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  • database datatype size

    - by yeeen
    Just to clarify by specifying sth like VARCHAR(45) means it can take up to max 45 characters? I rmb I heard from someone a few years ago that the number in the parathesis doesn't refer to the no of characters, then the person tried to explain to me sth quite complicated which i don't understand n forgot alr. And what is the difference btn CHAR and VARCHAR? I did search ard a bit and see that CHAR gives u the max of the size of the column and it is better to use it if ur data has a fix size and use VARCHAR if ur data size varies. But if it gives u the max of the size of the column of all the data of this col, isn't it better to use it when ur data size varies? Esp if u don't know how big is ur data size gg to be. VARCHAR needs to specify the size (CHAR don't really need right?), isn't it more troublesome?

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  • Quoting of decimal values

    - by Peter
    I am storing a running total in a Decimal(10,2) field and adding to it as items are processed. update foo set bar = bar + '3.15' About 20% of the times a warning is issued "Data truncated for column 'bar' at row 4" This warning is never issued if the update value is not quoted. Should decimal values be quoted?

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  • optimizing an sql query using inner join and order by

    - by Sergio B
    I'm trying to optimize the following query without success. Any idea where it could be indexed to prevent the temporary table and the filesort? EXPLAIN SELECT SQL_NO_CACHE `groups`.* FROM `groups` INNER JOIN `memberships` ON `groups`.id = `memberships`.group_id WHERE ((`memberships`.user_id = 1) AND (`memberships`.`status_code` = 1 AND `memberships`.`manager` = 0)) ORDER BY groups.created_at DESC LIMIT 5;` +----+-------------+-------------+--------+--------------------------+---------+---------+---------------------------------------------+------+----------------------------------------------+ | id | select_type | table | type | possible_keys | key | key_len | ref | rows | Extra | +----+-------------+-------------+--------+--------------------------+---------+---------+---------------------------------------------+------+----------------------------------------------+ | 1 | SIMPLE | memberships | ref | grp_usr,grp,usr,grp_mngr | usr | 5 | const | 5 | Using where; Using temporary; Using filesort | | 1 | SIMPLE | groups | eq_ref | PRIMARY | PRIMARY | 4 | sportspool_development.memberships.group_id | 1 | | +----+-------------+-------------+--------+--------------------------+---------+---------+---------------------------------------------+------+----------------------------------------------+ 2 rows in set (0.00 sec) +--------+------------+-----------------------------------+--------------+-----------------+-----------+-------------+----------+--------+------+------------+---------+ | Table | Non_unique | Key_name | Seq_in_index | Column_name | Collation | Cardinality | Sub_part | Packed | Null | Index_type | Comment | +--------+------------+-----------------------------------+--------------+-----------------+-----------+-------------+----------+--------+------+------------+---------+ | groups | 0 | PRIMARY | 1 | id | A | 6 | NULL | NULL | | BTREE | | | groups | 1 | index_groups_on_name | 1 | name | A | 6 | NULL | NULL | YES | BTREE | | | groups | 1 | index_groups_on_privacy_setting | 1 | privacy_setting | A | 6 | NULL | NULL | YES | BTREE | | | groups | 1 | index_groups_on_created_at | 1 | created_at | A | 6 | NULL | NULL | YES | BTREE | | | groups | 1 | index_groups_on_id_and_created_at | 1 | id | A | 6 | NULL | NULL | | BTREE | | | groups | 1 | index_groups_on_id_and_created_at | 2 | created_at | A | 6 | NULL | NULL | YES | BTREE | | +--------+------------+-----------------------------------+--------------+-----------------+-----------+-------------+----------+--------+------+------------+---------+ +-------------+------------+----------------------------------------------------------+--------------+-------------+-----------+-------------+----------+--------+------+------------+---------+ | Table | Non_unique | Key_name | Seq_in_index | Column_name | Collation | Cardinality | Sub_part | Packed | Null | Index_type | Comment | +-------------+------------+----------------------------------------------------------+--------------+-------------+-----------+-------------+----------+--------+------+------------+---------+ | memberships | 0 | PRIMARY | 1 | id | A | 2 | NULL | NULL | | BTREE | | | memberships | 0 | grp_usr | 1 | group_id | A | 2 | NULL | NULL | YES | BTREE | | | memberships | 0 | grp_usr | 2 | user_id | A | 2 | NULL | NULL | YES | BTREE | | | memberships | 1 | grp | 1 | group_id | A | 2 | NULL | NULL | YES | BTREE | | | memberships | 1 | usr | 1 | user_id | A | 2 | NULL | NULL | YES | BTREE | | | memberships | 1 | grp_mngr | 1 | group_id | A | 2 | NULL | NULL | YES | BTREE | | | memberships | 1 | grp_mngr | 2 | manager | A | 2 | NULL | NULL | YES | BTREE | | | memberships | 1 | complex_index | 1 | group_id | A | 2 | NULL | NULL | YES | BTREE | | | memberships | 1 | complex_index | 2 | user_id | A | 2 | NULL | NULL | YES | BTREE | | | memberships | 1 | complex_index | 3 | status_code | A | 2 | NULL | NULL | YES | BTREE | | | memberships | 1 | complex_index | 4 | manager | A | 2 | NULL | NULL | YES | BTREE | | | memberships | 1 | index_memberships_on_user_id_and_status_code_and_manager | 1 | user_id | A | 2 | NULL | NULL | YES | BTREE | | | memberships | 1 | index_memberships_on_user_id_and_status_code_and_manager | 2 | status_code | A | 2 | NULL | NULL | YES | BTREE | | | memberships | 1 | index_memberships_on_user_id_and_status_code_and_manager | 3 | manager | A | 2 | NULL | NULL | YES | BTREE | | +-------------+------------+----------------------------------------------------------+--------------+-------------+-----------+-------------+----------+--------+------+------------+---------+

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  • Update/Increment a single column on multiple rows at once

    - by Jordan Feldstein
    I'm trying to add rows to a column, keeping the order of the newest column set to one, and all other rows counting up from there. In this case, I add a new row with order=0, then use this query to update all the rows by one. "UPDATE favorits SET order = order+1" However, what happens is that all the rows are updated to the same value. I get a stack of favorites, all with order 6 for example, when it should be one with 1, the next with 2 and so on. How do I update these rows in a way that orders them the way they should be? Thanks, ~Jordan

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  • Query results taking too long on 200K database, speed up tips?

    - by colorfulgrayscale
    I have a sql statement where I'm joining about 4 tables, each with 200K rows. The query runs, but keeps freezing. When I do a join on 3 tables instead, it returns the rows (takes about 10secs). Any suggestion why? suggestions to speed up? Thanks! Code SELECT * FROM equipment, tiremap, workreference, tirework WHERE equipment.tiremap = tiremap.`TireID` AND tiremap.`WorkMap` = workreference.`aMap` AND workreference.`bMap` = tirework.workmap LIMIT 5 p.s and if it helps any, I'm using sql alchemy to generate this code, the sqlalchemy code for this is query = session.query(equipment, tiremap, workreference, tirework) query = query.filter(equipment.c.tiremap == tiremap.c.TireID) query = query.filter(tiremap.c.WorkMap==workreference.c.aMap) query = query.filter(workreference.c.bMap == tirework.c.workmap) query = query.limit(5) query.all()

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  • database splitting; multiple tables

    - by Ben
    I am coding a classifieds ad web app. What is the optimal way to structure the database for this? Because of the high repeatability, would it be faster (in terms of searching/indexing) to have a separate table in the database for each city? Or would it be okay to just have one table for every city (it would have a lot of rows..). The classifieds table has id, user_id, city_name, category,[description and detail fields].

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  • Help with SQL query (add 5% to users with conditions)

    - by Mestika
    Hi everyone, I’m having some difficulties with a query which purpose is to give users with more than one thread (called CS) in current year a 5% point “raise”. My relational schema looks like this: Thread = (threadid, threadname, threadLocation) threadoffering = (threadid, season, year, user) user = (name, points) Then, what I need is to check: WHERE thread.threadid = threadoffering.threadid AND where threadoffering.year AND threadoffering.season = currentDate AND where threadoffering.User 1 GIVE 5 % raise TO user.points I hope it is explained thoroughly but otherwise here it is in short text: Give a 5 % “point raise” to all users who has more than 1 thread in threadLocation CS in the current year and season (always dynamic, so for example now is year = 2010 and season is = spring). I am looking forward to your answer Sincerely, Emil

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  • How to mark posts as edited?

    - by user156814
    I would like to have questions marked as "Edited", but I dont know what the best way to do this would be. Users post a question, people answer/comment on the question, and if necessary the user edits/updates the question (just like SO). I would like to note that the user edited the question, but I'm not sure of the best way to do this. I was going to add a last_edited column in the table (because thats all thats really important to me), but I'm not sure if I should just split the edit times (and whatever else) into another table and record everytime the question gets edited.

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  • Suggest me solution to track the change in test DB and replicate in Another DB

    - by Parth
    Suggest me solution to track the change in test DB and replicate in Another DB... My Client need a script or any solution, if he has two Database, One Test DB in which he tests his data on test portal and if he find it appropriate he can use those changes to be done in main DB to display on Live site.. Fior this he needs the solution to record or track all updation/deletion/insertion, so that he can do the same in main DB if found appropriate, ** NOTE: ** we hav only on server, no separate server, hence binary log replication doesnt seems to be working for my case...

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  • optimize query: get al votes from user's item

    - by Toni Michel Caubet
    hi there! i did it my way because i'm very bad getting results from two tables... Basically, first i get all the id items that correspond to the user, and then i calculate the ratings of each item. But, there is two different types of object item, so i do this 2 times: show you: function votos_usuario($id){ $previa = "SELECT id FROM preguntas WHERE id_usuario = '$id'"; $r_previo = mysql_query($previa); $ids_p = '0, '; while($items_previos = mysql_fetch_array($r_previo)){ $ids_p .= $items_previos['id'].", "; //echo "ids pregunta usuario: ".$items_previos['id']."<br>"; } $ids = substr($ids_p,0,-2); //echo $ids; $consulta = "SELECT valor FROM votos_pregunta WHERE id_pregunta IN ( $ids )"; //echo $consulta; $resultado = mysql_query($consulta); $votos_preguntas = 0; while($voto = mysql_fetch_array($resultado)){ $votos_preguntas = $votos_preguntas + $voto['valor']; } $previa_r = "SELECT id FROM recetas WHERE id_usuario = '$id'"; $r_previo_r = mysql_query($previa_r); $ids_r = '0, '; while($items_previos_r = mysql_fetch_array($r_previo_r)){ $ids_r .= $items_previos_r['id'].", "; //echo "ids pregunta usuario: ".$items_previos['id']."<br>"; } $ids = substr($ids_r,0,-2); $consulta_b = "SELECT valor FROM votos_receta WHERE id_receta IN ( $ids )"; //echo $consulta; $resultado_b = mysql_query($consulta_b); $votos_recetas = 0; while($voto_r = mysql_fetch_array($resultado_b)){ $votos_recetas = $votos_recetas + $voto_r['valor']; } $total = $votos_preguntas + $votos_recetas; return $total; } As you can si this is two much.. O(n^2) Feel like thinking? thanks!

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  • Slug ID's with the same name?

    - by James Jeffery
    I want to create slug URL's from a users title in my system. If a user types "The best way's to get slim; period!", then I want the slug to be "the-best-ways-to-get-slim-period". Also, if someone has already created a page with that title I want the slug to be "the-best-ways-to-get-slim-period-1". My question is how can I check the database before a record is created? Ok, obviously I am going to have to perform a check in the database, and then a write. That's 2 queries. Is this the normal way to do it? Also, are there any conventional regular expressions for filtering non alpha/number characters and replacing spaces with hyphens? Any help is much appreciated. Thanks.

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  • Database Schema for survey polling application with a default choice.

    - by user156814
    I have a survey application, where users can create surveys and give choices for every survey. Other users can choose their answers for the aurvey and then polls are taken to get the results of the survey. I already have the database schema for this Questions id, user_id, category_id, question_text, date_started Answers id, user_id, question_id, choice_id, explanation, date_added Choices id, question_id, choice_text As for now, users can choose their own choice answers to their surveys... but I want to be able to add a default "I dont care" or "I dont know" choice to every survey for people who simply dont care about the topic to take sides or who cant choose. So lets say theres a survey that asks who was a better president, George W. Bush, Bill Clinton, Ronald Reagon, or Richard Nixon... I want to be able to add a default "I dont care" option. I was thinking to just add that extra choice EVERY TIME a user creates a survey, but then I wouldn't have much control over the text for that choice after that survey has been created, and I want to know if theres a better way to do this, like create another table or something Thanks

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  • SQL Join to only the maximum row puzzle

    - by Billy ONeal
    Given the following example data: Users +--------------------------------------------------+ | ID | First Name | Last Name | Network Identifier | +--------------------------------------------------+ | 1 | Billy | O'Neal | bro4 | +----+------------+-----------+--------------------+ | 2 | John | Skeet | jsk1 | +----+------------+-----------+--------------------+ Hardware +----+-------------------+---------------+ | ID | Hardware Name | Serial Number | +----+-------------------+---------------+ | 1 | Latitude E6500 | 5555555 | +----+-------------------+---------------+ | 2 | Latitude E6200 | 2222222 | +----+-------------------+---------------+ HardwareAssignments +---------+-------------+-------------+ | User ID | Hardware ID | Assigned On | +---------+-------------+-------------+ | 1 | 1 | April 1 | +---------+-------------+-------------+ | 1 | 2 | April 10 | +---------+-------------+-------------+ | 2 | 2 | April 1 | +---------+-------------+-------------+ | 2 | 1 | April 11 | +---------+-------------+-------------+ I'd like to write a SQL query which would give the following result: +--------------------+------------+-----------+----------------+---------------+-------------+ | Network Identifier | First Name | Last Name | Hardware Name | Serial Number | Assigned On | +--------------------+------------+-----------+----------------+---------------+-------------+ | bro4 | Billy | O'Neal | Latitude E6200 | 2222222 | April 10 | +--------------------+------------+-----------+----------------+---------------+-------------+ | jsk1 | John | Skeet | Latitude E6500 | 5555555 | April 11 | +--------------------+------------+-----------+----------------+---------------+-------------+ My trouble is that the maximum "Assigned On" date for each user needs to be selected for each individual user and used for the actual join ... Is there a clever way accomplish this in SQL?

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  • Best practice for storing global data in PHP?

    - by user281434
    Hi I'm running a web application that allows a user to log in. The user can add/remove content to his/her 'library' which is displayed on a page called "library.php". Instead of querying the database for the contents of the users library everytime they load "library.php", I want to store it globally for PHP when the user logs in, so that the query is only run once. Is there a best practice for doing this? fx. storing their library in an array in a session? Thanks for your time

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  • DataMapper is only returning the last part of this query

    - by Josh K
    So I'm using the following: $r = new Record(); $r->select('ip, count(*) as ipcount'); $r->group_by('ip'); $r->order_by('ipcount', 'desc'); $r->limit(5); $r->get(); foreach($r->all as $record) { echo($record->ip." "); echo($record->ipcount." <br />"); } And I only get the last (fifth) record echo'ed out and no ipcount echoed. Is there a different way to go around this? I'm working on learning DataMapper (hence the questions) and need to figure some of this out. I haven't quite wrapped my head around the whole ORM thing.

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  • what is called KEY

    - by Bharanikumar
    CREATE TABLE `ost_staff` ( `staff_id` int(11) unsigned NOT NULL auto_increment, `group_id` int(10) unsigned NOT NULL default '0', `dept_id` int(10) unsigned NOT NULL default '0', `username` varchar(32) collate latin1_german2_ci NOT NULL default '', `firstname` varchar(32) collate latin1_german2_ci default NULL, `lastname` varchar(32) collate latin1_german2_ci default NULL, `passwd` varchar(128) collate latin1_german2_ci default NULL, `email` varchar(128) collate latin1_german2_ci default NULL, `phone` varchar(24) collate latin1_german2_ci NOT NULL default '', `phone_ext` varchar(6) collate latin1_german2_ci default NULL, `mobile` varchar(24) collate latin1_german2_ci NOT NULL default '', `signature` varchar(255) collate latin1_german2_ci NOT NULL default '', `isactive` tinyint(1) NOT NULL default '1', `isadmin` tinyint(1) NOT NULL default '0', `isvisible` tinyint(1) unsigned NOT NULL default '1', `onvacation` tinyint(1) unsigned NOT NULL default '0', `daylight_saving` tinyint(1) unsigned NOT NULL default '0', `append_signature` tinyint(1) unsigned NOT NULL default '0', `change_passwd` tinyint(1) unsigned NOT NULL default '0', `timezone_offset` float(3,1) NOT NULL default '0.0', `max_page_size` int(11) NOT NULL default '0', `created` datetime NOT NULL default '0000-00-00 00:00:00', `lastlogin` datetime default NULL, `updated` datetime NOT NULL default '0000-00-00 00:00:00', PRIMARY KEY (`staff_id`), UNIQUE KEY `username` (`username`), KEY `dept_id` (`dept_id`), **KEY `issuperuser` (`isadmin`),** **KEY `group_id` (`group_id`,`staff_id`)** ) ENGINE=MyISAM AUTO_INCREMENT=35 DEFAULT CHARSET=latin1 COLLATE=latin1_german2_ci; Hi the above query is the osticket open source one, i know primary key , foreign key , unique but AM NOT SURE WHAT IS THIS KEY group_id (group_id,staff_id) Please tell me, this constraints name....

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  • SQL: Daily Average of Logins Per Hour

    - by jerrygarciuh
    This query is producing counts of logins per hour: SELECT DATEADD(hour, DATEDIFF(hour, 0, EVENT_DATETIME), 0), COUNT(*) FROM EVENTS_ALL_RPT_V1 WHERE EVENT_NAME = 'Login' AND EVENT_DATETIME >= CONVERT(DATETIME, '2010-03-17 00:00:00', 120) AND EVENT_DATETIME <= CONVERT(DATETIME, '2010-03-24 00:00:00', 120) GROUP BY DATEADD(hour, DATEDIFF(hour, 0, EVENT_DATETIME), 0) ORDER BY DATEADD(hour, DATEDIFF(hour, 0, EVENT_DATETIME), 0) ...with lots of results like this: Datetime COUNT(*) ---------------------------------- 2010-03-17 12:00:00.000 135 2010-03-17 13:00:00.000 129 2010-03-17 14:00:00.000 147 What I need to figure out is how to query the average logins per hour for a given day. Any help?

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  • Retrieve object from jquery in php script

    - by majc
    I'm trying to rebuild an object encoded with Json but i'm not getting any value. JQuery: $.post("views/insert_tasks.php",{ clickedRows : clickrows , <?php echo "tasks:'" . json_encode($tasks) . "'"; ?> }, function(data) { }); this is the PHPcode to retrieve the object: $tasks = json_decode(stripslashes($_POST['tasks']), true); $tasks is empty after execute the code above. This is what I'm getting with the $_POST['tasks']: [{"task_id":"1","description":"<p>Fazer heroi</p>","createdat":"Saturday 22nd of May 2010 11:37:37 PM","createdby":"Miguel Cardoso","max_requests":"2","max_duration":"5","job_id":"Concept Artist"},{"task_id":"2","description":"<p>teste2</p>","createdat":"Sunday 23rd of May 2010 11:23:55 AM","createdby":"Miguel Cardoso","max_requests":"2","max_duration":"5","job_id":"3D Modeller"},{"task_id":"3","description":"<p>teste3</p>","createdat":"Sunday 23rd of May 2010 11:45:39 AM","createdby":"Miguel Cardoso","max_requests":"1","max_duration":"10","job_id":"Writer"}] What I'm doing wrong?

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  • Query for props list with or without values

    - by vitto
    Hi, I'm trying to make a SELECT on three relational tables like these ones: table_materials -> material_id - material_name table_props -> prop_id - prop_name table_materials_props - row_id -> material_id -> prop_id - prop_value On my page, I'd like to get a result like this one but i have some problem with the query: material prop A prop B prop C prop D prop E wood 350 NULL NULL 84 16 iron NULL 17 NULL NULL 201 copper 548 285 99 NULL NULL so the query should return something like: material prop_name prop_value wood prop A 350 wood prop B NULL wood prop C NULL wood prop D 84 wood prop E 16 // and go on with others rows i thought to use something like: SELECT * FROM table_materials AS m INNER JOIN table_materials_props AS mp ON m.material_id = mp.material_id INNER JOIN table_materials_props AS p ON mp.prop_id = p.prop_id ORDER BY p.prop_name the problem is the query doesn't return the NULL values, and I need the same prop order for all the materials regardless of prop values are NULL or not I hope this example is clear!

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