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  • no ocijdbc10 in java.library.path

    - by B.Z.B
    Hey all, So I've been plagued by this issue, whenever I try to run my app in eclipse, I get this error. 2011-02-23 09:55:08,388 ERROR (com.xxxxx.services.factory.ServiceInvokerLocal:21) - java.lang.UnsatisfiedLinkError: no ocijdbc10 in java.library.path I've tried following the steps I found here with no luck. I've tried this on a XP VM as well as windows 7 (although in win 7 I get a different error, below) java.lang.UnsatisfiedLinkError: no ocijdbc9 in java.library.path I've made sure my oracle client was ok (by running TOAD) and I also re-added the classes12.jar / ojdbc14.jars to my WEB-INF/lib folder taken directly from my %ORACLE_HOME% folder (also re-added them to the lib path). I've also tried just adding the ojdbc14.jar without the classes12.jar. Any suggestions appreciated. In the XP VM I have my PATH variable set to C:\Program Files\Java\jdk1.6.0_24\bin;C:\ORACLE\product\10.2.0.1\BIN. I'm using Tomcat server 5.0

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  • In SQL, what does Group By mean without Count(*), or Sum(), Max(), avg(), ..., and what are some use

    - by Jian Lin
    In SQL, if we use Group By without Count(*) or Sum(), etc, then the result is as follows: mysql> select * from sentGifts; +--------+------------+--------+------+---------------------+--------+ | sentID | whenSent | fromID | toID | trytryWhen | giftID | +--------+------------+--------+------+---------------------+--------+ | 1 | 2010-04-24 | 123 | 456 | 2010-04-24 01:52:20 | 100 | | 2 | 2010-04-24 | 123 | 4568 | 2010-04-24 01:56:04 | 100 | | 3 | 2010-04-24 | 123 | NULL | NULL | 1 | | 4 | 2010-04-24 | NULL | 111 | 2010-04-24 03:10:42 | 2 | | 5 | 2010-03-03 | 11 | 22 | 2010-03-03 00:00:00 | 6 | | 6 | 2010-04-24 | 11 | 222 | 2010-04-24 03:54:49 | 6 | | 7 | 2010-04-24 | 1 | 2 | 2010-04-24 03:58:45 | 6 | +--------+------------+--------+------+---------------------+--------+ 7 rows in set (0.00 sec) mysql> select *, count(*) from sentGifts group by whenSent; +--------+------------+--------+------+---------------------+--------+----------+ | sentID | whenSent | fromID | toID | trytryWhen | giftID | count(*) | +--------+------------+--------+------+---------------------+--------+----------+ | 5 | 2010-03-03 | 11 | 22 | 2010-03-03 00:00:00 | 6 | 1 | | 1 | 2010-04-24 | 123 | 456 | 2010-04-24 01:52:20 | 100 | 6 | +--------+------------+--------+------+---------------------+--------+----------+ 2 rows in set (0.00 sec) mysql> select * from sentGifts group by whenSent; +--------+------------+--------+------+---------------------+--------+ | sentID | whenSent | fromID | toID | trytryWhen | giftID | +--------+------------+--------+------+---------------------+--------+ | 5 | 2010-03-03 | 11 | 22 | 2010-03-03 00:00:00 | 6 | | 1 | 2010-04-24 | 123 | 456 | 2010-04-24 01:52:20 | 100 | +--------+------------+--------+------+---------------------+--------+ 2 rows in set (0.00 sec) Only 1 row is returned per "group". What does it mean when there is no "Count(*)", etc when using "Group By", and what are it uses? thanks.

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  • Temporary intermediate table

    - by user289429
    In our project to generate massive reports in oracle we use some permanent table to hold intermediate results. For example to generate one report we run few queries and populate the table, at the final step we join the intermediate table with huge application tables. These intermediate tables are cleared for next report run. We have few concerns in performance areas. These intermediate tables are transactional and don't have statistics. Is it good idea to join these with application tables which are partitioned and have up to date statistics. We need these results stored in the intermediate tables to be available across requests from UI hence we are not in a position to use oracle provided temporary tables. Any thoughts on what could be done would be appreciated.

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  • Better data-structure design

    - by Tempname
    Currently in my application I have a single table that is giving me a bit of trouble. The issue at hand is I have a value object that is mapped to this table. When the data is returned to me as an array of value objects, I have to then loop through this array and begin my recursion by matching the ParentID to parent ObjectID's. The column ParentID is either null (acts a parent) or it holds the value of an ObjectID. I know there has to be a better way to create this data structure so that I do not have to do recursive loops to match ParentID's with their ObjectID's. Any help with this is greatly appreciated. Here is the table in describe form: +----------------+------------------+------+-----+---------------------+-----------------------------+ | Field | Type | Null | Key | Default | Extra | +----------------+------------------+------+-----+---------------------+-----------------------------+ | ObjectID | int(11) unsigned | NO | PRI | NULL | auto_increment | | ObjectHeight | decimal(6,2) | NO | | NULL | | | ObjectWidth | decimal(6,2) | NO | | NULL | | | ObjectX | decimal(6,2) | NO | | NULL | | | ObjectY | decimal(6,2) | NO | | NULL | | | ObjectLabel | varchar(255) | NO | | NULL | | | TemplateID | int(11) unsigned | NO | MUL | NULL | | | ObjectTypeID | int(11) unsigned | NO | MUL | NULL | | | ParentID | int(11) unsigned | YES | MUL | NULL | | | CreationDate | datetime | YES | | 0000-00-00 00:00:00 | | | LastModifyDate | timestamp | YES | | NULL | on update CURRENT_TIMESTAMP | +----------------+------------------+------+-----+---------------------+-----------------------------+e

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  • Need help in setting application name with JPA (EclipseLink)

    - by enrique
    hello everybody i am using JPA with EclipseLink and oracle as DB and i need to set the property v$session of jdbc4 it allows to set an identification name to the application for auditing purposes but i had no lucky setting it up....i have been trying through entitiyManager following the example in this page: http://wiki.eclipse.org/Configuring_a_EclipseLink_JPA_Application_(ELUG) it does not show any error but does not set the application name at all... when i see the audit in oracle it is not being audited with the name i set by code "Customers" but with OS_program_name=JDBC Thin Client it means that the property in the code is not being set properly and i have no idea where the issue is, the code i am using is the following : emProperties.put("v$session.program","Customers"); factory=Persistence.createEntityManagerFactory("clients",emProperties); em=factory.createEntityManager(emProperties); em.merge(clients); does anybody know how to do it or any idea.... thanks.-

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  • Two Instances of Sql Server (2005 and 2008)

    - by Felipe
    Hi All, I installed Visual Studio 2008 Professional in my machine and It had installed SQL Server Express 2005 database in machine, and I use it very fine! I installed SQL Managment Studio and works great. So, in this week I Installed Visual Studio 2010 Pro in machine and the setup installed the SQL Server express 2008 and it overwrite the instance of my SQL Server Express 2005. All right, Now, I'd like to know how can I have two instances of the SQL Server Express in my Machine, Express 2005 and Express 2008. I can not access the 2005 , only 2008 :( and my projects uses 2005.. Somebody Help me! thanks Bye

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  • MySQL: Combining multiple where conditions

    - by Karl
    I'm working on a menu system that takes a url and then queries the db to build the menu. My menu table is: +---------+--------------+------+-----+---------+----------------+ | Field | Type | Null | Key | Default | Extra | +---------+--------------+------+-----+---------+----------------+ | id | int(11) | NO | PRI | NULL | auto_increment | | node_id | int(11) | YES | | NULL | | | parent | int(11) | YES | | NULL | | | weight | int(11) | YES | | NULL | | | title | varchar(250) | YES | | NULL | | | alias | varchar(250) | YES | | NULL | | | exclude | int(11) | YES | | NULL | | +---------+--------------+------+-----+---------+----------------+ The relevant columns for my question are alias, parent and node_id. So for a url like: http://example.com/folder1/folder2/filename Alias would potentially = "filename", "folder1", "folder2" Parent = the node_id of the parent folder. What I know is how to split the url up into an array and check the alias for a match to each part. What I don't know is how to have it then filter by parent whose alias matches "folder2" and whose parent alias matches "folder1". I'm imagining a query like so: select * from menu where alias='filename' and where parent = node_id where alias='folder2' and parent = node_id where alias='folder1' Except I know that the above is wrong. I'm hoping this can be done in a single query. Thanks for any help in advance!

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  • How can you access two identically-named columns in a MySQL LEFT JOIN query?

    - by George Edison
    I have two tables. table_x: id INT(11) tag INT(11) table_tags: id INT(11) name VARCHAR(255) Then I use PHP to perform the following query: SELECT * FROM table_x LEFT JOIN table_tags ON table_x.tag = table_tags.id The only problem is: how do I access table_x.id and table_tags.id in the results? Here is the PHP code: $query = "SELECT * FROM table_x LEFT JOIN table_tags ON table_x.tag = table_tags.id"; $results = mysql_query($query); while($row = mysql_fetch_array($results)) { // how do I now access table_x.id and table_tags.id ??? }

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  • 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.

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  • 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.

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  • 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...

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  • Selecting a sequence NEXTVAL for multiple rows

    - by stringpoet
    I am building a SQL Server job to pull data from SQL Server into an Oracle database through a linked server. The table I need to populate has a sequence for the name ID, which is my primary key. I'm having trouble figuring out a way to do this simply, without some lengthy code. Here's what I have so far for the SELECT portion (some actual names obfuscated): SELECT (SELECT NEXTVAL FROM OPENQUERY(MYSERVER, 'SELECT ORCL.NAME_SEQNO.NEXTVAL FROM DUAL')), psn.BirthDate, psn.FirstName, psn.MiddleName, psn.LastName, c.REGION_CODE FROM Person psn LEFT JOIN MYSERVER..ORCL.COUNTRY c ON c.COUNTRY_CODE = psn.Country MYSERVER is the linked Oracle server, ORCL is obviously the schema. Person is a local table on the SQL Server database where the query is being executed. When I run this query, I get the same exact value for all records for the NEXTVAL. What I need is for it to generate a new value for each returned record. I found this similar question, with its answers, but am unsure how to apply it to my case (if even possible): Query several NEXTVAL from sequence in one satement

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  • 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?

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  • How to have partial incremental synchronizations based on a GUID?

    - by Gonçalo Veiga
    I need to synchronize an SQL Server database to Oracle through an Oracle Transparent Gateway. The synchronization is performed in batches, so I need to get the next set of data from the point where I left off. The problem I'm having is that the only field I have in the source, to help me, is a GUID. If it were a number I could just order by it, keep the last one processed and restart the process by getting the records which are my recorded number. This won't work with a GUID. Any ideas?

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  • 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

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  • How to display multiple categories and products underneath each category?

    - by shin
    Generally there is a category menu and each link to a category page where shows all the items under that category. Now I need to show all the categories and products underneath with PHP/MySQL in the same page. So it will be like this. Category 1 description of category 1 item 1 item 2 .. Category 2 description of category 2 item 5 item 6 .. Category 3 description of category 3 item 8 item 9 ... ... I have category and product table in my database. But I am not sure how to proceed. CREATE TABLE IF NOT EXISTS `omc_product` ( `id` int(11) NOT NULL AUTO_INCREMENT, `name` varchar(255) NOT NULL, `shortdesc` varchar(255) NOT NULL, `longdesc` text NOT NULL, `thumbnail` varchar(255) NOT NULL, `image` varchar(255) NOT NULL, `product_order` int(11) DEFAULT NULL, `class` varchar(255) DEFAULT NULL, `grouping` varchar(16) DEFAULT NULL, `status` enum('active','inactive') NOT NULL, `category_id` int(11) NOT NULL, `featured` enum('none','front','webshop') NOT NULL, `other_feature` enum('none','most sold','new product') NOT NULL, `price` float(7,2) NOT NULL, PRIMARY KEY (`id`) ) ENGINE=MyISAM DEFAULT CHARSET=latin1 AUTO_INCREMENT=2 ; CREATE TABLE IF NOT EXISTS `omc_category` ( `id` int(11) NOT NULL AUTO_INCREMENT, `name` varchar(255) NOT NULL, `shortdesc` varchar(255) NOT NULL, `longdesc` text NOT NULL, `status` enum('active','inactive') NOT NULL, `parentid` int(11) NOT NULL, PRIMARY KEY (`id`) ) ENGINE=MyISAM DEFAULT CHARSET=latin1 AUTO_INCREMENT=2 ; I will appreciate your help. Thanks in advance.

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  • 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?

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  • Tunneling through SSH for 1521 port access?

    - by A T
    I am developing locally on my computer, using my own Apache server with PHP configured. My database however is remotely located on an Oracle 11g Database Server. We were also given a separate remote server for hosting our .html and .php files, however only FTP access has been provided there. Development is far too slow waiting for the FTP connection to push. So I decided to develop locally, but still use the remote DB server. Unfortunately that gives me an error. Not sure how—or where—to integrate tunnelling. Do I add something to the oci_connect HOST in my PHP file, or do I encapsulate my whole environment over SSH?

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  • 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?

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  • Significance of date 02/31/2157?

    - by dpatchery
    I work in a large scale IT support environment. Twice now we have seen an invalid date of 02/31/2157 being inserted in an Oracle DATE column. So far I have not been able to reproduce this problem, but it appears to be happening occasionally when a user attempts to save '00/00/0000' into the column. I believe the value is originating from a PowerBuilder DataWindow update. The application uses myriad libraries for all sorts of technologies, so this question may be a bit vague, but... Has anyone seen the date 02/31/2157 in some established library that Oracle could be defaulting to when some other invalid date is entered? Perhaps an end-of-time concept analogous to the beginning-of-time date of 1/1/1970?

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  • Query on Triggers..

    - by RBA
    Created One Trigger in Oracle.. SQL> CREATE OR REPLACE TRIGGER student_after_insert 2 AFTER INSERT 3 ON student 4 FOR EACH ROW 5 BEGIN 6 @hello.pl 9 END student_after_insert; 10 / Contents of hello.pl are:- BEGIN DBMS_OUTPUT.PUT_LINE('hello world'); END; And.. the result is pretty good, as the content of hello.pl is displayed on screen while inserting a record.. Now, the query is -- When i change the content of the hello.pl file, after exiting from oracle, and then logging again, It doesn't shows the updated contents, instead it shows the previous content.. I noticed that, if i drop the trigger and create it again, then it is working fine.. Why is it happening so.. And what is the solution to this problem..

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  • 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?

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  • 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 ;

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