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  • Recalculate Counter Cache of 120k Records [Rails / ActiveRecord]

    - by Sebastian
    The following situation: I have a poi model, which has many pictures (1:n). I want to recalculate the counter_cache column, because the values are inconsistent. I've tried to iterate within ruby over each record, but this takes much too long and quits sometimes with some "segmentation fault" bugs. So i wonder, if its possible to do this with a raw sql query?

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  • Search SQL Question Between Related Two Tables

    - by mTuran
    Hi, I am writing some kind of search engine for my web application and i have a problem. I have 2 tables first of is projects table: PROJECTS TABLE id int(11) NO PRI NULL auto_increment employer_id int(11) NO MUL NULL project_title varchar(100) NO MUL NULL project_description text NO NULL project_budget int(11) NO NULL project_allowedtime int(11) NO NULL project_deadline datetime NO NULL total_bids int(11) NO NULL average_bid int(11) NO NULL created datetime NO MUL NULL active tinyint(1) NO MUL NULL PROJECTS_SKILLS TABLE project_id int(11) NO MUL NULL skill_id int(11) NO MUL NULL For example: I want ask this query to database: 1-) Skills are 5 and 7. 2-) Order results by created 3-) project title contains "php" word. 4-) Returned rows should contain projects.* columuns. 5-) Projects should be distinct(i don't want same projects in return of query). Please write sql query that ensure these conditions. Thank You.

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  • How to get the answers version and use it in IF loop

    - by sai
    delimiter // DROP PROCEDURE `getData`// CREATE DEFINER=`root`@`localhost` PROCEDURE `getData`(IN templateName VARCHAR(45),IN templateVersion VARCHAR(45),IN userId VARCHAR(45)) BEGIN set @version = CONCAT("SELECT `saveOEMsData_answersVersion` FROM `saveOEMsData` WHERE `saveOEMsData_templateName` = '",templateName,"' AND `saveOEMsData_templateVersion` = ",templateVersion," AND `saveOEMsData_userId`= ",userId); PREPARE s1 from @version; EXECUTE S1; END // delimiter ; I am retreiving saveOEMsData_answersVersion, but I have to use it in an IF loop, as in if the version == 1, then I would use a query, else I would use something else. But I am not able to use the version. Could someone help with this?? I am only able to print but not able to use the version.

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  • how to combine these queries

    - by mmcgrail
    and get meaningful results. Currently I am running these three queries: SELECT t.type,t.id,s.title FROM db1.tags t INNER JOIN db1.st s ON s.id=t.id WHERE id LIKE '%%' AND t.tag='foo' AND t.type='s' ORDER BY tag desc LIMIT 0, 19 SELECT t.type,t.id,v.title FROM db1.tags t INNER JOIN db1.vi v ON v.id=t.id WHERE id LIKE '%%' AND t.tag='foo' AND t.type='v' ORDER BY tag desc LIMIT 0, 19 SELECT t.type,t.id,i.ca AS title FROM db1.tags t INNER JOIN db2.tablename i ON i.id=t.id WHERE id LIKE '%%' AND t.tag='foo' AND t.type='i' ORDER BY tag desc LIMIT 0, 19 then trying to combine the data results but what I would really prefer is if I could combine them into a single query. Any thoughts?

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  • SQL Query Math Gymnastics

    - by keruilin
    I have two tables of concern here: users and race_weeks. User has many race_weeks, and race_week belongs to User. Therefore, user_id is a fk in the race_weeks table. I need to perform some challenging math on fields in the race_weeks table in order to return users with the most all-time points. Here are the fields that we need to manipulate in the race_weeks table. races_won (int) races_lost (int) races_tied (int) points_won (int, pos or neg) recordable_type(varchar, Robots can race, but we're only concerned about type 'User') Just so that you fully understand the business logic at work here, over the course of a week a user can participate in many races. The race_week record represents the summary results of the user's races for that week. A user is considered active for the week if races_won, races_lost, or races_tied is greater than 0. Otherwise the user is inactive. So here's what we need to do in our query in order to return users with the most points won (actually net_points_won): Calculate each user's net_points_won (not a field in the DB). To calculate net_points, you take (1000 * count_of_active_weeks) - sum(points__won). (Why 1000? Just imagine that every week the user is spotted a 1000 points to compete and enter races. We want to factor-out what we spot the user because the user could enter only one race for the week for 100 points, and be sitting on 900, which we would skew who actually EARNED the most points.) This one is a little convoluted, so let me know if I can clarify further.

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  • Dealing with a badly formatted CSV file

    - by Josh K
    I have an exceptionally bad CSV file. Although I "solved" the problem in the end by manually writing scripts to process and reprocess this specific file I wanted to know if there were any other solutions out there. You have a CSV file that has all the fields terminated by | (pipe) characters. Running a quick check shows you that there are 53 fields in the file. The person who gave you the file claims there there are only 28 fields. Not all of the fields have information in them. For example there are five custom_field_{num} fields which may or may not have data. How would you get this into a database nicely? The ideal solution (and one I searched high and low for) would be to just throw it all into a table with no column names or specifications. Then remove any columns that were completely blank and then give them titles and specifications.

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  • In SQL, can we always write an inner join statement as a main query and subquery if we only want to

    - by Jian Lin
    In SQL, can we always write an inner join statement as a main query and subquery or vice versa if we only want to find the intersection? For example, select * from gifts g where g.giftID in (select giftID from sentGifts); can do a join and show the gifts sent in the sentGifts table, but it won't be able to show the sentTime because that is inside the subquery. But if all we care is to find the intersection, without caring what is being displayed, then we can always convert one to the other?

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  • 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();

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  • Letting users trial your web app before sign-up: sessions or temp db?

    - by Mat
    I've seen a few instances now where web applications are letting try them out without you having to sign-up (though to save you need to of course). example: trial at http://minutedock.com/ I'm wondering about doing this for my own web app and the fundamental question is whether to store their info into sessions or into a temp user table? The temp user table would allow logging and potentially be less of a hit on the server correct? Is there a best practice here?

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  • How can I get the rank of rows relative to total number of rows based on a field?

    - by Arms
    I have a scores table that has two fields: user_id score I'm fetching specific rows that match a list of user_id's. How can I determine a rank for each row relative to the total number of rows, based on score? The rows in the result set are not necessarily sequential (the scores will vary widely from one row to the next). I'm not sure if this matters, but user_id is a unique field. Edit @Greelmo I'm already ordering the rows. If I fetch 15 rows, I don't want the rank to be 1-15. I need it to be the position of that row compared against the entire table by the score property. So if I have 200 rows, one row's rank may be 3 and another may be 179 (these are arbitrary #'s for example only). Edit 2 I'm having some luck with this query, but I actually want to avoid ties SELECT s.score , s.created_at , u.name , u.location , u.icon_id , u.photo , (SELECT COUNT(*) + 1 FROM scores WHERE score > s.score) AS rank FROM scores s LEFT JOIN users u ON u.uID = s.user_id ORDER BY s.score DESC , s.created_at DESC LIMIT 15 If two or more rows have the same score, I want the latest one (or earliest - I don't care) to be ranked higher. I tried modifying the subquery with AND id > s.id but that ended up giving me an unexpected result set and different ties.

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  • Remove redundant SQL code

    - by Dave Jarvis
    Code The following code calculates the slope and intercept for a linear regression against a slathering of data. It then applies the equation y = mx + b against the same result set to calculate the value of the regression line for each row. Can the two separate sub-selects be joined so that the data and its slope/intercept are calculated without executing the data gathering part of the query twice? SELECT AVG(D.AMOUNT) as AMOUNT, Y.YEAR * ymxb.SLOPE + ymxb.INTERCEPT as REGRESSION_LINE, Y.YEAR as YEAR, MAKEDATE(Y.YEAR,1) as AMOUNT_DATE FROM CITY C, STATION S, YEAR_REF Y, MONTH_REF M, DAILY D, (SELECT ((avg(t.AMOUNT * t.YEAR)) - avg(t.AMOUNT) * avg(t.YEAR)) / (stddev( t.AMOUNT ) * stddev( t.YEAR )) as CORRELATION, ((sum(t.YEAR) * sum(t.AMOUNT)) - (count(1) * sum(t.YEAR * t.AMOUNT))) / (power(sum(t.YEAR), 2) - count(1) * sum(power(t.YEAR, 2))) as SLOPE, ((sum( t.YEAR ) * sum( t.YEAR * t.AMOUNT )) - (sum( t.AMOUNT ) * sum(power(t.YEAR, 2)))) / (power(sum(t.YEAR), 2) - count(1) * sum(power(t.YEAR, 2))) as INTERCEPT FROM ( SELECT AVG(D.AMOUNT) as AMOUNT, Y.YEAR as YEAR, MAKEDATE(Y.YEAR,1) as AMOUNT_DATE FROM CITY C, STATION S, YEAR_REF Y, MONTH_REF M, DAILY D WHERE $X{ IN, C.ID, CityCode } AND SQRT( POW( C.LATITUDE - S.LATITUDE, 2 ) + POW( C.LONGITUDE - S.LONGITUDE, 2 ) ) < $P{Radius} AND S.STATION_DISTRICT_ID = Y.STATION_DISTRICT_ID AND Y.YEAR BETWEEN 1900 AND 2009 AND M.YEAR_REF_ID = Y.ID AND M.CATEGORY_ID = $P{CategoryCode} AND M.ID = D.MONTH_REF_ID AND D.DAILY_FLAG_ID <> 'M' GROUP BY Y.YEAR ) t ) ymxb WHERE $X{ IN, C.ID, CityCode } AND SQRT( POW( C.LATITUDE - S.LATITUDE, 2 ) + POW( C.LONGITUDE - S.LONGITUDE, 2 ) ) < $P{Radius} AND S.STATION_DISTRICT_ID = Y.STATION_DISTRICT_ID AND Y.YEAR BETWEEN 1900 AND 2009 AND M.YEAR_REF_ID = Y.ID AND M.CATEGORY_ID = $P{CategoryCode} AND M.ID = D.MONTH_REF_ID AND D.DAILY_FLAG_ID <> 'M' GROUP BY Y.YEAR Question How do I execute the duplicate bits only once per query, instead of twice? The duplicate bit is the WHERE clause: $X{ IN, C.ID, CityCode } AND SQRT( POW( C.LATITUDE - S.LATITUDE, 2 ) + POW( C.LONGITUDE - S.LONGITUDE, 2 ) ) < $P{Radius} AND S.STATION_DISTRICT_ID = Y.STATION_DISTRICT_ID AND Y.YEAR BETWEEN 1900 AND 2009 AND M.YEAR_REF_ID = Y.ID AND M.CATEGORY_ID = $P{CategoryCode} AND M.ID = D.MONTH_REF_ID AND D.DAILY_FLAG_ID <> 'M' Related http://stackoverflow.com/questions/1595659/how-to-eliminate-duplicate-calculation-in-sql Thank you!

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  • friendship database schema

    - by Daniel Hertz
    I'm creating a db schema that involves users that can be friends, and I was wondering what the best way to model the ability for these friends to have friendships. Should it be its own table that simply has two columns that each represent a user? Thanks!

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  • PHP PDO - Num Rows

    - by Ian
    PDO apparently has no means to count the number of rows returned from a select query (mysqli has the num_rows variable). Is there a way to do this, short of using count($results->fetchAll()) ?

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  • 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..............

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  • Trouble creating a SQL query

    - by JoBu1324
    I've been thinking about how to compose this SQL query for a while now, but after thinking about it for a few hours I thought I'd ask the SO community to see if they have any ideas. Here is a mock up of the relevant portion of the tables: contracts id date ar (yes/no) term payments contract_id payment_date The object of the query is to determine, per month, how many payments we expect, vs how many payments we received. conditions for expecting a payment Expected payments begin on contracts.term months after contracts.date, if contracts.ar is "yes". Payments continue to be expected until the month after the first missed payment. There is one other complication to this: payments might be late, but they need to show up as if they were paid on the date expected. The data is all there, but I've been having trouble wrapping my head around the SQL query. I am not an SQL guru - I merely have a decent amount of experience handling simpler queries. I'd like to avoid filtering the results in code, if possible - but without your help that may be what I have to do. Expected Output Month Expected Payments Received Payments January 500 450 February 498 478 March 234 211 April 987 789 ... SQL Fiddle I've created an SQL Fiddle: http://sqlfiddle.com/#!2/a2c3f/2

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  • 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.

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  • SQL hidden techniques?

    - by AlexRednic
    What are those pro/subtle techniques that SQL provides and not many know about which also cut code and improve performance? eg: I have just learned how to use CASE statements inside aggregate functions and it totally changed my approach on things. Are there others?

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  • How to get the value of a field in PHP?

    - by user272899
    I need to get the value of a field; I think I am along the right lines but not quite sure this is the proper code. The "Delete Movie" button is where I am trying to get the value of that row like so: value="'.$row['id'].'" Can you help? <?php //connect to database mysql_connect($mysql_hostname,$mysql_user,$mysql_password); @mysql_select_db($mysql_database) or die("<b>Unable to connect to specified database</b>"); //query databae $query = "select * from movielist"; $result=mysql_query($query) or die('Error, insert query failed'); $row=0; $numrows=mysql_num_rows($result); echo "<table border=1>"; echo "<tr> <td>ID</td> <td>Type</td> <td>Title</td> <td>Description</td> <td>Imdb URL</td> <td>Year</td> <td>Genre</td> <td>Actions</td> </tr>"; while($row<$numrows) { $id=mysql_result($result,$row,"id"); $type=mysql_result($result,$row,"type"); $title=mysql_result($result,$row,"title"); $description=mysql_result($result,$row,"description"); $imdburl=mysql_result($result,$row,"imdburl"); $year=mysql_result($result,$row,"year"); $genre=mysql_result($result,$row,"genre"); ?> <tr> <td><?php echo $id; ?></td> <td><?php echo $type; ?></td> <td><?php echo $title; ?></td> <td><?php echo $description; ?></td> <td><?php echo $imdburl; ?></td> <td><?php echo $year; ?></td> <td><?php echo $genre; ?></td> <td> <!-- Delete Movie Button --> <form style="display: inline;" action="delete/" method="post" onsubmit="return movie_delete()"> <input type="hidden" name="moviedeleteid" value="'.$row['id'].'"> <button type="submit" class="tooltip table-button ui-state-default ui-corner-all" title="Delete trunk"><span class="ui-icon ui-icon-trash"></span></button> </form> </td> </tr> <?php $row++; } echo "</table>"; ?>

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  • Connecting to 3rd party databse in Joomla!?

    - by Michael
    I need to connect to another database in Joomla! that's on another server. This is for a plugin and I need to pull some data from a table. Now what I don't want is to use this database to run Joomla!, I already have Joomla! installed and running on its own database on its server but I want to connect to another database (ON TOP of the current one) to pull some data, then disconnect from that 3rd party database - all while keeping the original Joomla database connection in tact.

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  • Better way to do SELECT with GROUP BY

    - by Luca Romagnoli
    Hi i've wrote a query that works: SELECT `comments`.* FROM `comments` RIGHT JOIN (SELECT MAX( id ) AS id, core_id, topic_id FROM comments GROUP BY core_id, topic_id order by id desc) comm ON comm.id = comments.id LIMIT 10 I want know if it is possible (and how) to rewrite it to get better performance. Thanks

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  • How to generate a user role grid

    - by Svish
    I have the following tables: users (id, username, ... ) roles (id, name) roles_users (user_id, role_id) I am wondering how I can create a nice sort of user-role-grid from that which an admin can use to administer roles to users in a clear way. What I would like is basically a table full of checkboxes sort of like this: Login Editor Admin Alice ¦ ¦ ¦ Bob ¦ ? ? Carol ¦ ¦ ? [Apply] Generating the table isn't too much of a deal, but I am very unsure how to handle it when it comes to how to name all the checkboxes and especially how to read and update the database in a not too clumsy way. Does anyone have any good advice or pointers on how to do this in a mostly clean way? I'm using the Kohana 3 framework, if there is anything there that can make this even easier, but I of course welcome any answer.

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  • function and class to show image not working in php

    - by Gully
    i am trying to get get the following working nothing is happen when i use the function i am trying to get it to display images class ItemRes { //items DB var $img=""; } function ShowItemImage($index,$res_a){ if(sizeof($res_a) > $index){ if($res_a[$index] != NULL) { $cimg = $res_a[$index]->img; return "<img src='$cimg' width='70' height='70' style='cursor:pointer'></img>"; } }else{ return "<center class='whitetxt'><strong>Empty</strong></center>"; } } $res_array = array(); $idx=0; $result21 = mysql_query("SELECT * FROM photos WHERE eid='$eid' ORDER BY id DESC") or die (mysql_error()); while ($row21 = mysql_fetch_array($result21)) { $img_path = $row21['path']; $obj = new ItemRes(); $obj->img = $img_path; $res_array[$idx] = $obj; $idx++; } ShowItemImage(0,$res_array) ShowItemImage(1,$res_array)

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  • 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?

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