars SET user = 'bob' WHERE user IS NULL LIMIT 1 SELECT * FROM cars WHERE user IS bob that way, i claim the resource atomically and then i can see which row i just claimed. this doesn't work when 'bob' can claim multiple cars. i realize i can get a list of cars already claimed by bob, claim another one, and then SELECT again to see what's changed, but that feels hackish. What I'm wondering is, is there some way to see which rows i just updated with my last UPDATE? failing that, is there some other trick to atomically claiming a row? i really want to avoid using SERIALIZABLE isolation level. If I do something like this: 1 SELECT id FROM cars WHERE user IS NULL 2 <here, my PHP or whatever picks a car id> 3 UPDATE cars SET user = 'bob' WHERE id = <the one i picked> would REPEATABLE READ be sufficient here? in other words, could i be guaranteed that some other transactions won't claim the row my software has picked during step 2?

    Read the article

  • Mysql SELECT nested query, very complicated?

    - by smartbear
    Okay, first following are my tables: Table house: id | items_id | 1 | 1,5,10,20 | Table items: id | room_name | refer 1 | kitchen | 3 5 | room1 | 10 Table kitchen: id | detail_name | refer 3 | spoon | 4 5 | fork | 10 Table spoon: id | name | color | price | quantity_available | 4 | spoon_a | white | 50 | 100 | 5 | spoon_b | black | 30 | 200 | How to do a nested select statement, where I want to select id, name, color, price and quantity_available column, from the each value inside the 'items_id' column in 'house' table? This is very challenging!! EDIT: after read robin's answer Table house: id | items_id | house1 | 1 | house1 | 5 | house1 | 10 | house2 | 20 | If this it the house table, how to do the nested, join, or whatever select statement??

    Read the article

  • Script working with mysql and php into a textarea and back

    - by Tribalcomm
    I am trying to write a custom script that will keep a list of strings in a textarea. Each line of the textarea will be a row from a table. The problem I have is how to work the script to allow for adding, updating, or deleting rows based on a submit. So, for instance, I currently have 3 rows in the database: john sue mark I want to be able to delete sue and add richard and it will delete the row with sue and insert a row for richard. My code so far is as follows: To query the db and list it in the textarea: $basearray = mysql_query("SELECT name FROM mytable ORDER BY name"); <textarea name="names" cols=6 rows=12>'); <?php foreach($basearray as $base){ echo $base->name."\n"; } ?> </textarea> After the submit, I have: <?php $namelist = $_REQUEST[names]; $newarray = explode("\n", $namelist); foreach($newarray as $name) { if (!in_array($name, $basearray)) { mysql_query(DELETE FROM mytable WHERE word='$name'"); } elseif (in_array($name, $basearray)) { ; } else { mysql_query("INSERT INTO mytable (name) VALUES ("$name")"); } } ?> Please tell me what I am doing wrong. I am not getting any functions to work when I edit the contents of the textarea. Thanks!

    Read the article

  • MySQL optimized sentence

    - by Ivan
    I have a simple table where I have to extract some records. The problem is that the evaluation function is a very time-consuming stored procedure so I shouldn't to call it twice like in this sentence: SELECT *, slow_sp(row) FROM table WHERE slow_sp(row)>0 ORDER BY dist DESC LIMIT 10 First I thought in optimize like this: SELECT *, slow_sp(row) AS value FROM table WHERE value>0 ORDER BY dist DESC LIMIT 10 But it doesn't works due "value" is not processed when the WHERE clause is evaluated. Any idea to optimize this sentence? Thanks.

    Read the article

  • mySQL - Separate Lastname,Firstname and CompanyName entries from a single column

    - by Decalmo
    I've got a column in a database which contains company names, and customer names all in one field... what I'd like to do is keep the CompanyName column completely intact, but wherever there is a comma in the CompanyName I'd like to take that information and populate it into a FirstName and LastName field. So that basically... Before: CompanyName: Big Company Inc Smith, John Sue, Maggie After: CompanyName: Big Company Inc Smith, John Sue, Maggie LastName: Smith Sue FirstName: John Maggie This one is pretty dang tricky for me... Any help is greatly appreciated!

    Read the article

  • Select from mysql by day with different timezones (php)

    - by Adam
    I'm storing leads in a database, and each lead has a datetime field with a PST timezone based date & time. I want my user to be able to display all leads from a certain date (e.g. today, yseterday), and choose the timezone. E.g. if I want to see all leads that were generated yesterday in EST timezone, I need to first convert (or read) all the datetime values to EST, and then only select those who are in the right range (yesterday). What would be the best way to do that?

    Read the article

  • MySQL - How do I inner join sorting the joined data

    - by Gary
    I'm trying to write a report which will join a person, their work, and their hourly wage at the time of work. I cannot seem to figure out the best way to join the person's cost when the date is less than the date of the work. Let's say a person cost $30 per hour at the start of the year then got a $10 raise o Feb 5 and another on Mar 1. 01/01/2010 $30.00 (per hour) 02/05/2010 $40.00 03/01/2010 $45.00 The person put in hours several days which span the rasies. 01/05/2010 10 hours (should be at $30/hr) 01/27/2010 5 hours (again at $30) 02/10/2010 10 hours (at $40/hr) 03/03/2010 5 hours (at $45/hr) I'm trying to write one SQL statement which will pull the hours, the cost per hour, and the hours*cost. The cost is the hourly rate last entered into the system so the cost date is less than the work date, ordered by cost date limit 1. SELECT person.id, person.name, work.hours, person_costs.value, work.hours * person_costs.value AS value FROM person INNER JOIN work ON (person.id = work.person_id) INNER JOIN person_costs ON (person.id = person_costs.person_id AND person_costs.date < work.date) WHERE person.id = 1234 ORDER BY work.date ASC The problem I'm having, the person_costs isn't ordered by date in descending order. It's pulling out "any" value (naturally sorted by record position) which matches the condition. How do I select the first person_cost value which is older than the work date? Thanks!

    Read the article

  • Simple MySQL Query taking 45 seconds (Gets a record and its "latest" child record)

    - by Brian Lacy
    I have a query which gets a customer and the latest transaction for that customer. Currently this query takes over 45 seconds for 1000 records. This is especially problematic because the script itself may need to be executed as frequently as once per minute! I believe using subqueries may be the answer, but I've had trouble constructing it to actually give me the results I need. SELECT customer.CustID, customer.leadid, customer.Email, customer.FirstName, customer.LastName, transaction.*, MAX(transaction.TransDate) AS LastTransDate FROM customer INNER JOIN transaction ON transaction.CustID = customer.CustID WHERE customer.Email = '".$email."' GROUP BY customer.CustID ORDER BY LastTransDate LIMIT 1000 I really need to get this figured out ASAP. Any help would be greatly appreciated!

    Read the article

  • Mysql issue with decimal

    - by azz0r
    Hello, I have two fields - amount (decimal (11, 2)) - gift_amount (decimal (11, 2)) When I do an update on either for a value equal to or below 999.99, it saves correctly. However, if I go over that, then it drops the value right back to down 1 - 10. Is this a known issue or am I going wrong using decimal? Heres some PHP code of what I'm doing just to make it clearer (although I'm 100% its not the PHP's fault. if ($total_balance >= $cost) { if ($this->user->balance->gift_amount > 0) { $total_to_be_paid = number_format($cost, 2) - number_format($this->user->balance->gift_amount, 2);//figure out how much is left after the gift total $this->user->balance->gift_amount -= number_format($cost, 2); //deduct from the gift balance $this->user->balance->gift_amount = (number_format($this->user->balance->gift_amount, 2) < 0) ? number_format(00.00, 2) : number_format($this->user->balance->gift_amount, 2); //if the gift balance went below 0, lets set it to 0 if ($total_to_be_paid > 0) { $this->user->balance->amount = number_format($this->user->balance->amount, 2) - number_format($total_to_be_paid, 2); } } else { $this->user->balance->amount = number_format($this->user->balance->amount, 2) - number_format($cost, 2); } if ($object = Model_ClipBought::create(array('clip_id' => $clip->id, 'user_id' => $this->user->id, 'currency_name' => $user_currency, 'cost' => $cost, 'downloads' => $clip->downloads, 'expires' => time() + ($clip->expires * 86400)))) { $this->user->balance->save(); $download = new Model_Download(ROOT_PATH."/public/files/Clip/$clip->file_url"); $download->execute(); } else { throw new exception('We could not finish the purchase, this has been reported, sorry for the inconvenience.'); } } else { throw new exception('You dont have enough money in your account todo this'); } exit; }

    Read the article

  • How to use MySql date_add in Nhibernate?

    - by jalchr
    This really puzzled for hours, I searched all over the internet, but got no working solution. Can someone point where the problem is ... thanks ! I created my own dialect class public class MySQLDialectExtended : MySQLDialect { public MySQLDialectExtended() { RegisterFunction("date_add_interval", new SQLFunctionTemplate(NHibernateUtil.Date, "date_add(?1, INTERVAL ?2 ?3)")); } } Then I try to use it as follows: query.Append( " ( date_add_interval(D.ApprovalDate, 1, YEAR) < current_timestamp() < date_add_interval(D.RenewalDate, -1, YEAR) )"); It fails with following exception: NHibernate.Hql.Ast.ANTLR.QuerySyntaxException : Exception of type 'Antlr.Runtime.NoViableAltException' was thrown. near line 1, column 677 where the column number is at the end of the first 'YEAR' word. Edit: here is my configuration <property name="dialect">MyCompanyName.MySQLDialectExtended, MyCompanyName</property> <property name="hbm2ddl.keywords">none</property>

    Read the article

  • Setting up MySQL database

    - by mathew
    I do have single database and near about 11 tables. while my web page is opening informations from these 11 tables will be accessed same time. according to my current settings what I did now is for each table database is opening and closing. say I had given username and password to open databse for each table and close after retrieving information from that table. Is this the right way to do it?? I feel because of this the database is opeing and closing 11 times!!!! Am I right?? is this the right way to do that?? Oh well I do have some tables which update date is differ from others... THanks Mathew

    Read the article

  • Import excel files with image in php/mysql

    - by Marcel
    Hi all! I want to make an import script which allows users to upload their excel file (extension not important) to my php application. The application should reconize a list of items (so far so good). The difficulty in this case is that the excel files contain images...I've read information about phpexcel library but it does not say anything about images. Anybody ideas? Regards, Marcel

    Read the article

  • mysql view performance

    - by vamsivanka
    I have a table for about 100,000 users in it. First Case: explain select * from users where state = 'ca' when i do an explain plan for the above query i got the cost as 5200 Second Case: Create or replace view vw_users as select * from users Explain select * from vw_users where state = 'ca' when i do an explain plan on the second query i got the cost as 100,000. How does the where clause in the view work ?? Is the where clause applied after the view retrieves all the rows. Please let know, how can i fix this issue. Thanks

    Read the article

  • Mysql - help me optimize this query

    - by sandeepan-nath
    About the system: -The system has a total of 8 tables - Users - Tutor_Details (Tutors are a type of User,Tutor_Details table is linked to Users) - learning_packs, (stores packs created by tutors) - learning_packs_tag_relations, (holds tag relations meant for search) - tutors_tag_relations and tags and orders (containing purchase details of tutor's packs), order_details linked to orders and tutor_details. For a more clear idea about the tables involved please check the The tables section in the end. -A tags based search approach is being followed.Tag relations are created when new tutors register and when tutors create packs (this makes tutors and packs searcheable). For details please check the section How tags work in this system? below. Following is a simpler representation (not the actual) of the more complex query which I am trying to optimize:- I have used statements like explanation of parts in the query select SUM(DISTINCT( t.tag LIKE "%Dictatorship%" )) as key_1_total_matches, SUM(DISTINCT( t.tag LIKE "%democracy%" )) as key_2_total_matches, td., u., count(distinct(od.id_od)), if (lp.id_lp > 0) then some conditional logic on lp fields else 0 as tutor_popularity from Tutor_Details AS td JOIN Users as u on u.id_user = td.id_user LEFT JOIN Learning_Packs_Tag_Relations AS lptagrels ON td.id_tutor = lptagrels.id_tutor LEFT JOIN Learning_Packs AS lp ON lptagrels.id_lp = lp.id_lp LEFT JOIN `some other tables on lp.id_lp - let's call learning pack tables set (including Learning_Packs table)` LEFT JOIN Order_Details as od on td.id_tutor = od.id_author LEFT JOIN Orders as o on od.id_order = o.id_order LEFT JOIN Tutors_Tag_Relations as ttagrels ON td.id_tutor = ttagrels.id_tutor JOIN Tags as t on (t.id_tag = ttagrels.id_tag) OR (t.id_tag = lptagrels.id_tag) where some condition on Users table's fields AND CASE WHEN ((t.id_tag = lptagrels.id_tag) AND (lp.id_lp 0)) THEN `some conditions on learning pack tables set` ELSE 1 END AND CASE WHEN ((t.id_tag = wtagrels.id_tag) AND (wc.id_wc 0)) THEN `some conditions on webclasses tables set` ELSE 1 END AND CASE WHEN (od.id_od0) THEN od.id_author = td.id_tutor and some conditions on Orders table's fields ELSE 1 END AND ( t.tag LIKE "%Dictatorship%" OR t.tag LIKE "%democracy%") group by td.id_tutor HAVING key_1_total_matches = 1 AND key_2_total_matches = 1 order by tutor_popularity desc, u.surname asc, u.name asc limit 0,20 ===================================================================== What does the above query do? Does AND logic search on the search keywords (2 in this example - "Democracy" and "Dictatorship"). Returns only those tutors for which both the keywords are present in the union of the two sets - tutors details and details of all the packs created by a tutor. To make things clear - Suppose a Tutor name "Sandeepan Nath" has created a pack "My first pack", then:- Searching "Sandeepan Nath" returns Sandeepan Nath. Searching "Sandeepan first" returns Sandeepan Nath. Searching "Sandeepan second" does not return Sandeepan Nath. ====================================================================================== The problem The results returned by the above query are correct (AND logic working as per expectation), but the time taken by the query on heavily loaded databases is like 25 seconds as against normal query timings of the order of 0.005 - 0.0002 seconds, which makes it totally unusable. It is possible that some of the delay is being caused because all the possible fields have not yet been indexed, but I would appreciate a better query as a solution, optimized as much as possible, displaying the same results ========================================================================================== How tags work in this system? When a tutor registers, tags are entered and tag relations are created with respect to tutor's details like name, surname etc. When a Tutors create packs, again tags are entered and tag relations are created with respect to pack's details like pack name, description etc. tag relations for tutors stored in tutors_tag_relations and those for packs stored in learning_packs_tag_relations. All individual tags are stored in tags table. ==================================================================== The tables Most of the following tables contain many other fields which I have omitted here. CREATE TABLE IF NOT EXISTS users ( id_user int(10) unsigned NOT NULL AUTO_INCREMENT, name varchar(100) NOT NULL DEFAULT '', surname varchar(155) NOT NULL DEFAULT '', PRIMARY KEY (id_user) ) ENGINE=InnoDB DEFAULT CHARSET=utf8 AUTO_INCREMENT=636 ; CREATE TABLE IF NOT EXISTS tutor_details ( id_tutor int(10) NOT NULL AUTO_INCREMENT, id_user int(10) NOT NULL DEFAULT '0', PRIMARY KEY (id_tutor), KEY Users_FKIndex1 (id_user) ) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=51 ; CREATE TABLE IF NOT EXISTS orders ( id_order int(10) unsigned NOT NULL AUTO_INCREMENT, PRIMARY KEY (id_order), KEY Orders_FKIndex1 (id_user), ) ENGINE=InnoDB DEFAULT CHARSET=utf8 AUTO_INCREMENT=275 ; ALTER TABLE orders ADD CONSTRAINT Orders_ibfk_1 FOREIGN KEY (id_user) REFERENCES users (id_user) ON DELETE NO ACTION ON UPDATE NO ACTION; CREATE TABLE IF NOT EXISTS order_details ( id_od int(10) unsigned NOT NULL AUTO_INCREMENT, id_order int(10) unsigned NOT NULL DEFAULT '0', id_author int(10) NOT NULL DEFAULT '0', PRIMARY KEY (id_od), KEY Order_Details_FKIndex1 (id_order) ) ENGINE=InnoDB DEFAULT CHARSET=utf8 AUTO_INCREMENT=284 ; ALTER TABLE order_details ADD CONSTRAINT Order_Details_ibfk_1 FOREIGN KEY (id_order) REFERENCES orders (id_order) ON DELETE NO ACTION ON UPDATE NO ACTION; CREATE TABLE IF NOT EXISTS learning_packs ( id_lp int(10) unsigned NOT NULL AUTO_INCREMENT, id_author int(10) unsigned NOT NULL DEFAULT '0', PRIMARY KEY (id_lp), KEY Learning_Packs_FKIndex2 (id_author), KEY id_lp (id_lp) ) ENGINE=InnoDB DEFAULT CHARSET=utf8 AUTO_INCREMENT=23 ; CREATE TABLE IF NOT EXISTS tags ( id_tag int(10) unsigned NOT NULL AUTO_INCREMENT, tag varchar(255) DEFAULT NULL, PRIMARY KEY (id_tag), UNIQUE KEY tag (tag), KEY id_tag (id_tag), KEY tag_2 (tag), KEY tag_3 (tag) ) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=3419 ; CREATE TABLE IF NOT EXISTS tutors_tag_relations ( id_tag int(10) unsigned NOT NULL DEFAULT '0', id_tutor int(10) DEFAULT NULL, KEY Tutors_Tag_Relations (id_tag), KEY id_tutor (id_tutor), KEY id_tag (id_tag) ) ENGINE=InnoDB DEFAULT CHARSET=latin1; ALTER TABLE tutors_tag_relations ADD CONSTRAINT Tutors_Tag_Relations_ibfk_1 FOREIGN KEY (id_tag) REFERENCES tags (id_tag) ON DELETE NO ACTION ON UPDATE NO ACTION; CREATE TABLE IF NOT EXISTS learning_packs_tag_relations ( id_tag int(10) unsigned NOT NULL DEFAULT '0', id_tutor int(10) DEFAULT NULL, id_lp int(10) unsigned DEFAULT NULL, KEY Learning_Packs_Tag_Relations_FKIndex1 (id_tag), KEY id_lp (id_lp), KEY id_tag (id_tag) ) ENGINE=InnoDB DEFAULT CHARSET=latin1; ALTER TABLE learning_packs_tag_relations ADD CONSTRAINT Learning_Packs_Tag_Relations_ibfk_1 FOREIGN KEY (id_tag) REFERENCES tags (id_tag) ON DELETE NO ACTION ON UPDATE NO ACTION; =================================================================================== Following is the exact query (this includes classes also - tutors can create classes and search terms are matched with classes created by tutors):- select count(distinct(od.id_od)) as tutor_popularity, CASE WHEN (IF((wc.id_wc 0), ( wc.wc_api_status = 1 AND wc.wc_type = 0 AND wc.class_date '2010-06-01 22:00:56' AND wccp.status = 1 AND (wccp.country_code='IE' or wccp.country_code IN ('INT'))), 0)) THEN 1 ELSE 0 END as 'classes_published', CASE WHEN (IF((lp.id_lp 0), (lp.id_status = 1 AND lp.published = 1 AND lpcp.status = 1 AND (lpcp.country_code='IE' or lpcp.country_code IN ('INT'))),0)) THEN 1 ELSE 0 END as 'packs_published', td . * , u . * from Tutor_Details AS td JOIN Users as u on u.id_user = td.id_user LEFT JOIN Learning_Packs_Tag_Relations AS lptagrels ON td.id_tutor = lptagrels.id_tutor LEFT JOIN Learning_Packs AS lp ON lptagrels.id_lp = lp.id_lp LEFT JOIN Learning_Packs_Categories AS lpc ON lpc.id_lp_cat = lp.id_lp_cat LEFT JOIN Learning_Packs_Categories AS lpcp ON lpcp.id_lp_cat = lpc.id_parent LEFT JOIN Learning_Pack_Content as lpct on (lp.id_lp = lpct.id_lp) LEFT JOIN Webclasses_Tag_Relations AS wtagrels ON td.id_tutor = wtagrels.id_tutor LEFT JOIN WebClasses AS wc ON wtagrels.id_wc = wc.id_wc LEFT JOIN Learning_Packs_Categories AS wcc ON wcc.id_lp_cat = wc.id_wp_cat LEFT JOIN Learning_Packs_Categories AS wccp ON wccp.id_lp_cat = wcc.id_parent LEFT JOIN Order_Details as od on td.id_tutor = od.id_author LEFT JOIN Orders as o on od.id_order = o.id_order LEFT JOIN Tutors_Tag_Relations as ttagrels ON td.id_tutor = ttagrels.id_tutor JOIN Tags as t on (t.id_tag = ttagrels.id_tag) OR (t.id_tag = lptagrels.id_tag) OR (t.id_tag = wtagrels.id_tag) where (u.country='IE' or u.country IN ('INT')) AND CASE WHEN ((t.id_tag = lptagrels.id_tag) AND (lp.id_lp 0)) THEN lp.id_status = 1 AND lp.published = 1 AND lpcp.status = 1 AND (lpcp.country_code='IE' or lpcp.country_code IN ('INT')) ELSE 1 END AND CASE WHEN ((t.id_tag = wtagrels.id_tag) AND (wc.id_wc 0)) THEN wc.wc_api_status = 1 AND wc.wc_type = 0 AND wc.class_date '2010-06-01 22:00:56' AND wccp.status = 1 AND (wccp.country_code='IE' or wccp.country_code IN ('INT')) ELSE 1 END AND CASE WHEN (od.id_od0) THEN od.id_author = td.id_tutor and o.order_status = 'paid' and CASE WHEN (od.id_wc 0) THEN od.can_attend_class=1 ELSE 1 END ELSE 1 END AND 1 group by td.id_tutor order by tutor_popularity desc, u.surname asc, u.name asc limit 0,20 Please note - The provided database structure does not show all the fields and tables as in this query

    Read the article

< Previous Page | 188 189 190 191 192 193 194 195 196 197 198 199  | Next Page >