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  • SQL LEFT JOIN help

    - by Stolz
    My scenario: There are 3 tables for storing tv show information; season, episode and episode_translation. My data: There are 3 seasons, with 3 episodes each one, but there is only translation for one episode. My objetive: I want to get a list of all the seasons and episodes for a show. If there is a translation available in a specified language, show it, otherwise show null. My attempt to get serie 1 information in language 1: SELECT season_number AS season,number AS episode,name FROM season NATURAL JOIN episode NATURAL LEFT JOIN episode_trans WHERE id_serie=1 AND id_lang=1 ORDER BY season_number,number result: +--------+---------+--------------------------------+ | season | episode | name | +--------+---------+--------------------------------+ | 3 | 3 | Episode translated into lang 1 | +--------+---------+--------------------------------+ expected result +-----------------+--------------------------------+ | season | episode| name | +-----------------+--------------------------------+ | 1 | 1 | NULL | | 1 | 2 | NULL | | 1 | 3 | NULL | | 2 | 1 | NULL | | 2 | 2 | NULL | | 2 | 3 | NULL | | 3 | 1 | NULL | | 3 | 2 | NULL | | 3 | 3 | Episode translated into lang 1 | +--------+--------+--------------------------------+ Full DB dump http://pastebin.com/Y8yXNHrH

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  • Optimize master-detail insert statements

    - by Dave Jarvis
    Quest After a day of running (against nearly 1 GB of data), a set of statements are tumbling down to 40 inserts per second. I am looking to increase that by an order of magnitude or two. SQL Code The code to insert the information comes in two parts: a master record and detail records. The master record: INSERT INTO MONTH_REF (DISTRICT_ID, STATION_ID, CATEGORY_ID, YEAR, MONTH) VALUES ('101', '0066', '010', 1984, 07); The detail records: INSERT INTO DAILY (MONTH_REF_ID, AMOUNT, DAILY_FLAG_ID, DAY) VALUES ((SELECT ID FROM MONTH_REF M WHERE M.DISTRICT_ID = '101' AND M.STATION_ID = '0066' AND M.CAT EGORY_ID = '010' AND M.YEAR = 1984 AND M.MONTH = 07), 0, ' ', 1); INSERT INTO DAILY (MONTH_REF_ID, AMOUNT, DAILY_FLAG_ID, DAY) VALUES ((SELECT ID FROM MONTH_REF M WHERE M.DISTRICT_ID = '101' AND M.STATION_ID = '0066' AND M.CAT EGORY_ID = '010' AND M.YEAR = 1984 AND M.MONTH = 07), 0.5, ' ', 2); INSERT INTO DAILY (MONTH_REF_ID, AMOUNT, DAILY_FLAG_ID, DAY) VALUES ((SELECT ID FROM MONTH_REF M WHERE M.DISTRICT_ID = '101' AND M.STATION_ID = '0066' AND M.CAT EGORY_ID = '010' AND M.YEAR = 1984 AND M.MONTH = 07), 0, 'T', 3); Proposed Solution INSERT INTO MONTH_REF (DISTRICT_ID, STATION_ID, CATEGORY_ID, YEAR, MONTH) VALUES ('101', '0066', '010', 1984, 07); SET @month_ref_id := (SELECT LAST_INSERT_ID()); INSERT INTO DAILY (MONTH_REF_ID, AMOUNT, DAILY_FLAG_ID, DAY) VALUES (@month_ref_id, 0, ' ', 1); INSERT INTO DAILY (MONTH_REF_ID, AMOUNT, DAILY_FLAG_ID, DAY) VALUES (@month_ref_id, 0.5, ' ', 2); INSERT INTO DAILY (MONTH_REF_ID, AMOUNT, DAILY_FLAG_ID, DAY) VALUES (@month_ref_id, 0, 'T', 3); Constraints The MONTH_REF table has an AUTO_INCREMENT primary key and is indexed on it. The DAILY table has no index and no primary key. A primary key can be added to the DAILY table, if it would help. Question Is there a more efficient way to execute the (billion or so) insert statements than the proposed solution? Thank you!

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  • Performance with timestamp conditions

    - by Tim Whitlock
    Which of the following is faster, or are they equivalent? (grabbing recent most records from a TIMESTAMP COLUMN) SELECT UNIX_TIMESTAMP(`modified`) stamp FROM `some_table` HAVING stamp > 127068799 ORDER BY stamp DESC or SELECT UNIX_TIMESTAMP(`modified`) stamp FROM `some_table` WHERE UNIX_TIMESTAMP(`modified`) > 127068799 ORDER BY `modified` DESC or even another combination?

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  • How to select latest change done in the given Table structure?

    - by OM The Eternity
    I have a Table structure as id, trackid, table_name, operation, oldvalue, newvalue, field, changedonetime Now if I have 3 rows for the same "trackid" same "field", then how can i select the latest out of the three? i.e. for e.g.: id = 100 trackid = 152 table_name = jos_menu operation= UPDATE oldvalue = IPL newvalue = IPLcccc field = name live = 0 changedonetime = 2010-04-30 17:54:39 and id = 101 trackid = 152 table_name = jos_menu operation= UPDATE oldvalue = IPLcccc newvalue = IPL2222 field = name live = 0 changedonetime = 2010-04-30 18:54:39 As u can see above the secind entry is the latest change, Now what query I shoud use to get the only one and Latest row out of many such rows...

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  • How can I get columns name from select query in php?

    - by Farshad Mehrvarzan
    I want to execute a SELECT query but I don't how many columns to select. Like: select name, family from persons; How can I know which columns to select? "I am currently designing a site for the execute query by users. So when the user executes this query, I won't know which columns selected. But when I want to show the results and draw a table for the user I should know which columns selected."

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  • jQuery: Sorting hierarchical data?

    - by Industrial
    Hi everybody, I have tried for some time to work out a way of sorting nested categories with jQuery. I failed to build my own plugin to do this, so I tried to find something that were available already. Tried a few hours now with this one, http://www.jordivila.net/code/js/jquery/ui-widgetTreeList_inheritance/widgetTreeListSample.aspx and cant get it to work. What are the alternatives of creating a jQuery / jQuery UI script that handles sorting children and parent categories in a way that can be combined with a AJAX PHP backend to handle the actual sorting in the database? Thanks!

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  • Exploring search options for PHP

    - by Joshua
    I have innoDB table using numerous foreign keys, but we just want to look up some basic info out of it. I've done some research but still lost. 1) How can I tell if my host has Sphinx installed already? I don't see it as an option for table storage method (i.e. innodb, myisam). 2) Zend_Search_Lucene, responsive enough for AJAX functionality of millions of records? 3) Mirror my innoDB with a myisam? Make every innodb transaction end with a write to the myisam, then use 1:1 lookups? How would I do this automagically? This should make MyISAM ACID-compliant and free(er) from corruption no? 4) PostgreSQL fulltext queries don't even look like SQL to me wtf, I don't have time to learn a new SQL syntax I need noob options 5) ???????????????????? This is high volume site on a decently-equipped VPS Thanks very much for any ideas.

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  • Table format for Click stats

    - by Francesc
    Hello. I'm currently developing an URL shortening service. I want to allow users to see the stats for their URLs. How has to be the table. First, it has to be the url ID, but then, how I can sort the clicks per day?

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  • SELECT(IF(IN query.

    - by Harold
    There are 3 tables. Products, Options and Prod_Opts_relations. The latter holds product_id and option_id so i should be able to figure out which options are selected for any given product. Now i want to retrieve all options from the options table where the value of an extra alias field should hold checked or unchecked depending on the existance of a mathing record in the relations table for a give product id. Thus far i came up with this: SELECT IF(IN(SELECT id_option FROM prod_opt_relations WHERE id_product='18'),'y','n') AS booh ,optionstable.id AS parent_id ,optionstable.name_en AS parent_english ,optionstable.name_es AS parent_spanish FROM product_options AS optionstable WHERE 1 resulting in syntax errors. Alas i just cannot figure out where things go wrong here

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  • How to make an add friend/defriend function in PHP?

    - by user300371
    I have created a site where people can create a profile. But I am trying to figure out how to start on making an add friend button so users can have friends. In my user table, i have user_id, first_name, last_name, email, etc. Should I somehow relate the user_id of the user and the friend in a friend table? I am a novice to programming, so these things are still new to me. Thanks!

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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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  • Getting a table's values into a tree

    - by Jason
    So, I have a table like such: id|root|kw1|kw2|kw3|kw4|kw5|name 1| A| B| C| D| E| F|fileA 2| A| B| | | | |fileB 3| B| C| D| E| | |fileC 4| A| B| | | | |fileD (several hundred rows...) And I need to get it into a tree like the following: *A *B -fileB -fileD *C *D *E *F -fileA *B *C *D *E -fileC I'm pretty sure the table is laid out poorly but it's what I have to live with. I've read a little about Adjacency List Model & Modified Preorder Tree Traversal but I don't think my data is laid out correctly. I think this requires a recursive function, but I'm not at all sure how to go about that. I'm open to any ideas of how to get this done even if it means extracting the data into a new table just to process this. Are there any good options available to me or any good ways to do this? (Examples are a bonus of course)

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  • Is it necessary to write ROLLBACK if queries fail?

    - by Donator
    I write mysql_query("SET AUTOCOMMIT=0"); mysql_query("START TRANSACTION"); before I write all queries. Then check if all of them are true and then write: mysql_query("COMMIT"); But if one of query fails, I just pass COMMIT query. So do I really need ROLLBACK function if one of the queries fail? Because without ROLLBACK it also works. Thanks.

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  • how to join 3 relational tables

    - by orioncabbar
    Hello there, how to join 3 relational tables with the structure: t1 | id t2 | id | rating t3 | source_id | relation t3 stores the data of a field which t1 and t2 uses both. so source_id field can be t1's id or t2's id. input : t1 id output : t2 rating an example: **t1** id | --------- 42 | **t2** id | rating ------------- 37 | 9.2 **t3** id | source_id -------------- 42 | 1 37 | 1 26 | 2 23 | 1 what i want is to get 9.2 output with 42 input. can you do that in one sql query?

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  • problem in creating a php tree menu

    - by Mac Taylor
    hi mates im writing a tree menu for my categories in php and i wonder how can i code it correctly ! this is my table in database " |----topicid------topicname--------parent | |---- 1 ------ News -------- 0 | |---- 2 ------ sport -------- 1 | |---- 3 ------ games -------- 1 | |---- 4 ------ PES -------- 3 | so now for showing it like a tree i did try but not worked : $result = mysql_query("SELECT * FROM Topics ORDER BY topicid"); while ($row = mysql_fetchrow($result)) { $id = intval($row['topicid']); $title = filter($row['topicname'], "nohtml"); $parent = $row['parent'] ; if ($parent==0) { $menu_item .= "<li><span class='folder'><a title = \"$alt\" href=\"modules.php?name=News&amp;new_topic=$id\">$title</a></span></li>"; }else { $result = mysql_query("SELECT * FROM ".$prefix."_Topics where parent='$id' ORDER BY topicid"); while ($row = mysql_fetchrow($result)) { $id = intval($row['topicid']); $title = filter($row['topicname'], "nohtml"); $parent = $row['parent'] ; $menu_item .= " <ul><li><span class='file'><a title = \"$alt\" href=\"modules.php?name=News&amp;new_topic=$id\">$title</a></span></li></ul>"; } } i dont know how to solve this

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  • Returning multiple aggregate functions as rows

    - by SDLFunTimes
    I need some help formulating a select statement. I need to select the total quantity shipped for each part with a distinct color. So the result should be a row with the color name and the total. Here's my schema: create table s ( sno char(5) not null, sname char(20) not null, status smallint, city char(15), primary key (sno) ); create table p ( pno char(6) not null, pname char(20) not null, color char(6), weight smallint, city char(15), primary key (pno) ); create table sp ( sno char(5) not null, pno char(6) not null, qty integer not null, primary key (sno, pno) );

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