Search Results

Search found 10067 results on 403 pages for 'pdp 11'.

Page 65/403 | < Previous Page | 61 62 63 64 65 66 67 68 69 70 71 72  | Next Page >

  • A hard Question ?

    - by the-ifl
    Hi Guys , I try To find a solution to a question .... we have a number , example : 20 ... and we have 6 number :{ a ,b , c , d , e , f} < 20 , t try to find all values of these numbers , but only if we can combinate (whit + or -) whit 2 of this numbers and getting all the value below to 20 : for example we choose 31 : a = 22 b = 21 c = 14 d = 11 e = 9 f = 5 we have : 22 - 21 = 1 ; 11 - 9 = 2 ; 14 - 11 = 3 ; 9 - 5 = 4 ; f = 5 ; 11 - 5 = 6 ; 21 - 14 = 7 ; .... .... .... .... .... 21 + 9 = 30 ; 9 + 22 = 31 ;

    Read the article

  • How do I use Perl to parse the output of the sqlplus command?

    - by benjamin button
    I have an SQL file which will give me an output like below: 10|1 10|2 10|3 11|2 11|4 . . . I am using this in a Perl script like below: my @tmp_cycledef = `sqlplus -s $connstr \@DLCycleState.sql`; after this above statement, since @tmp_cycledef has all the output of the SQL query, I want to show the output as: 10 1,2,3 11 2,4 How could I do this using Perl?

    Read the article

  • Query to sum duplicated fields

    - by g0sha
    Here is mysql data id usr good quant delayed cart_ts ------------------------------------------------------ 14 4 1 1 0 20100601235348 13 4 11 1 0 20100601235345 12 4 4 1 0 20100601235335 11 4 1 1 0 20100601235051 10 4 11 1 0 20100601235051 9 4 4 1 0 20100601235051 15 4 2 1 0 20100601235350 16 4 7 1 0 20100602000537 17 4 3 1 0 20100602000610 18 4 3 1 0 20100602000616 19 4 8 1 0 20100602000802 20 4 8 1 0 20100602000806 21 4 8 1 0 20100602000828 22 4 8 1 0 20100602000828 23 4 8 1 0 20100602000828 24 4 8 1 0 20100602000828 25 4 8 1 0 20100602000828 26 4 8 1 0 20100602000829 27 4 8 1 0 20100602000829 28 4 9 1 0 20100602001045 29 4 10 1 0 20100602001046 I need to group fields in witch usr & good has duplicated values with summing quant field for getting smth like this: id usr good quant delayed cart_ts ------------------------------------------------------ 14 4 1 2 0 20100601235348 13 4 11 2 0 20100601235345 12 4 4 2 0 20100601235335 15 4 2 1 0 20100601235350 16 4 7 1 0 20100602000537 17 4 3 2 0 20100602000610 19 4 8 9 0 20100602000802 28 4 9 1 0 20100602001045 29 4 10 1 0 20100602001046 Which MySQL query I need to do to have this effect?

    Read the article

  • How do I add values from two separate queries in SQL

    - by fishhead
    Below is my attempt at adding two values from separate select statements...it's not working, and I can't see why. I'm looking for some direction. select (v1.Value + v2.Value) as total from ( (Select Max(Value) as [Value] from History WHERE Datetime>='Apr 11 2010 6:05AM' and Datetime<='Apr 11 2010 6:05PM' and Tagname ='RWQ272017DTD' ) as v1 (Select Max(Value) as [Value] from History WHERE Datetime>='Apr 11 2010 6:05AM' and Datetime<='Apr 11 2010 6:05PM' and Tagname ='RU282001DTD' ) as v2 ) EDIT: Boy do I feel foolish...I asked the same question a few days ago...now I can't delete this.

    Read the article

  • Why this query is so slow?

    - by Silver Light
    This query appears in mysql slow query log: it takes 11 seconds. INSERT INTO record_visits ( record_id, visit_day ) VALUES ( '567', NOW() ); The table has 501043 records and it's structure looks like this: CREATE TABLE IF NOT EXISTS `record_visits` ( `id` int(11) NOT NULL AUTO_INCREMENT, `record_id` int(11) DEFAULT NULL, `visit_day` date DEFAULT NULL, `visit_cnt` bigint(20) DEFAULT '1', PRIMARY KEY (`id`), UNIQUE KEY `record_id_visit_day` (`record_id`,`visit_day`) ) ENGINE=MyISAM DEFAULT CHARSET=utf8 ; What could be wrong? Why this INSERT takes so long?

    Read the article

  • Relating categories with tags using SQL

    - by Pablo
    I want be able to find tags of items under the a certain category. Following is example of my database design: images +----------+-----+-------------+-----+ | image_id | ... | category_id | ... | +----------+-----+-------------+-----+ | 1 | ... | 11 | ... | +----------+-----+-------------+-----+ | 2 | ... | 12 | ... | +----------+-----+-------------+-----+ | 3 | ... | 11 | ... | +----------+-----+-------------+-----+ | 4 | ... | 11 | ... | +----------+-----+-------------+-----+ images_tags +----------+--------+ | image_id | tag_id | +----------+--------+ | 1 | 53 | +----------+--------+ | 3 | 54 | +----------+--------+ | 2 | 55 | +----------+--------+ | 1 | 56 | +----------+--------+ | 4 | 57 | +----------+--------+ tags and categories each have their own table relating the id to an actual name(text). So my question is how will i find out that images with category_id=11 have have the tag_id 53 54 55 56 57. In other words how to find the tags that images in certain category have?

    Read the article

  • Cannot add or update a child row: a foreign key constraint fails

    - by Tom
    table 1 +----------+-------------+------+-----+---------+----------------+ | Field | Type | Null | Key | Default | Extra | +----------+-------------+------+-----+---------+----------------+ | UserID | int(11) | NO | PRI | NULL | auto_increment | | Password | varchar(20) | NO | | | | | Username | varchar(25) | NO | | | | | Email | varchar(60) | NO | | | | +----------+-------------+------+-----+---------+----------------+ table2 +------------------+--------------+------+-----+---------+----------------+ | Field | Type | Null | Key | Default | Extra | +------------------+--------------+------+-----+---------+----------------+ | UserID | int(11) | NO | MUL | | | | PostID | int(11) | NO | PRI | NULL | auto_increment | | Title | varchar(50) | NO | | | | | Summary | varchar(500) | NO | | | | +------------------+--------------+------+-----+---------+----------------+ com.mysql.jdbc.exceptions.MySQLIntegrityConstraintViolationException: Cannot add or update a child row: a foreign key constraint fails (myapp/table2, CONSTRAINT table2_ibfk_1 FOREIGN KEY (UserID) REFERENCES table1 (UserID)) What have I done wrong? I read this: http://www.w3schools.com/Sql/sql_foreignkey.asp and i don't see whats wrong. :S

    Read the article

  • Slow query. Wrong database structure?

    - by Tin
    I have a database with table that contains tasks. Tasks have a lifecycle. The status of the task's lifecycle can change. These state transitions are stored in a separate table tasktransitions. Now I wrote a query to find all open/reopened tasks and recently changed tasks but I already see with a rather small number of tasks (<1000) that execution time has becoming very long (0.5s). Tasks +-------------+---------+------+-----+---------+----------------+ | Field | Type | Null | Key | Default | Extra | +-------------+---------+------+-----+---------+----------------+ | taskid | int(11) | NO | PRI | NULL | auto_increment | | description | text | NO | | NULL | | +-------------+---------+------+-----+---------+----------------+ Tasktransitions +------------------+-----------+------+-----+-------------------+----------------+ | Field | Type | Null | Key | Default | Extra | +------------------+-----------+------+-----+-------------------+----------------+ | tasktransitionid | int(11) | NO | PRI | NULL | auto_increment | | taskid | int(11) | NO | MUL | NULL | | | status | int(11) | NO | MUL | NULL | | | description | text | NO | | NULL | | | userid | int(11) | NO | | NULL | | | transitiondate | timestamp | NO | | CURRENT_TIMESTAMP | | +------------------+-----------+------+-----+-------------------+----------------+ Query SELECT tasks.taskid,tasks.description,tasklaststatus.status FROM tasks LEFT OUTER JOIN ( SELECT tasktransitions.taskid,tasktransitions.transitiondate,tasktransitions.status FROM tasktransitions INNER JOIN ( SELECT taskid,MAX(transitiondate) AS lasttransitiondate FROM tasktransitions GROUP BY taskid ) AS tasklasttransition ON tasklasttransition.lasttransitiondate=tasktransitions.transitiondate AND tasklasttransition.taskid=tasktransitions.taskid ) AS tasklaststatus ON tasklaststatus.taskid=tasks.taskid WHERE tasklaststatus.status IS NULL OR tasklaststatus.status=0 or tasklaststatus.transitiondate>'2013-09-01'; I'm wondering if the database structure is best choice performance wise. Could adding indexes help? I already tried to add some but I don't see great improvements. +-----------------+------------+----------------+--------------+------------------+-----------+-------------+----------+--------+------+------------+---------+---------------+ | Table | Non_unique | Key_name | Seq_in_index | Column_name | Collation | Cardinality | Sub_part | Packed | Null | Index_type | Comment | Index_comment | +-----------------+------------+----------------+--------------+------------------+-----------+-------------+----------+--------+------+------------+---------+---------------+ | tasktransitions | 0 | PRIMARY | 1 | tasktransitionid | A | 896 | NULL | NULL | | BTREE | | | | tasktransitions | 1 | taskid_date_ix | 1 | taskid | A | 896 | NULL | NULL | | BTREE | | | | tasktransitions | 1 | taskid_date_ix | 2 | transitiondate | A | 896 | NULL | NULL | | BTREE | | | | tasktransitions | 1 | status_ix | 1 | status | A | 3 | NULL | NULL | | BTREE | | | +-----------------+------------+----------------+--------------+------------------+-----------+-------------+----------+--------+------+------------+---------+---------------+ Any other suggestions?

    Read the article

  • Regular Expression to parse SQL Structure

    - by user351429
    I am trying to parse the MySQL data types returned by "DESCRIBE [TABLE]". It returns strings like: int(11) float varchar(200) int(11) unsigned float(6,2) I've tried to do the job using regular expressions but it's not working. PHP CODE: $string = "int(11) numeric"; $regex = '/(\w+)\s*(\w+)/'; var_dump( preg_split($regex, $string) );

    Read the article

  • How can I read a continuously updating log file in Perl?

    - by Octopus
    I have a application generating logs in every 5 sec. The logs are in below format. 11:13:49.250,interface,0,RX,0 11:13:49.250,interface,0,TX,0 11:13:49.250,interface,1,close,0 11:13:49.250,interface,4,error,593 11:13:49.250,interface,4,idle,2994215 and so on for other interfaces... I am working to convert these into below CSV format: Time,interface.RX,interface.TX,interface.close.... 11:13:49,0,0,0,.... Simple as of now but the problem is, I have to get the data in CSV format online, i.e as soon the log file updated the CSV should also be updated. What I have tried to read the output and make the header is: #!/usr/bin/perl -w use strict; use File::Tail; my $head=["Time"]; my $pos={}; my $last_pos=0; my $current_event=[]; my $events=[]; my $file = shift; $file = File::Tail->new($file); while(defined($_=$file->read)) { next if $_ =~ some filters; my ($time,$interface,$count,$eve,$value) = split /[,\n]/, $_; my $key = $interface.".".$eve; if (not defined $pos->{$eve_key}) { $last_pos+=1; $pos->{$eve_key}=$last_pos; push @$head,$eve; } print join(",", @$head) . "\n"; } Is there any way to do this using Perl?

    Read the article

  • Need MYSQL query for finding lowest score per game player

    - by Chris Barnhill
    I have a game on Facebook called Rails Across Europe. I have a Best Scores page where I show the players with the best 20 scores, which in game terms refers to the lowest winning turn. The problem is that there are a small number of players who play frequently, and their scores dominate the page. I'd like to make the scores page open to more players. So I thought that I could display the single lowest winning turn for each player instead of displaying all of the lowest winning turns for all players. The problem is that the query for this eludes me. So I hope that one of you brilliant StackOverflow folks can help me with this. I have included the relevant MYSQL table schemas below. Here are the the table relationships: player_stats contains statistics for either a game in progress or a completed game. If a game is in progress, winning_turn is zero (which means that games with a winning_turn of zero should not be included in the query). player_stats has a game_player table id reference. game_player contains data describing games currently in progress. game_player has a player table id reference. player contains data describing a person who plays the game. Here's the query I'm currently using: 'SELECT p.fb_user_id, ps.winning_turn, gp.difficulty_level, c.name as city_name, g.name as goods_name, d.cost FROM game_player as gp, player as p, player_stats as ps, demand as d, city as c, goods as g WHERE p.status = "ACTIVE" AND gp.player_id = p.id AND ps.game_player_id = gp.id AND d.id = ps.highest_demand_id AND c.id = d.city_id AND g.id = d.goods_id AND ps.winning_turn > 0 ORDER BY ps.winning_turn ASC, d.cost DESC LIMIT '.$limit.';'; Here are the relevant table schemas: -- -- Table structure for table `player_stats` -- CREATE TABLE IF NOT EXISTS `player_stats` ( `id` int(11) NOT NULL auto_increment, `game_player_id` int(11) NOT NULL, `winning_turn` int(11) NOT NULL, `highest_demand_id` int(11) NOT NULL, PRIMARY KEY (`id`), KEY `game_player_id` (`game_player_id`,`highest_demand_id`) ) ENGINE=InnoDB DEFAULT CHARSET=utf8 AUTO_INCREMENT=3814 ; -- -- Table structure for table `game_player` -- CREATE TABLE IF NOT EXISTS `game_player` ( `id` int(10) unsigned NOT NULL auto_increment, `game_id` int(10) unsigned NOT NULL, `player_id` int(10) unsigned NOT NULL, `player_number` int(11) NOT NULL, `funds` int(10) unsigned NOT NULL, `turn` int(10) unsigned NOT NULL, `difficulty_level` enum('STANDARD','ADVANCED','MASTER','ULTIMATE') NOT NULL, `date_last_used` datetime NOT NULL, PRIMARY KEY (`id`), KEY `game_id` (`game_id`,`player_id`) ) ENGINE=InnoDB DEFAULT CHARSET=utf8 AUTO_INCREMENT=3814 ; -- -- Table structure for table `player` -- CREATE TABLE IF NOT EXISTS `player` ( `id` int(11) NOT NULL auto_increment, `fb_user_id` char(255) NOT NULL, `fb_proxied_email` text NOT NULL, `first_name` char(255) NOT NULL, `last_name` char(255) NOT NULL, `birthdate` date NOT NULL, `date_registered` datetime NOT NULL, `date_last_logged_in` datetime NOT NULL, `status` enum('ACTIVE','SUSPENDED','CLOSED') NOT NULL, PRIMARY KEY (`id`), KEY `fb_user_id` (`fb_user_id`) ) ENGINE=InnoDB DEFAULT CHARSET=utf8 AUTO_INCREMENT=1646 ;

    Read the article

  • How to make MySQL utilize available system resources, or find "the real problem"?

    - by anonymous coward
    This is a MySQL 5.0.26 server, running on SuSE Enterprise 10. This may be a Serverfault question. The web user interface that uses these particular queries (below) is showing sometimes 30+, even up to 120+ seconds at the worst, to generate the pages involved. On development, when the queries are run alone, they take up to 20 seconds on the first run (with no query cache enabled) but anywhere from 2 to 7 seconds after that - I assume because the tables and indexes involved have been placed into ram. From what I can tell, the longest load times are caused by Read/Update Locking. These are MyISAM tables. So it looks like a long update comes in, followed by a couple 7 second queries, and they're just adding up. And I'm fine with that explanation. What I'm not fine with is that MySQL doesn't appear to be utilizing the hardware it's on, and while the bottleneck seems to be the database, I can't understand why. I would say "throw more hardware at it", but we did and it doesn't appear to have changed the situation. Viewing a 'top' during the slowest times never shows much cpu or memory utilization by mysqld, as if the server is having no trouble at all - but then, why are the queries taking so long? How can I make MySQL use the crap out of this hardware, or find out what I'm doing wrong? Extra Details: On the "Memory Health" tab in the MySQL Administrator (for Windows), the Key Buffer is less than 1/8th used - so all the indexes should be in RAM. I can provide a screen shot of any graphs that might help. So desperate to fix this issue. Suffice it to say, there is legacy code "generating" these queries, and they're pretty much stuck the way they are. I have tried every combination of Indexes on the tables involved, but any suggestions are welcome. Here's the current Create Table statement from development (the 'experimental' key I have added, seems to help a little, for the example query only): CREATE TABLE `registration_task` ( `id` varchar(36) NOT NULL default '', `date_entered` datetime NOT NULL default '0000-00-00 00:00:00', `date_modified` datetime NOT NULL default '0000-00-00 00:00:00', `assigned_user_id` varchar(36) default NULL, `modified_user_id` varchar(36) default NULL, `created_by` varchar(36) default NULL, `name` varchar(80) NOT NULL default '', `status` varchar(255) default NULL, `date_due` date default NULL, `time_due` time default NULL, `date_start` date default NULL, `time_start` time default NULL, `parent_id` varchar(36) NOT NULL default '', `priority` varchar(255) NOT NULL default '9', `description` text, `order_number` int(11) default '1', `task_number` int(11) default NULL, `depends_on_id` varchar(36) default NULL, `milestone_flag` varchar(255) default NULL, `estimated_effort` int(11) default NULL, `actual_effort` int(11) default NULL, `utilization` int(11) default '100', `percent_complete` int(11) default '0', `deleted` tinyint(1) NOT NULL default '0', `wf_task_id` varchar(36) default '0', `reg_field` varchar(8) default '', `date_offset` int(11) default '0', `date_source` varchar(10) default '', `date_completed` date default '0000-00-00', `completed_id` varchar(36) default NULL, `original_name` varchar(80) default NULL, PRIMARY KEY (`id`), KEY `idx_reg_task_p` (`deleted`,`parent_id`), KEY `By_Assignee` (`assigned_user_id`,`deleted`), KEY `status_assignee` (`status`,`deleted`), KEY `experimental` (`deleted`,`status`,`assigned_user_id`,`parent_id`,`date_due`) ) ENGINE=MyISAM DEFAULT CHARSET=latin1 And one of the ridiculous queries in question: SELECT users.user_name assigned_user_name, registration.FIELD001 parent_name, registration_task.status status, registration_task.date_modified date_modified, registration_task.date_due date_due, registration.FIELD240 assigned_wf, if(LENGTH(registration_task.description)>0,1,0) has_description, registration_task.* FROM registration_task LEFT JOIN users ON registration_task.assigned_user_id=users.id LEFT JOIN registration ON registration_task.parent_id=registration.id where (registration_task.status != 'Completed' AND registration.FIELD001 LIKE '%' AND registration_task.name LIKE '%' AND registration.FIELD060 LIKE 'GN001472%') AND registration_task.deleted=0 ORDER BY date_due asc LIMIT 0,20; my.cnf - '[mysqld]' section. [mysqld] port = 3306 socket = /var/lib/mysql/mysql.sock skip-locking key_buffer = 384M max_allowed_packet = 100M table_cache = 2048 sort_buffer_size = 2M net_buffer_length = 100M read_buffer_size = 2M read_rnd_buffer_size = 160M myisam_sort_buffer_size = 128M query_cache_size = 16M query_cache_limit = 1M EXPLAIN above query, without additional index: +----+-------------+-------------------+--------+--------------------------------+----------------+---------+------------------------------------------------+---------+-----------------------------+ | id | select_type | table | type | possible_keys | key | key_len | ref | rows | Extra | +----+-------------+-------------------+--------+--------------------------------+----------------+---------+------------------------------------------------+---------+-----------------------------+ | 1 | SIMPLE | registration_task | ref | idx_reg_task_p,status_assignee | idx_reg_task_p | 1 | const | 1067354 | Using where; Using filesort | | 1 | SIMPLE | registration | eq_ref | PRIMARY,gbl | PRIMARY | 8 | sugarcrm401.registration_task.parent_id | 1 | Using where | | 1 | SIMPLE | users | ref | PRIMARY | PRIMARY | 38 | sugarcrm401.registration_task.assigned_user_id | 1 | | +----+-------------+-------------------+--------+--------------------------------+----------------+---------+------------------------------------------------+---------+-----------------------------+ EXPLAIN above query, with 'experimental' index: +----+-------------+-------------------+--------+-----------------------------------------------------------+------------------+---------+------------------------------------------------+--------+-----------------------------+ | id | select_type | table | type | possible_keys | key | key_len | ref | rows | Extra | +----+-------------+-------------------+--------+-----------------------------------------------------------+------------------+---------+------------------------------------------------+--------+-----------------------------+ | 1 | SIMPLE | registration_task | range | idx_reg_task_p,status_assignee,NewIndex1,tcg_experimental | tcg_experimental | 259 | NULL | 103345 | Using where; Using filesort | | 1 | SIMPLE | registration | eq_ref | PRIMARY,gbl | PRIMARY | 8 | sugarcrm401.registration_task.parent_id | 1 | Using where | | 1 | SIMPLE | users | ref | PRIMARY | PRIMARY | 38 | sugarcrm401.registration_task.assigned_user_id | 1 | | +----+-------------+-------------------+--------+-----------------------------------------------------------+------------------+---------+------------------------------------------------+--------+-----------------------------+

    Read the article

  • How to turn this simple 10 digit hex number back into 8 digits?

    - by Babil
    The algorithm to convert input 8 digit hex number into 10 digit are following: Given that the 8 digit number is: '12 34 56 78' x1 = 1 * 16^8 * 2^3 x2 = 2 * 16^7 * 2^2 x3 = 3 * 16^6 * 2^1 x4 = 4 * 16^4 * 2^4 x5 = 5 * 16^3 * 2^3 x6 = 6 * 16^2 * 2^2 x7 = 7 * 16^1 * 2^1 x8 = 8 * 16^0 * 2^0 Final 10 digit hex is: = x1 + x2 + x3 + x4 + x5 + x6 + x7 + x8 = '08 86 42 98 E8' The problem is - how to go back to 8 digit hex from a given 10 digit hex (for example: 08 86 42 98 E8 to 12 34 56 78) Some sample input and output are following: input output 11 11 11 11 08 42 10 84 21 22 22 33 33 10 84 21 8C 63 AB CD 12 34 52 D8 D0 88 64 45 78 96 32 21 4E 84 98 62 FF FF FF FF 7B DE F7 BD EF

    Read the article

  • ASUS N550 laptop not charging on ubuntu after updating kernel

    - by OBY Trance
    I'm using Ubuntu 13.10 on a pretty new ASUS N550jv laptop. When I was using kernel linux-image-3.11.0-11 everything was quite well (kinda..), however, when I updated the kernel using the automatic update, the kernels which were released afterwards (12+) were faulty on my machine, and caused the battery not to be charged and only stay at their current charging level (even when the machine was off!) The only fix I had was to roll-back to the '..11' version (on boot screen - advanced options), and hard-reboot (AC cord disconnected and reconnected) but now version 14 was released and "pushed away" the good old version 11. How can I fix that?? Please help me...

    Read the article

  • Algorithm: Find smallest subset containing K 0's

    - by Vishal
    I have array of 1's and 0's only. Now I want to find contiguous subset/subarray which contains at least K 0's. Example Array is 1 1 0 1 1 0 1 1 0 0 0 0 1 0 1 1 0 0 0 1 1 0 0 1 0 0 0 and K(6) should be 0 0 1 0 1 1 0 0 0 or 0 0 0 0 1 0 1 1 0.... My Solution Array: 1 1 0 1 1 0 1 1 0 0 0 0 1 0 1 1 0 0 0 1 1 0 0 Index: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 Sum: 1 2 2 3 4 4 5 6 6 6 6 6 7 7 8 9 9 9 9 10 11 11 11 Diff(I-S): 0 0 1 1 1 2 2 2 3 4 5 6 6 7 7 7 8 9 10 10 10 11 12 For K(6) Start with 9-15 = Store difference in diff. Next increase difference 8-15(Difference in index) 8-14(Compare Difference in index) So on keep moving to find element with least elements... I am looking for better algorithm for this solution.

    Read the article

  • 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

    Read the article

  • Optimize SQL query (Facebook-like application)

    - by fabriciols
    My application is similar to Facebook, and I'm trying to optimize the query that get user records. The user records are that he as src ou dst. The src is in usermuralentry directly, the dst list are in usermuralentry_user. So, a entry can have one src and many dst. I have those tables: mysql> desc usermuralentry ; +-----------------+------------------+------+-----+---------+----------------+ | Field | Type | Null | Key | Default | Extra | +-----------------+------------------+------+-----+---------+----------------+ | id | int(11) | NO | PRI | NULL | auto_increment | | user_src_id | int(11) | NO | MUL | NULL | | | private | tinyint(1) | NO | | NULL | | | content | longtext | NO | | NULL | | | date | datetime | NO | | NULL | | | last_update | datetime | NO | | NULL | | +-----------------+------------------+------+-----+---------+----------------+ 10 rows in set (0.10 sec) mysql> desc usermuralentry_user ; +-------------------+---------+------+-----+---------+----------------+ | Field | Type | Null | Key | Default | Extra | +-------------------+---------+------+-----+---------+----------------+ | id | int(11) | NO | PRI | NULL | auto_increment | | usermuralentry_id | int(11) | NO | MUL | NULL | | | userinfo_id | int(11) | NO | MUL | NULL | | +-------------------+---------+------+-----+---------+----------------+ 3 rows in set (0.00 sec) And the following query to retrieve information from two users. mysql> explain SELECT * FROM usermuralentry AS a , usermuralentry_user AS b WHERE a.user_src_id IN ( 1, 2 ) OR ( a.id = b.usermuralentry_id AND b.userinfo_id IN ( 1, 2 ) ); +----+-------------+-------+------+-------------------------------------------------------------------------------------------+------+---------+------+---------+------------------------------------------------+ | id | select_type | table | type | possible_keys | key | key_len | ref | rows | Extra | +----+-------------+-------+------+-------------------------------------------------------------------------------------------+------+---------+------+---------+------------------------------------------------+ | 1 | SIMPLE | b | ALL | usermuralentry_id,usermuralentry_user_bcd7114e,usermuralentry_user_6b192ca7 | NULL | NULL | NULL | 147188 | | | 1 | SIMPLE | a | ALL | PRIMARY | NULL | NULL | NULL | 1371289 | Range checked for each record (index map: 0x1) | +----+-------------+-------+------+-------------------------------------------------------------------------------------------+------+---------+------+---------+------------------------------------------------+ 2 rows in set (0.00 sec) but it is taking A LOT of time... Some tips to optimize? Can the table schema be better in my application?

    Read the article

  • Use SQL to clone data in two tables that have a 1-1 relationship with each other

    - by AmoebaMan17
    Using MS SQL 2005, Table 1 ID | T1Value | T2ID | GroupID ---------------------------------- 1 | a | 10 | 1 2 | b | 11 | 1 3 | c | 12 | 1 4 | a | 22 | 2 Table 2 ID | T2Value ---------------- 10 | H 11 | J 12 | K 22 | H I want to clone the data for GroupID == 1 into a new GroupID so that I result with the following: Table 1 ID | T1Value | T2ID | GroupID ---------------------------------- 1 | a | 10 | 1 2 | b | 11 | 1 3 | c | 12 | 1 4 | a | 22 | 2 5 | a | 23 | 3 6 | b | 24 | 3 7 | c | 25 | 3 Table 2 ID | T2Value ---------------- 10 | H 11 | J 12 | K 22 | H 23 | H 24 | J 25 | K I've found some SQL clone patterns that allow me to clone data in the same table well... but as I start to deal with cloning data in two tables at the same time and then linking up the new rows correctly... that's just not something I feel like I have a good grasp of. I thought I could do some self-joins to deal with this, but I am worried in the cases where the non-key fields have the same data in multiple rows.

    Read the article

  • mysql count(*) left join group by - the number of files in a folder

    - by Flavius
    Hi I have the following tables CREATE TABLE `files` ( `fileid` int(11) NOT NULL AUTO_INCREMENT, `filename` varchar(255) NOT NULL, `filesize` int(11) NOT NULL, `folder` int(11) NOT NULL, PRIMARY KEY (`fileid`) ) ENGINE=MyISAM DEFAULT CHARSET=utf8; CREATE TABLE `folders` ( `directoryid` int(11) NOT NULL AUTO_INCREMENT, `directoryname` varchar(255) NOT NULL, PRIMARY KEY (`directoryid`) ) ENGINE=MyISAM DEFAULT CHARSET=utf8; How to get a list of all folders and the number of files they hold, including folders who have no (zero) files? Recursion must not be taken into account. Found it select folders.directoryid, folders.directoryname, count(files.fileid) as no_files from folders left join files on files.folder = folders.directoryid group by folders.directoryid, folders.directoryname I hope it will help someone.

    Read the article

  • How to let the UITable sorted by number?

    - by Tattat
    I loaded a plist from the UITable, but there is something wrong. my plist file is some data like that: 3 3uyyhuhu34 ..... 5 5uyyhuhu34 ..... 11 11uyyhuhu34 ..... I found that the list is sorted from 11, then 3, after that is 5. But I want to sorted from 3, 5 to 11. What can I do to change the behaviors? thx

    Read the article

  • Oracle date / order by question

    - by user561793
    I want to select a date from oracle table formatted like select (to_char(req_date,'MM/YYYY')) but I also want to order the result set on this date format. I want them to be ordered like dates not strings. Like this 09/2009 10/2009 11/2009 12/2009 01/2010 02/2010 03/2010 04/2010 05/2010 06/2010 07/2010 08/2010 09/2010 10/2010 11/2010 12/2010 Not like 01/2010 02/2010 03/2010 04/2010 05/2010 06/2010 07/2010 08/2010 09/2009 09/2010 10/2009 10/2010 11/2009 11/2010 12/2009 12/2010 Any way to do this in sql? full sql is select (to_char(req_date,'MM/YYYY')) as monthYear,count(req_id) as count from REQUISITION_CURRENT t group by to_char(req_date,'MM/YYYY') Thanks

    Read the article

  • Use SQL to clone data in two tables that have a 1-1 relationship in each table

    - by AmoebaMan17
    Using MS SQL 2005, Table 1 ID | T1Value | T2ID | GroupID ---------------------------------- 1 | a | 10 | 1 2 | b | 11 | 1 3 | c | 12 | 1 4 | a | 22 | 2 Table 2 ID | T2Value ---------------- 10 | H 11 | J 12 | K 22 | H I want to clone the data for GroupID == 1 into a new GroupID so that I result with the following: Table 1 ID | T1Value | T2ID | GroupID ---------------------------------- 1 | a | 10 | 1 2 | b | 11 | 1 3 | c | 12 | 1 4 | a | 22 | 2 5 | a | 23 | 3 6 | b | 24 | 3 7 | c | 25 | 3 Table 2 ID | T2Value ---------------- 10 | H 11 | J 12 | K 22 | H 23 | H 24 | J 25 | K I've found some SQL clone patterns that allow me to clone data in the same table well... but as I start to deal with cloning data in two tables at the same time and then linking up the new rows correctly... that's just not something I feel like I have a good grasp of. I thought I could do some self-joins to deal with this, but I am worried in the cases where the non-key fields have the same data in multiple rows.

    Read the article

  • How can I maintain a sorted hash in Perl?

    - by srk
    @aoh =( { 3 => 15, 4 => 8, 5 => 9, }, { 3 => 11, 4 => 25, 5 => 6, }, { 3 => 5, 4 => 18, 5 => 5, }, { 0 => 16, 1 => 11, 2 => 7, }, { 0 => 21, 1 => 13, 2 => 31, }, { 0 => 11, 1 => 14, 2 => 31, }, ); I want the hashes in each array index sorted in reverse order based on values.. @sorted = sort { ........... please fill this..........} @aoh; expected output @aoh =( { 4 => 8, 5 => 9, 3 => 15, }, { 5 => 6, 3 => 11, 4 => 25, }, { 5 => 5, 3 => 5, 4 => 18, }, { 2 => 7, 1 => 11, 0 => 16, }, { 1 => 13, 0 => 21, 2 => 31, }, { 0 => 11, 1 => 14, 2 => 31, }, ); Please help.. Thanks in advance.. Stating my request again: I only want the hashes in each array index to be sorted by values.. i dont want the array to be sorted..

    Read the article

  • [SQL] Select 3 lastest order for each customer

    - by Ratiug
    Hi Here is my table CusOrder that collect customer order OrderID Cus_ID Product_ID NumberOrder OrderDate 1 0000000001 9 1 6/5/2553 0:00:00 2 0000000001 10 1 6/5/2553 0:00:00 3 0000000004 9 2 13/4/2553 0:00:00 4 0000000004 9 1 17/3/2553 0:00:00 5 0000000002 9 1 22/1/2553 0:00:00 7 0000000005 9 1 16/12/2552 0:00:00 8 0000000003 9 3 13/12/2552 0:00:00 10 0000000001 9 2 19/11/2552 0:00:00 11 0000000003 9 2 10/11/2552 0:00:00 12 0000000002 9 1 23/11/2552 0:00:00 I need to select 3 lastest order for each customer and I need all customer so it will show each customer and his/her 3 lastest order how can I do it sorry for my bad english

    Read the article

  • Collaborate 2010 Recap: A lot of Excitement for Oracle Content Management 11g

    - by [email protected]
    Normal 0 false false false EN-US X-NONE X-NONE MicrosoftInternetExplorer4 /* Style Definitions */ table.MsoNormalTable {mso-style-name:"Table Normal"; mso-tstyle-rowband-size:0; mso-tstyle-colband-size:0; mso-style-noshow:yes; mso-style-priority:99; mso-style-qformat:yes; mso-style-parent:""; mso-padding-alt:0in 5.4pt 0in 5.4pt; mso-para-margin-top:0in; mso-para-margin-right:0in; mso-para-margin-bottom:10.0pt; mso-para-margin-left:0in; line-height:115%; mso-pagination:widow-orphan; font-size:11.0pt; font-family:"Calibri","sans-serif"; mso-ascii-font-family:Calibri; mso-ascii-theme-font:minor-latin; mso-fareast-font-family:"Times New Roman"; mso-fareast-theme-font:minor-fareast; mso-hansi-font-family:Calibri; mso-hansi-theme-font:minor-latin;} Collaborate brought me to Las Vegas last week and what a week it was.  Each day was jam packed with Oracle Content Management sessions, and almost every session I attended was full.  Across the 35+ sessions that were given by my Oracle peers, Oracle partners, and Oracle customers, the majority of the discussion and questions that were asked had to do with the release of Oracle Content Management 11g.  Just to bring everyone up-to-speed, the first wave of Oracle Content Management 11g releases happened this past January as Oracle Imaging & Process Management and Oracle Information Rights Management went GA.  The next wave, which should be released soon, includes Oracle Universal Content Management and Oracle Universal Records Management. Normal 0 false false false EN-US X-NONE X-NONE MicrosoftInternetExplorer4 /* Style Definitions */ table.MsoNormalTable {mso-style-name:"Table Normal"; mso-tstyle-rowband-size:0; mso-tstyle-colband-size:0; mso-style-noshow:yes; mso-style-priority:99; mso-style-qformat:yes; mso-style-parent:""; mso-padding-alt:0in 5.4pt 0in 5.4pt; mso-para-margin-top:0in; mso-para-margin-right:0in; mso-para-margin-bottom:10.0pt; mso-para-margin-left:0in; line-height:115%; mso-pagination:widow-orphan; font-size:11.0pt; font-family:"Calibri","sans-serif"; mso-ascii-font-family:Calibri; mso-ascii-theme-font:minor-latin; mso-fareast-font-family:"Times New Roman"; mso-fareast-theme-font:minor-fareast; mso-hansi-font-family:Calibri; mso-hansi-theme-font:minor-latin;} Andy MacMillan and Roel Stalman kicked off these discussions last Monday, as they presented Oracle Content Management's product strategy and roadmap.  It seemed that the attendees liked what they heard regarding the strategy and future direction, but the question that seems to always come up after roadmap presentations is "when will the product be released"?  This is a question that none of us have the power to answer, but soon customers will be able to enjoy these new product capabilities: Normal 0 false false false EN-US X-NONE X-NONE MicrosoftInternetExplorer4 /* Style Definitions */ table.MsoNormalTable {mso-style-name:"Table Normal"; mso-tstyle-rowband-size:0; mso-tstyle-colband-size:0; mso-style-noshow:yes; mso-style-priority:99; mso-style-qformat:yes; mso-style-parent:""; mso-padding-alt:0in 5.4pt 0in 5.4pt; mso-para-margin-top:0in; mso-para-margin-right:0in; mso-para-margin-bottom:10.0pt; mso-para-margin-left:0in; line-height:115%; mso-pagination:widow-orphan; font-size:11.0pt; font-family:"Calibri","sans-serif"; mso-ascii-font-family:Calibri; mso-ascii-theme-font:minor-latin; mso-fareast-font-family:"Times New Roman"; mso-fareast-theme-font:minor-fareast; mso-hansi-font-family:Calibri; mso-hansi-theme-font:minor-latin;} Unified content repository across ECMCentralized installation, access, administration & monitoringCertified application integrations with solution templatesOpen Web Content Management Normal 0 false false false EN-US X-NONE X-NONE MicrosoftInternetExplorer4 /* Style Definitions */ table.MsoNormalTable {mso-style-name:"Table Normal"; mso-tstyle-rowband-size:0; mso-tstyle-colband-size:0; mso-style-noshow:yes; mso-style-priority:99; mso-style-qformat:yes; mso-style-parent:""; mso-padding-alt:0in 5.4pt 0in 5.4pt; mso-para-margin-top:0in; mso-para-margin-right:0in; mso-para-margin-bottom:10.0pt; mso-para-margin-left:0in; line-height:115%; mso-pagination:widow-orphan; font-size:11.0pt; font-family:"Calibri","sans-serif"; mso-ascii-font-family:Calibri; mso-ascii-theme-font:minor-latin; mso-fareast-font-family:"Times New Roman"; mso-fareast-theme-font:minor-fareast; mso-hansi-font-family:Calibri; mso-hansi-theme-font:minor-latin;} Stay tuned for more news about the release of Oracle Universal Content Management and Oracle Records Management.  There are a lot of new assets currently being built that will help get everyone up-to-speed quickly. Normal 0 false false false EN-US X-NONE X-NONE MicrosoftInternetExplorer4 /* Style Definitions */ table.MsoNormalTable {mso-style-name:"Table Normal"; mso-tstyle-rowband-size:0; mso-tstyle-colband-size:0; mso-style-noshow:yes; mso-style-priority:99; mso-style-qformat:yes; mso-style-parent:""; mso-padding-alt:0in 5.4pt 0in 5.4pt; mso-para-margin-top:0in; mso-para-margin-right:0in; mso-para-margin-bottom:10.0pt; mso-para-margin-left:0in; line-height:115%; mso-pagination:widow-orphan; font-size:11.0pt; font-family:"Calibri","sans-serif"; mso-ascii-font-family:Calibri; mso-ascii-theme-font:minor-latin; mso-fareast-font-family:"Times New Roman"; mso-fareast-theme-font:minor-fareast; mso-hansi-font-family:Calibri; mso-hansi-theme-font:minor-latin;} Outside of the sessions that were presented, there were a lot of other activities that took place at Collaborate.  The Enterprise 2.0 solutions demo pod was busy, and attendees were anxious to see demonstrations of Oracle's end-to-end document imaging solution, WebCenter Spaces, and web site creation using Oracle Universal Content Management.   I also want to thank our partners (Fishbowl Solutions, Redstone Content Solutions, Bezzotech, Team Informatics, and DTI) for their efforts in creating detailed, insightful presentations.  Also, special thanks are in order to Thomas Feldmeier and Markus Neubauer of Silbury IT-Beratung GmbH for their participation.  It seems that Thomas and Markus were doomed to be stranded in Frankfurt after the Icelandic ash storm.  They couldn't get a flight out of their native Germany, and with fear that they would miss Collaborate, they rented a car and drove to Rome - some 800 miles (1,200 kilometers).  Anyway, they made it safe and sound to Las Vegas, and although probably a bit tired, they gave 2 Oracle Content Management presentations.  Talk about commitment. Normal 0 false false false EN-US X-NONE X-NONE MicrosoftInternetExplorer4 /* Style Definitions */ table.MsoNormalTable {mso-style-name:"Table Normal"; mso-tstyle-rowband-size:0; mso-tstyle-colband-size:0; mso-style-noshow:yes; mso-style-priority:99; mso-style-qformat:yes; mso-style-parent:""; mso-padding-alt:0in 5.4pt 0in 5.4pt; mso-para-margin-top:0in; mso-para-margin-right:0in; mso-para-margin-bottom:10.0pt; mso-para-margin-left:0in; line-height:115%; mso-pagination:widow-orphan; font-size:11.0pt; font-family:"Calibri","sans-serif"; mso-ascii-font-family:Calibri; mso-ascii-theme-font:minor-latin; mso-fareast-font-family:"Times New Roman"; mso-fareast-theme-font:minor-fareast; mso-hansi-font-family:Calibri; mso-hansi-theme-font:minor-latin;} Finally, a very special thanks to Al Hoof and Dave Chaffee of the Oracle Content Management Special Interest Group (SIG).  Al and Dave did most of the heavy lifting for Collaborate, including the coordination of all the sessions.  The Independent Oracle Users Group presented Al with the Chris Wooldridge award, recognizing him as the volunteer of the year.  Here is Al with his award: Normal 0 false false false EN-US X-NONE X-NONE MicrosoftInternetExplorer4 /* Style Definitions */ table.MsoNormalTable {mso-style-name:"Table Normal"; mso-tstyle-rowband-size:0; mso-tstyle-colband-size:0; mso-style-noshow:yes; mso-style-priority:99; mso-style-qformat:yes; mso-style-parent:""; mso-padding-alt:0in 5.4pt 0in 5.4pt; mso-para-margin-top:0in; mso-para-margin-right:0in; mso-para-margin-bottom:10.0pt; mso-para-margin-left:0in; line-height:115%; mso-pagination:widow-orphan; font-size:11.0pt; font-family:"Calibri","sans-serif"; mso-ascii-font-family:Calibri; mso-ascii-theme-font:minor-latin; mso-fareast-font-family:"Times New Roman"; mso-fareast-theme-font:minor-fareast; mso-hansi-font-family:Calibri; mso-hansi-theme-font:minor-latin;} I hope to see you next year at Collaborate as the show returns to Orlando.

    Read the article

< Previous Page | 61 62 63 64 65 66 67 68 69 70 71 72  | Next Page >