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  • How to store coordinates in a database

    - by Tim
    Hello all! I have a Flex GUI where I have to place quadrate elements. The position of these elements need to be stored into a database. So I can create two integer fields in the db table x and y. Also I need an angle, because the user can rotate these elements, so I can also make a int (int is okay, I do not need a double value therefore). As a ORM, I use Hibernate. But the question is, if creating three integer fields is the best way to handle this. Perhaps someone can tell me if this will be okay or if there are better ways? Thanks a lot in advance & Best Regards.

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  • Wordpress SQL_CALC fix causes PHP error

    - by ok1ha
    I'm looking for some followup on an older topic for Wordpress where SQL_CALC was found to slow things down when you have a large DB in Wordpress. I have been using the code, at the bottom of this post, to get around it but it does generate an error in my error log. How would I prevent this error? PHP Warning: Division by zero in /var/www/vhosts/domain.com/httpdocs/wp-content/themes/greatTheme/functions.php on line 19 The original thread: http://wordpress.org/support/topic/slow-queries-sql_calc_found_rows-bringing-down-site?replies=25 The code in my functions.php: add_filter('pre_get_posts', 'optimized_get_posts', 100); function optimized_get_posts() { global $wp_query, $wpdb; $wp_query->query_vars['no_found_rows'] = 1; $wp_query->found_posts = $wpdb->get_var( "SELECT COUNT(*) FROM wp_posts WHERE 1=1 AND wp_posts.post_type = 'post' AND (wp_posts.post_status = 'publish' OR wp_posts.post_status = 'private')" ); $wp_query->found_posts = apply_filters_ref_array( 'found_posts', array( $wp_query->found_posts, &$wp_query ) ); $wp_query->max_num_pages = ceil($wp_query->found_posts / $wp_query->query_vars['posts_per_page']); return $wp_query; }

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  • How can I remove this query from within a loop?

    - by Chris
    I am currently designing a forum as a personal project. One of the recurring issues I've come across is database queries in loops. I've managed to avoid doing that so far by using table joins or caching of data in arrays for later use. Right now though I've come across a situation where I'm not sure how I can write the code in such a way that I can use either of those methods easily. However I'd still prefer to do at most 2 queries for this operation rather than 1 + 1 per group of forums, which so far has resulted in 5 per page. So while 5 isn't a huge number (though it will increase for each forum group I add) it's the principle that's important to me here, I do NOT want to write queries in loops What I'm doing is displaying forum index groupings (eg admin forums, user forums etc) and then each forum within that group on a single page index, it's the combination of both in one page that's causing me issue. If it had just been a single group per page, I'd use a table join and problem solved. But if I use a table join here, although I can potentially get all the data I need it'll be in one mass of results and it needs displaying properly. Here's the code (I've removed some of the html for clarity) <?php $sql= "select * from forum_groups"; //query 1 $result1 = $database->query($sql); while($group = mysql_fetch_assoc($result1)) //first loop {?> <table class="threads"> <tr> <td class="forumgroupheader"> <?php echo $group['group_name']; ?> </td> </tr> <tr> <td class="forumgroupheader2"> <?php echo $group['group_desc']; ?> </td> </tr> </table> <table> <tr> <th class="thforum"> Forum Name</th> <th class="thforum"> Forum Decsription</th> <th class="thforum"> Last Post </th> <tr> <?php $group_id = $group['id']; $sql = "SELECT forums.id, forums.forum_group_id, forums.forum_name, forums.forum_desc, forums.visible_rank, forums.locked, forums.lock_rank, forums.topics, forums.posts, forums.last_post, forums.last_post_id, users.username FROM forums LEFT JOIN users on forums.last_post_id=users.id WHERE forum_group_id='{$group_id}'"; //query 2 $result2 = $database->query($sql); while($forum = mysql_fetch_assoc($result2)) //second loop {?> So how can I either a) write the SQL in such a way as to remove the second query from inside the loop or b) combine the results in an array either way I need to be able to access the data as an when so I can format it properly for the page output, ie within the loops still.

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  • Simple PHP query question: LIKE

    - by pg
    When I replace $ordering = "apples, bananas, cranberries, grapes"; with $ordering = "apples, bananas, grapes"; I no longer want cranberries to be returned by my query, which I've written out like this: $query = "SELECT * from dbname where FruitName LIKE '$ordering'"; Of Course this doesn't work, because I used LIKE wrong. I've read through various manuals that describe how to use LIKE and it doesn't quite make sense to me. If I change the end of the db to "LIKE "apples"" that works for limiting it to just apples. Do I have to explode the ordering on the ", " or is there a way to do this in the query?

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  • indexing question

    - by user522962
    I have a table w/ 3 columns: name, phone, date. I have 3 indexes: 1 on phone, 1 on date and 1 on phone and date. I have the following statement: SELECT * FROM ( SELECT * FROM people WHERE phone IS NOT NULL ORDER BY date DESC) as t GROUP BY phone Basically, I want to get all unique phone numbers ordered by date. This table has about 2.5 million rows but takes forever to execute....are my indexes right? UPDATE: My EXPLAIN statement comes back with 2 rows: 1 for primary table and 1 for derived table. It says I am using temporary and using filesort for my primary table. For my derived table, it says my possible keys are (phone), and (phone, date) but it is using filesort.

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  • I can't delete record in Codeigniter

    - by jomblo
    I'm learning CRUD in codeigniter. I have table name "posting" and the coloumns are like this (id, title, post). I successed to create a new post (both insert into database and display in the view). But I have problem when I delete my post in the front-end. Here is my code: Model Class Post_Model extends CI_Model{ function index(){ //Here is my homepage code } function delete_post($id) { $this->db->where('id', $id); $this->db->delete('posting'); } } Controller Class Post extends CI_Controller{ function delete() { $this->load->model('Post_Model'); $this->Post_Model->delete_post("id"); redirect('Post/index/', 'refresh'); } } After click "delete" in the homepage, there was nothing happens. While I'm looking into my database, my records still available. Note: (1) to delete record, I'm following the codeigniter manual / user guide, (2) I found a message error (Undefined variable: id) after hiting the "delete" button in the front-end Any help or suggestion, please

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  • differentiating results of sql right join

    - by Sourabh
    Hi I have a below SQL query SELECT `User`.`username` , Permalink.perma_link_id, Permalink.locale, Permalink.title, DATEDIFF( CURDATE( ) , Permalink.created ) AS dtdiff, `TargetSegment`.segment_text, TargetSegment.source_segment_id ,TargetSegment.perma_link_id ,TargetSegment.created ,TargetSegment.updated, DATEDIFF( CURDATE( ) , TargetSegment.updated ) AS datediff FROM `users` AS `User` RIGHT JOIN perma_links AS `PermaLink` ON ( `PermaLink`.`username` = `User`.`username` ) RIGHT JOIN target_segments AS `TargetSegment` ON ( `TargetSegment`.`username` = `User`.`username` ) RIGHT JOIN source_segments AS `SourceSegment` ON ( `SourceSegment`.`source_detail_id` = `PermaLink`.`source_detail_id` ) LEFT JOIN source_details AS `SourceDetail` ON ( `SourceSegment`.`source_detail_id` = `SourceDetail`.`id` ) WHERE `TargetSegment`.`username` = "xxxx" AND `TargetSegment`.`segment_text` <> "" AND `Permalink`.`perma_link_id` = `TargetSegment`.`perma_link_id` AND `TargetSegment`.`source_segment_id` = `SourceSegment`.`id` AND `Permalink`.`source_detail_id` = `SourceDetail`.`id` ORDER BY `TargetSegment`.`updated` DESC LIMIT 0 , 10 This SQL is fetching correct results for me.I want to identify from which table each row if from , to be specific which result is due to PermaLink table and which is from TargetSegment table. is this achievable ?

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  • How can I fake sql data while preserving statements without commenting my server-side code?

    - by Fedor
    I have to use hardcoded values for certain fields because at this moment we don't have access to the real data. When we do get access, I don't want to go through a lot of work uncommenting. Is it possible to keep this statement the way it is, except use '25' as the alias for ratecode? IF(special.ratecode IS NULL, br.ratecode, special.ratecode) AS ratecode, I have about 8 or so IF statements similar to this and I'm just too lazy ( even with vim ) to re-append while commenting out each if statement line by line. I would have to do this: $sql = 'SELECT u.*,'; // IF ( special.ratecode IS NULL, br.ratecode, special.ratecode) AS ratecode $sql.= '25 AS ratecode';

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  • Problem with joining to an empty table

    - by Imran Omar Bukhsh
    I use the following query: select * from A LEFT JOIN B on ( A.t_id != B.t_id) to get all the records in A that are not in B. The results are fine except when table B is completely empty, but then I do not get any records, even from table A. Later It wont work yet! CREATE TABLE IF NOT EXISTS T1 ( id int(11) unsigned NOT NULL AUTO_INCREMENT, title varchar(50) CHARACTER SET utf8 COLLATE utf8_unicode_ci NOT NULL, t_id int(11) NOT NULL, PRIMARY KEY (id) ) ENGINE=MyISAM DEFAULT CHARSET=latin1 AUTO_INCREMENT=3 ; -- -- Dumping data for table T1 INSERT INTO T1 (id, title, t_id) VALUES (1, 'apple', 1), (2, 'orange', 2); -- -- Table structure for table T2 CREATE TABLE IF NOT EXISTS T2 ( id int(11) NOT NULL AUTO_INCREMENT, title varchar(50) CHARACTER SET utf8 COLLATE utf8_unicode_ci NOT NULL, t_id int(11) NOT NULL, PRIMARY KEY (id) ) ENGINE=MyISAM DEFAULT CHARSET=latin1 AUTO_INCREMENT=2 ; -- -- Dumping data for table T2 INSERT INTO T2 (id, title, t_id) VALUES (1, 'dad', 2); Now I want to get all records in T1 that do not have a corresponding records in T2 I try SELECT * FROM T1 LEFT OUTER JOIN T2 ON T1.t_id != T2.t_id and it won't work

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  • table column accepting "0" as a member Id

    - by user682417
    I have two tables one is members table with columns member id , member first name, member last name. I have another table guest passes with columns guest pass id and member id and issue date . I have a list view that will displays guest passes details (I.e) like member name and issue date and I have two text boxes those are for entering member name and issue date . member name text box is auto complete text box that working fine.... but the problem is when I am entering the name that is not in member table at this time it will accept and displays a blank field in list view in member name column and member id is stored as "0" in guest pass table ...... I don't want to display the member name empty blank and I don t want to store "0" in guest pass table and this is the insert statement sql2 = @"INSERT INTO guestpasses(member_Id,guestPass_IssueDate)"; sql2 += " VALUES("; sql2 += "'" + tbCGuestPassesMemberId.Text + "'"; sql2 += ",'" + tbIssueDate.Text + "'"; guestpassmemberId = memberid is there any validation that need to be done can any one suggestions on this pls... and this is the auto complete text box statement sql = @"SELECT member_Id FROM members WHERE concat(member_Firstname,'',member_Lastname) ='" + tbMemberName.Text+"'"; if (dt != null) { if (dt.Rows.Count > 0) { tbCGuestPassesMemberId.Text = Convert.ToInt32(dt.Rows[0] ["member_Id"]).ToString(); } } can any one help me on this ... is there any type of validation with sql query pls help me .....

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  • Use where clause with Like in codeigniter

    - by user2524013
    I am working on a project. I am implementing the Search functionality in my System. I will have to show the search record from two tables base on the current use login. I have tried the following code: function searchActivity($limit,$offset,$keyword1,$keyword2,$recruiter_id) { $q=$this->db->select('*')->from('tbl_activity')->limit($limit,$offset); $this->db->join('tbl_job', 'tbl_job.job_id = tbl_activity.job_id_fk', 'left outer'); $this->db->order_by("activity_id", "ASC"); $this->db->like('job_title',$keyword1,'both'); $this->db->or_like('job_title',$keyword2,'both'); $this->db->or_like('activity_subject',$keyword1,'both'); $this->db->or_like('activity_subject',$keyword2,'both'); $this->db->or_like('activity_details',$keyword1,'both'); $this->db->or_like('activity_details',$keyword2,'both'); $this->db->where('tbl_activity.recruiter_id_fk',$recruiter_id); $ret['rows']=$q->get()->result(); return $ret; } I want to show search results based on the current user id, which is currently store in $recruiter. Thanks in advance.

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  • How to structure data... Sequential or Hierarchical?

    - by Ryan
    I'm going through the exercise of building a CMS that will organize a lot of the common documents that my employer generates each time we get a new sales order. Each new sales order gets a 5 digit number (12222,12223,122224, etc...) but internally we have applied a hierarchy to these numbers: + 121XX |--01 |--02 + 122XX |--22 |--23 |--24 In my table for sales orders, is it better to use the 5 digital number as an ID and populate up or would it be better to use the hierarchical structure that we use when referring to jobs in regular conversation? The only benefit to not populating sequentially seems to be formatting the data later on in my view, but that doesn't sound like a good enough reason to go through the extra work. Thanks

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  • Is this SQL is valid?

    - by Beck
    UPDATE polls_options SET `votes`=`votes`+1, `percent`=ROUND((`votes`+1) / (SELECT voters FROM polls WHERE poll_id=? LIMIT 1) * 100,1) WHERE option_id=? AND poll_id=? Don't have table data yet, to test it properly. :) And by the way, in what type % integers should be stored in database? Thanks for the help!

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  • Problem with sending "SetCookie" first in php code

    - by Camran
    According to this manual: http://us2.php.net/setcookie I have to set the cookie before anything else. Here is my cookie code: if (isset($_COOKIE['watched_ads'])){ $expir = time()+1728000; //20 days $ad_arr = unserialize($_COOKIE['watched_ads']); $arr_elem = count($ad_arr); if (in_array($ad_id, $ad_arr) == FALSE){ if ($arr_elem>10){ array_shift($ad_arr); } $ad_arr[]=$ad_id; setcookie('watched_ads', serialize($ad_arr), $expir, '/'); } } else { $expir = time()+1728000; //20 days $ad_arr[] = $ad_id; setcookie('watched_ads', serialize($ad_arr), $expir, '/'); } As you can see I am using variables in setting the cookie. The variables comes from a mysql_query and I have to do the query first. But then, if I do, I will get an error message: Cannot modify header information - headers already sent by ... The error points to the line where I set the cookie above. What should I do?a

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  • Does UNIQ constraint mean also an index on that field(s)?

    - by Gremo
    As title, should i defined a separate index on email column (for searching purposes) or the index is "automatically" added along with UNIQ_EMAIL_USER constraint? CREATE TABLE IF NOT EXISTS `customer` ( `id` int(11) NOT NULL AUTO_INCREMENT, `user_id` int(11) NOT NULL, `first` varchar(255) NOT NULL, `last` varchar(255) NOT NULL, `slug` varchar(255) NOT NULL, `email` varchar(255) NOT NULL, `created_at` datetime NOT NULL, `updated_at` datetime NOT NULL, PRIMARY KEY (`id`), UNIQUE KEY `UNIQ_SLUG` (`slug`), UNIQUE KEY `UNIQ_EMAIL_USER` (`email`,`user_id`), KEY `IDX_USER` (`user_id`) ) ENGINE=InnoDB;

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  • Error during data INSERT in php

    - by nectar
    here my code- $sql = "INSERT INTO tblpin ('pinId', 'ownerId', 'usedby', 'status') VALUES "; for($i=0; $i0) { $sql .= ", "; } $sql .= "('$pin[$i]', '$ownerid', 'Free', '1')"; } $sql .= ";"; echo $sql; mysql_query($sql); if(mysql_affected_rows() 0) { echo "done"; } else { echo "Fail"; } output: ** INSERT INTO tblpin ('pinId', 'ownerId', 'usedby', 'status') VALUES ('13837927', 'admin', 'Free', '1'), ('59576082', 'admin', 'Free', '1'); Fail why it is not inserting values when $sql query is right?

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  • login/logout problem in PHP

    - by user356170
    i have a question. I have a website in which i am giving security like login id and password( as usual). No what i want is that, 1) I don't want to allow a single user to login in different machine at the same time. 2) For this i am using a column in database which is keeping the current status of user(i.e. loging/logout). I am allowing user to login only when has session has not closed and status is login. 3) So my problem is that when i am logging out manually. it is closing the session as well as updating the database with status "logout". 4) but when i am closing the window from Cross buttonat top right corner. it is closing the ssion but table data is still "login". so later on i can't be able to login into the same user. 5) So how could i solve this problem. Please help me!

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  • SQL query construction - separate data in a column into two columns

    - by Tommy
    I have a column that contains links. The problem is that the titles of the links are in the same column, so it looks like this: linktitle|-|linkurl I want link title and linkurl in separate columns. I've created a new column for the urls, so I'm looking for a way to extract them and update the linkurl column with them. Is there any clever way to construct a query that does this?

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  • Django database - how to add this column in raw SQL.

    - by alex
    Suppose I have my models set up already. class books(models.Model): title = models.CharField... ISBN = models.Integer... What if I want to add this column to my table? user = models.ForeignKey(User, unique=True) How would I write the raw SQL in my database so that this column works?

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  • Problem while redirecting user after registration

    - by Eternal Learner
    I am creating a simple website . My situation is like this. After registering an user, I want to redirect the user after say 3 seconds to a main page(if the registration succeeds) . The code I have now is as below $query = "INSERT INTO Privileges VALUES('$user','$password1','$role')"; $result = mysql_query($query, $dbcon) or die('Registration Failed: ' . mysql_error()); print 'Thanks for Registering , You will be redirected shortly'; ob_start(); echo "Test"; header("Location: http://www.php.net"); ob_flush() I get the error message Warning: Cannot modify header information - headers already sent by (output started at/home/srinivasa/public_html/ThanksForRegistering.php:27) in /home/srinivasa /public_html/ThanksForRegistering.php on line 35. What do I need to do now ?

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  • Insert into select and update in single query

    - by Ossi
    I have 4 tables: tempTBL, linksTBL and categoryTBL, extra on my tempTBL I have: ID, name, url, cat, isinserted columns on my linksTBL I have: ID, name, alias columns on my categoryTBL I have: cl_id, link_id,cat_id on my extraTBL I have: id, link_id, value How do I do a single query to select from tempTBL all items where isinsrted = 0 then insert them to linksTBL and for each record inserted, pickup ID (which is primary) and then insert that ID to categoryTBL with cat_id = 88. after that insert extraTBL ID for link_id and url for value. I know this is so confusing, put I'll post this anyhow... This is what I have so far: INSERT IGNORE INTO linksTBL (link_id,link_name,alias) VALUES(NULL,'tex2','hello'); # generate ID by inserting NULL INSERT INTO categoryTBL (link_id,cat_id) VALUES(LAST_INSERT_ID(),'88'); # use ID in second table I would like to add here somewhere that it only selects items where isinserted = 0 and iserts those records, and onse inserted, will change isinserted to 1, so when next time it runs, it will not add them again.

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  • How do you display a PHPmyadmin table onto a web browser/web page?

    - by user1390754
    Just a simple question, i've been searching around and for some reason i cannot find an answer to this. after creating a database/table in Phpmyadmin using xampp, what command do i need to put into my webpage (PHP) to show the table I made? I know the first step involves connecting to the database and i think i've done that properly. This is the code i found from somewhere about connecting (w3 tutorials I believe) $con = mysql_connect("localhost","xxxxxx,""); if (!$con) { die('Could not connect: ' . mysql_error()); }

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