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  • Multiple many-to-many JOINs in a single mysql query without Cartesian Product

    - by VWD
    At the moment I can get the results I need with two seperate SELECT statements SELECT COUNT(rl.refBiblioID) FROM biblioList bl LEFT JOIN refList rl ON bl.biblioID = rl.biblioID GROUP BY bl.biblioID SELECT GROUP_CONCAT( CONCAT_WS( ':', al.lastName, al.firstName ) ORDER BY al.authorID ) FROM biblioList bl LEFT JOIN biblio_author ba ON ba.biblioID = bl.biblioID JOIN authorList al ON al.authorID = ba.authorID GROUP BY bl.biblioID Combining them like this however SELECT GROUP_CONCAT( CONCAT_WS( ':', al.lastName, al.firstName ) ORDER BY al.authorID ), COUNT(rl.refBiblioID) FROM biblioList bl LEFT JOIN biblio_author ba ON ba.biblioID = bl.biblioID JOIN authorList al ON al.authorID = ba.authorID LEFT JOIN refList rl ON bl.biblioID = rl.biblioID GROUP BY bl.biblioID causes the author result column to have duplicate names. How can I get the desired results from one SELECT statement without using DISTINCT? With subqueries?

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  • Referencing outer query's tables in a subquery

    - by soulmerge
    Is it possible to reference an outer query in a subquery with MySQL? I know there are some cases where this is possible: SELECT * FROM table t1 WHERE t1.date = ( SELECT MAX(date) FROM table t2 WHERE t2.id = t1.id)` ); But I'm wondering if something like this could work: SELECT u.username, c._postCount FROM User u INNER JOIN ( SELECT p.user, COUNT(*) AS _postCount FROM Posting p --# This is the reference I would need: WHERE p.user = u.id ) c ON c.user = u.id WHERE u.joinDate < '2009-10-10'; I know I could achieve the same using a GROUP BY or by pulling the outer WHERE clause into the sub-query, but I need this for automatic SQL generation and cannot use either alternative for various other reasons.

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  • SUM of column with Left Outer Join

    - by Matt
    I am trying to get the Count of all records that have at least on person who is authorized on the record. Basically, a Record can have more than one person associated with it. I want to return the count of Total Records, a count of total Authorized Records where at least 1 person is authorized, and a count of total NotAuthorized records where no person associated with record is authorized. It doesn't matter if one person is authorized per Record or if 3 people are authorized for that record, that should add 1 to the Authorized counter. The current query is incrementing Auth and Non auth for each person added per record rather, than one per record. If no people are assigned to the record that should also count towards Not Auth. SELECT Count(DISTINCT Record.RecordID) AS TotalRecords, SUM(CASE WHEN People.PersonLevel = 1 THEN 1 ELSE 0 END) AS Authorized, SUM(CASE WHEN People.PersonLevel <> 1 THEN 1 ELSE 0 END) AS NotAuthorized FROM Record LEFT OUTER JOIN RecordPeople ON Record.RecordID = RecordPeople.RecordID LEFT OUTER JOIN People ON RecordPeople.PersonID = People.PersonID

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  • SQL SELECT multiple INNER JOINs

    - by Noam Smadja
    The SELECT statement includes a reserved word or an argument name that is misspelled or missing, or the punctuation is incorrect its Access database.. i have a Library table, where Autnm Topic Size Cover Lang are foreign Keys each record is actually a book which has its properties such as author and stuff. i am not quite sure i am even using the correct JOIN.. quite new with "complex" SQL :) SELECT Library.Bknm_Hebrew, Library.Bknm_English, Library.Bknm_Russian, Library.Note, Library.ISBN, Library.Pages, Library.PUSD, Author.ID AS [AuthorID], Author.Author_hebrew AS [AuthorHebrew], Author.Author_English AS [AuthorEnglish], Author.Author_Russian AS [AuthorRussian], Topic.ID AS [TopicID], Topic.Topic_Hebrew AS [TopicHebrew], Topic.Topic_English AS [TopicEnglish], Topic.Topic_Russian AS [TopicRussian], Size.Size AS [Size], Cover.ID AS [TopicID], Cover.Cvrtyp_Hebrew AS [CoverHebrew], Cover.Cvrtyp_English AS [TopicEnglish], Cover.Cvrtyp_Russian AS [CoverRussian], Lang.ID AS [LangID], Lang.Lang_Hebrew AS [LangHebrew], Lang.Lang_English AS [LangEnglish], FROM Library INNER JOIN Author ON Library.Autnm = Author.ID INNER JOIN Topic ON Library.Topic = Topic.ID INNER JOIN Size ON Library.Size = Size.ID INNER JOIN Cover ON Library.Cover = Cover.ID INNER JOIN Lang ON Library.Lang = Lang.ID Thx in advance

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  • How to join this table?

    - by pamella
    ads table img90.imageshack.us/img90/6295/adsvo.png phones table img194.imageshack.us/img194/3713/phones.png cars table img35.imageshack.us/img35/1035/carsm.png i have 3 tables ads,cars and phones. i want to join tables is based on category in ads table. and i tried this query but no luck,any helps? SELECT * FROM `ads` JOIN `ads.category` ON `ads.id` = `ads.category.id` ** i cant add comment any of your post,but i want it to be automatic based on category in ads table. for example :- if in table have phones category,i will automatic join phones table then SELECT * FROM `ads` JOIN `phone` ON `ads.id` = `phone.id` if in table have cars category,i will automatic join cars table SELECT * FROM `ads` JOIN `cars` ON `ads.id` = `cars.id`

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  • Symfony/Propel NestedSet left/right ID corruption/adjustment

    - by Mike Crowe
    Hi folks, I have a nested set application that seems to be getting corrupted. Here's what I'm seeing: We're using nested sets for a binary tree (any node can have 2 children). It appears to be working fine, but some event causes a discrepancy. For instance, when I do a getNumberOfDescendants() for the root node, it will slowly increase as this event happens. However, displaying the tree works fine, as does inserting (apparently). Has anybody seen anything like this before? For instance, my repair program shows this as the repairs that it makes: User pxxxxx left 0=>0, right 145=>129 User axxxxx left 1=>1, right 124=>106 User mxxxxx left 119=>117, right 120=>118 User fxxxxx left 125=>107, right 144=>128 User fxxxxx left 126=>108, right 131=>113 User rxxxxx left 127=>109, right 128=>110 User mxxxxx left 129=>111, right 130=>112 User mxxxxx left 132=>114, right 143=>127 User cxxxxx left 133=>115, right 142=>126 User gxxxxx left 134=>116, right 137=>121 User mxxxxx left 135=>119, right 136=>120 User jxxxxx left 138=>122, right 141=>125 User axxxxx left 139=>123, right 140=>124 I thought at first it was when I deleted a user, but it has since occurred w/o that event. Anybody know of a cause that might generate this? I've tested ad nauseum on my local machine, but I can't duplicate it. I do have an issue where my production box is PHP 5.2.0, whereas my test device is 5.2.10. Could that be an issue? TIA Mike

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  • Not getting the right expected output for my Mysql Query?

    - by user1878107
    i've 4 tables as shown below doctors id name ------------ 1 Mathew 2 Praveen 3 Rosie 4 Arjun 5 Denis doctors_appointments id doctors_id patient_name contact date status -------------------------------------------------------------------------------------- 1 5 Nidhin 9876543210 2012-12-10 15:39:41 Registered 2 5 Sunny 9876543210 2012-12-18 15:39:48 Registered 3 5 Mani 9876543210 2012-12-12 15:39:57 Registered 4 2 John 9876543210 2012-12-24 15:40:09 Registered 5 4 Raj 9876543210 2012-12-05 15:41:57 Registered 6 3 Samuel 9876543210 2012-12-14 15:41:33 Registered 7 2 Louis 9876543210 2012-12-24 15:40:23 Registered 8 1 Federick 9876543210 2012-12-28 15:41:05 Registered 9 2 Sam 9876543210 2012-12-12 15:40:38 Registered 10 4 Sita 9876543210 2012-12-12 15:41:00 Registered doctors_dutyplan id doctor_id weeks time no_of_patients ------------------------------------------------------------------ 1 1 3,6,7 9:00am-1:00pm 10 2 2 3,4,5 1:00pm-4:00pm 7 3 3 3,6,7 10:00am-2:00pm 10 4 4 3,4,5,6 8:30am-12:30pm 12 5 5 3,4,5,6,7 9:00am-4:00pm 30 emp_leave id empid leavedate -------------------------------- 1 2 2012-12-05 14:42:36 2 2 2012-12-03 14:42:59 3 3 2012-12-03 14:43:06 4 3 2012-12-06 14:43:14 5 5 2012-12-04 14:43:24 My task is to find all the days in a month in which the doctor is available excluding the leave dates. My query what is wrote is given below: SELECT DATE_ADD( '2012-12-01', INTERVAL ROW DAY ) AS Date, ROW +1 AS DayOfMonth FROM ( SELECT @row := @row +1 AS ROW FROM ( SELECT 0 UNION ALL SELECT 1 UNION ALL SELECT 3 UNION ALL SELECT 4 UNION ALL SELECT 5 UNION ALL SELECT 6 )t1, ( SELECT 0 UNION ALL SELECT 1 UNION ALL SELECT 3 UNION ALL SELECT 4 UNION ALL SELECT 5 UNION ALL SELECT 6 )t2, ( SELECT @row := -1 )t3 LIMIT 31 )b WHERE DATE_ADD( '2012-12-01', INTERVAL ROW DAY ) BETWEEN '2012-12-01' AND '2012-12-31' AND DAYOFWEEK( DATE_ADD( '2012-12-01', INTERVAL ROW DAY ) ) =2 AND DATE_ADD( '2012-12-01', INTERVAL ROW DAY ) NOT IN ( SELECT DATE_FORMAT( l.leavedate, '%Y-%m-%d' ) AS date FROM doctors_dutyplan d LEFT JOIN emp_leave AS l ON d.doctor_id = l.empid WHERE doctor_id =2 ) This works fine for all doctors who took any leave in a particular day in a month (here in the example it is Decemeber 2012). and the result is shown below: Date DayOfMonth ----------------------- 2012-12-10 10 2012-12-17 17 2012-12-24 24 2012-12-31 31 But on the other hand for the doctors who did'nt took any leave , for that my query is showing empty table, example for the doctor Mathew whose id is 1, my query returns an empty result can anyone please tell a solution for this problem. Thanks in advance.

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  • How do I create a self referential association (self join) in a single class using ActiveRecord in Rails?

    - by Daniel Chang
    I am trying to create a self join table that represents a list of customers who can refer each other (perhaps to a product or a program). I am trying to limit my model to just one class, "Customer". The schema is: create_table "customers", force: true do |t| t.string "name" t.integer "referring_customer_id" t.datetime "created_at" t.datetime "updated_at" end add_index "customers", ["referring_customer_id"], name: "index_customers_on_referring_customer_id" My model is: class Customer < ActiveRecord::Base has_many :referrals, class_name: "Customer", foreign_key: "referring_customer_id", conditions: {:referring_customer_id => :id} belongs_to :referring_customer, class_name: "Customer", foreign_key: "referring_customer_id" end I have no problem accessing a customer's referring_customer: @customer.referring_customer.name ... returns the name of the customer that referred @customer. However, I keep getting an empty array when accessing referrals: @customer.referrals ... returns []. I ran binding.pry to see what SQL was being run, given a customer who has a "referer" and should have several referrals. This is the SQL being executed. Customer Load (0.3ms) SELECT "customers".* FROM "customers" WHERE "customers"."id" = ? ORDER BY "customers"."id" ASC LIMIT 1 [["id", 2]] Customer Exists (0.2ms) SELECT 1 AS one FROM "customers" WHERE "customers"."referring_customer_id" = ? AND "customers"."referring_customer_id" = 'id' LIMIT 1 [["referring_customer_id", 3]] I'm a bit lost and am unsure where my problem lies. I don't think my query is correct -- @customer.referrals should return an array of all the referrals, which are the customers who have @customer.id as their referring_customer_id.

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  • Slow queries in Rails- not sure if my indexes are being used.

    - by Max Williams
    I'm doing a quite complicated find with lots of includes, which rails is splitting into a sequence of discrete queries rather than do a single big join. The queries are really slow - my dataset isn't massive, with none of the tables having more than a few thousand records. I have indexed all of the fields which are examined in the queries but i'm worried that the indexes aren't helping for some reason: i installed a plugin called "query_reviewer" which looks at the queries used to build a page, and lists problems with them. This states that indexes AREN'T being used, and it features the results of calling 'explain' on the query, which lists various problems. Here's an example find call: Question.paginate(:all, {:page=>1, :include=>[:answers, :quizzes, :subject, {:taggings=>:tag}, {:gradings=>[:age_group, :difficulty]}], :conditions=>["((questions.subject_id = ?) or (questions.subject_id = ? and tags.name = ?))", "1", 19, "English"], :order=>"subjects.name, (gradings.difficulty_id is null), gradings.age_group_id, gradings.difficulty_id", :per_page=>30}) And here are the generated sql queries: SELECT DISTINCT `questions`.id FROM `questions` LEFT OUTER JOIN `taggings` ON `taggings`.taggable_id = `questions`.id AND `taggings`.taggable_type = 'Question' LEFT OUTER JOIN `tags` ON `tags`.id = `taggings`.tag_id LEFT OUTER JOIN `subjects` ON `subjects`.id = `questions`.subject_id LEFT OUTER JOIN `gradings` ON gradings.question_id = questions.id WHERE (((questions.subject_id = '1') or (questions.subject_id = 19 and tags.name = 'English'))) ORDER BY subjects.name, (gradings.difficulty_id is null), gradings.age_group_id, gradings.difficulty_id LIMIT 0, 30 SELECT `questions`.`id` AS t0_r0 <..etc...> FROM `questions` LEFT OUTER JOIN `answers` ON answers.question_id = questions.id LEFT OUTER JOIN `quiz_questions` ON (`questions`.`id` = `quiz_questions`.`question_id`) LEFT OUTER JOIN `quizzes` ON (`quizzes`.`id` = `quiz_questions`.`quiz_id`) LEFT OUTER JOIN `subjects` ON `subjects`.id = `questions`.subject_id LEFT OUTER JOIN `taggings` ON `taggings`.taggable_id = `questions`.id AND `taggings`.taggable_type = 'Question' LEFT OUTER JOIN `tags` ON `tags`.id = `taggings`.tag_id LEFT OUTER JOIN `gradings` ON gradings.question_id = questions.id LEFT OUTER JOIN `age_groups` ON `age_groups`.id = `gradings`.age_group_id LEFT OUTER JOIN `difficulties` ON `difficulties`.id = `gradings`.difficulty_id WHERE (((questions.subject_id = '1') or (questions.subject_id = 19 and tags.name = 'English'))) AND `questions`.id IN (602, 634, 666, 698, 730, 762, 613, 645, 677, 709, 741, 592, 624, 656, 688, 720, 752, 603, 635, 667, 699, 731, 763, 614, 646, 678, 710, 742, 593, 625) ORDER BY subjects.name, (gradings.difficulty_id is null), gradings.age_group_id, gradings.difficulty_id SELECT count(DISTINCT `questions`.id) AS count_all FROM `questions` LEFT OUTER JOIN `answers` ON answers.question_id = questions.id LEFT OUTER JOIN `quiz_questions` ON (`questions`.`id` = `quiz_questions`.`question_id`) LEFT OUTER JOIN `quizzes` ON (`quizzes`.`id` = `quiz_questions`.`quiz_id`) LEFT OUTER JOIN `subjects` ON `subjects`.id = `questions`.subject_id LEFT OUTER JOIN `taggings` ON `taggings`.taggable_id = `questions`.id AND `taggings`.taggable_type = 'Question' LEFT OUTER JOIN `tags` ON `tags`.id = `taggings`.tag_id LEFT OUTER JOIN `gradings` ON gradings.question_id = questions.id LEFT OUTER JOIN `age_groups` ON `age_groups`.id = `gradings`.age_group_id LEFT OUTER JOIN `difficulties` ON `difficulties`.id = `gradings`.difficulty_id WHERE (((questions.subject_id = '1') or (questions.subject_id = 19 and tags.name = 'English'))) Actually, looking at these all nicely formatted here, there's a crazy amount of joining going on here. This can't be optimal surely. Anyway, it looks like i have two questions. 1) I have an index on each of the ids and foreign key fields referred to here. The second of the above queries is the slowest, and calling explain on it (doing it directly in mysql) gives me the following: +----+-------------+----------------+--------+---------------------------------------------------------------------------------+-------------------------------------------------+---------+------------------------------------------------+------+----------------------------------------------+ | id | select_type | table | type | possible_keys | key | key_len | ref | rows | Extra | +----+-------------+----------------+--------+---------------------------------------------------------------------------------+-------------------------------------------------+---------+------------------------------------------------+------+----------------------------------------------+ | 1 | SIMPLE | questions | range | PRIMARY,index_questions_on_subject_id | PRIMARY | 4 | NULL | 30 | Using where; Using temporary; Using filesort | | 1 | SIMPLE | answers | ref | index_answers_on_question_id | index_answers_on_question_id | 5 | millionaire_development.questions.id | 2 | | | 1 | SIMPLE | quiz_questions | ref | index_quiz_questions_on_question_id | index_quiz_questions_on_question_id | 5 | millionaire_development.questions.id | 1 | | | 1 | SIMPLE | quizzes | eq_ref | PRIMARY | PRIMARY | 4 | millionaire_development.quiz_questions.quiz_id | 1 | | | 1 | SIMPLE | subjects | eq_ref | PRIMARY | PRIMARY | 4 | millionaire_development.questions.subject_id | 1 | | | 1 | SIMPLE | taggings | ref | index_taggings_on_taggable_id_and_taggable_type,index_taggings_on_taggable_type | index_taggings_on_taggable_id_and_taggable_type | 263 | millionaire_development.questions.id,const | 1 | | | 1 | SIMPLE | tags | eq_ref | PRIMARY | PRIMARY | 4 | millionaire_development.taggings.tag_id | 1 | Using where | | 1 | SIMPLE | gradings | ref | index_gradings_on_question_id | index_gradings_on_question_id | 5 | millionaire_development.questions.id | 2 | | | 1 | SIMPLE | age_groups | eq_ref | PRIMARY | PRIMARY | 4 | millionaire_development.gradings.age_group_id | 1 | | | 1 | SIMPLE | difficulties | eq_ref | PRIMARY | PRIMARY | 4 | millionaire_development.gradings.difficulty_id | 1 | | +----+-------------+----------------+--------+---------------------------------------------------------------------------------+-------------------------------------------------+---------+------------------------------------------------+------+----------------------------------------------+ The query_reviewer plugin has this to say about it - it lists several problems: Table questions: Using temporary table, Long key length (263), Using filesort MySQL must do an extra pass to find out how to retrieve the rows in sorted order. To resolve the query, MySQL needs to create a temporary table to hold the result. The key used for the index was rather long, potentially affecting indices in memory 2) It looks like rails isn't splitting this find up in a very optimal way. Is it, do you think? Am i better off doing several find queries manually rather than one big combined one? Grateful for any advice, max

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  • Join us for Live Oracle Linux and Oracle VM Cloud Events in Europe

    - by Monica Kumar
    Join us for a series of live events and discover how Oracle VM and Oracle Linux offer an integrated and optimized infrastructure for quickly deploying a private cloud environment at lower cost. As one of the most widely deployed operating systems today, Oracle Linux delivers higher performance, better reliability, and stability, at a lower cost for your cloud environments. Oracle VM is an application-driven server virtualization solution fully integrated and certified with Oracle applications to deliver rapid application deployment and simplified management. With Oracle VM, you have peace of mind that the entire Oracle stack deployed is fully certified by Oracle. Register now for any of the upcoming events, and meet with Oracle experts to discuss how we can help in enabling your private cloud. Nov 20: Foundation for the Cloud: Oracle Linux and Oracle VM (Belgium) Nov 21: Oracle Linux & Oracle VM Enabling Private Cloud (Germany) Nov 28: Realize Substantial Savings and Increased Efficiency with Oracle Linux and Oracle VM (Luxembourg) Nov 29: Foundation for the Cloud: Oracle Linux and Oracle VM (Netherlands) Dec 5: MySQL Tech Tour, including Oracle Linux and Oracle VM (France) Hope to see you at one of these events!

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  • Join the Dark Side of Visual Studio 2010

    - by InfinitiesLoop
    Hard to believe it’s been so long, but it was almost 4 years ago when I published Join the Dark Side of Visual Studio . That was when a lot of people were still using VS2003, and importing and exporting environment settings required a custom add-in, VSStyler, which has since fallen off the planet and is hard to find (link, anyone? Let me know). Three versions of VS later, and I’m still using and loving the dark side. Pleased, I am (haha). In fact, that article for one reason or another is still one...(read more)

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  • Join Our Call: Sun Storage 2500-M2 Announcement

    - by user797911
    Oracle's Sun Storage 2500-M2 array brings together the latest Fibre Channel (FC) and SAS2 technologies with Oracle's Sun Storage Common Array software from Oracle to create a robust solution that’s equally adept in an entry-level storage area network (SAN) for the mid-size business and integrating into an existing storage network within the enterprise. The Sun Storage 2500-M2 replaces Sun's Storage 2500 array product line and is designed so that the customer may have a quick qualification time for fast and easy deployment in the traditional 2500 environments. Jun Jang, Oracle Principal Product Manager will be hosting this 1 hour live call (a recording will be available), please join us to find out more: Event Date: 24-JUN-11 Event Time: 08:00 am PST/PDT/4pm UK time Web Registration and Access: http://oukc.oracle.com/static09/opn/login/?t=livewebcast|c=1031672594 Access for Mobile Devices: http://my.oracle.com/content/web/cnt636926 Call Provider: Intercall International Participant Dial-In Number: 706-634-8508 Additional International Dial-In Numbers Link: http://www.intercall.com/national/oracleuniversity/gdnam.html Dial-In Passcode: 96395

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  • Join Our Call: Sun Storage 2500-M2 Announcement

    - by mseika
    Oracle's Sun Storage 2500-M2 array brings together the latest Fibre Channel (FC) and SAS2 technologies with Oracle's Sun Storage Common Array software from Oracle to create a robust solution that’s equally adept in an ! entry-level storage area network (SAN) for the mid-size business and integrating into an existing storage network within the enterprise. The Sun Storage 2500-M2 replaces Sun's Storage 2500 array product line and is designed so that the customer may have a quick qualification time for fast and easy deployment in the traditional 2500 environments. Jun Jang, Oracle Principal Product Manager will be hosting this 1 hour live call (a recording will be available), please join us to find out more:24. Jun 2011 08:00 am PST/PDT/4pm UK timeWeb Registration and AccessAccess for Mobile DevicesInternational Participant Dial-In Number: 706-634-8508Additional International Dial-In Numbers LinkDial-In Passcode: 6395

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  • mysql join with multiple values in one column

    - by CYREX
    I need to make a query that creates 3 columns that come from 2 tables which have the following relations: TABLE 1 has Column ID that relates to TABLE 2 with column ID2 In TABLE 1 there is a column called user In TABLE 2 there is a column called names There can be 1 unique user but there can be many names associated to that user. If i do the following i get all data BUT the user column repeats itself for each name it has associated. What i want is for use to appear unique but the names columns appear with all the names associated to the user column but separated by commas, like the following: select user,names from TABLE1 left join TABLE2 on TABLE1.id = TABLE2.id This will show the users repeated everytime a name appears for that user. what i want is to appear like this: USER - NAMES cyrex - pedrox, rambo, zelda homeboy - carmen, carlos, tom, sandra jerry - seinfeld, christine ninja - soloboy etc....

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  • Join Us for the Next Quarterly Customer Update Webcast

    - by michelle.huff
    Join us for the next Oracle Content Management Quarterly Customer Update Webcast scheduled for this coming June 30 / July 1 2010. Don't miss this chance to get an overview on the latest updates to Oracle Content Management. We'll be covering the latest ECM Suite 11g release - highlighting the Universal Content Management (UCM) and Universal Records Management releases. Register Today! Americas / EMEA time zones: Customer Update June 30, 2010 9:00am US PDT / 12:00pm US EDT / 16:00 GMT Length: 1 hour *Please use your corporate email address to register. Asia-Pacific time zones: Customer Update (Repeat Webcast) July 1, 2010 12:00pm Sydney AEST, 10:00am Singapore (June 30, 2010 @ 7:00pm US PDT) Length: 1 hour *Please use your corporate email address to register Please Note: If you have attended previous Quarterly Customer Update Webcasts, we are now using a new web conference system, WebEx, to host the meeting. Missed Previous Customer Quarterly Updates? Get caught up on Oracle & ECM news. View a recording or the presentation from previous Webcasts held since June 2008 (available from My Oracle Support).

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  • Join Us for the Next Quarterly Customer Update Webcast

    - by michelle.huff
    Join us for the next Oracle Content Management Quarterly Customer Update Webcast scheduled for this coming January 19 & 20, 2010. In this webcast we'll bring you up to speed on the latest updates and changes made available these past few months. Additionally, we'll cover the new features and certifications in the latest ODC & ODDC 10.1.3.5.1 release, as well as the upcoming Enterprise Content Management Suite 11gR1 PS3 (patch set 3) release. Register Today! Americas / EMEA time zones: Customer Update January 19, 2010 9:00am US PT / 12:00pm US ET / 17:00 London Length: 1 hour *Please use your corporate email address to register. Asia-Pacific time zones: Customer Update (Repeat Webcast) January 20, 2010 1:00pm Sydney AET, 10:00am Singapore (Jan 19, 2010 @ 6:00pm US PT) Length: 1 hour *Please use your corporate email address to register Missed Previous Customer Quarterly Updates? Get caught up on Oracle & ECM news. View a recording or the presentation from previous Webcasts held since June 2008 (available from My Oracle Support).

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  • Join the SPARC Go To Market Webinar on June 21st

    - by swalker
    Please join the World Wide webinar focused on SPARC, and designed to provide insights and selling guidance, at 5 p.m. CET on Thursday, June 21. The speaker, Bud Koch, Sr Principal Product Marketing Director will focus on SPARC / T4 Marketing: with a review of current assets and where we are going into FY13.  Details about the meeting can be found here. Please plan on joining 10 minutes before the scheduled start time. If you are not able to participate in real time, a replay will be available shortly afterward.

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  • Join the SPARC Go To Market Webinar on June 21st

    - by Cinzia Mascanzoni
    Please join the World Wide webinar focused on SPARC, and designed to provide insights and selling guidance, at 5 p.m. CET on Thursday, June 21. The speaker, Bud Koch, Sr Principal Product Marketing Director will focus on SPARC / T4 Marketing: with a review of current assets and where we are going into FY13.  Details about the meeting can be found here. Please plan on joining 10 minutes before the scheduled start time. If you are not able to participate in real time, a replay will be available shortly afterward.

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  • Joining Two MKV files in Ubuntu?

    - by Ryan McClure
    I have an opera that I'm ripping to my computer in MKV format with Handbrake. This opera is on two discs. Is there a way to join the resulting MKV's together? They will have the same bitrate, resolution, etc. If I do this, can I keep chapters from both MKV files organized? And, since I have subtitles in the file (not burnt in), will they still stay intact? I'm not too sure if this question is off-topic or not. If it is, feel more than free to delete it. :)

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  • Join Gretchen Alarcon In Person for an Oracle HCM Applications Strategy Updates

    - by jay.richey
    How can you benefit from staying current and moving to the latest release of your Oracle HCM applications? Where does Fusion HCM fit in and what do they mean to your existing investments? What does Oracle offer in terms of SaaS for HCM? What is Oracle doing to maintain excellence in your current applications portfolio while innovating in new and creative ways? Join us for an exclusive breakfast briefing where you will have the opportunity to hear about Oracle's current blockbuster releases for HCM: PeopleSoft 9.1 and E-Business Suite 12.1. Take this opportunity to hear about what the latest releases mean to you and learn how organizations like yours are successfully moving forward. Our featured speaker, Gretchen Alarcon, Oracle's Vice President of Fusion HCM Product Strategy will share how Oracle's latest HCM offerings - Fusion HCM and Fusion Talent Management On Demand - can work alongside your Oracle PeopleSoft, E-Business Suite, or JD Edwards HR foundation to show immediate business value. This event promises to provide you with an opportunity to share experiences, best practices, challenges, and successes with fellow business executives. Coming to: Chicago, Minneaoplis, St. Louis

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  • Join Companies in Web and Telecoms by Adopting MySQL Cluster

    - by Antoinette O'Sullivan
    Join Web and Telecom companies who have adopted MySQL Cluster to facilitate application in the following areas: Web: High volume OLTP eCommerce User profile management Session management and caching Content management On-line gaming Telecoms: Subscriber databases (HLR/HSS) Service deliver platforms VAS: VoIP, IPTV and VoD Mobile content delivery Mobile payments LTE access To come up to speed on MySQL Cluster, take the 3-day MySQL Cluster training course. Events already on the schedule include:  Location  Date  Delivery Language  Berlin, Germany  16 December 2013  German  Munich, Germany  2 December 2013  German  Budapest, Hungary  4 December 2013  Hungarian  Madrid, Spain  9 December 2013  Spanish  Jakarta Barat, Indonesia  27 January 2014  English  Singapore  20 December 2013  English  Bangkok, Thailand  28 January 2014  English  San Francisco, CA, United States  28 May 2014  English  New York, NY, United States  17 December 2013  English For more information about this course or to request an additional event, go to the MySQL Curriculum Page (http://education.oracle.com/mysql).

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  • Join us for 2 JCP sessions today + beer

    - by heathervc
    Remember to join the 2 JCP sessions at JavaOne this afternoon in the Hilton.  First up the JCP.Next panel with JCP EC Members, followed by the 101 Ways to Participate BOF.  Stop in to learn what's new and how you can make the future Java and enjoy a beer or 2.  We will also be in the OTN Java Demogrounds in the Hilton Grand Ballroom from 4:00 - 4:30 PM.  Hope to see you there. JCP.Next: Reinvigorating Java Standards Session ID: BOF6272 Location: Hilton San Francisco - Plaza A/B Date and Time: 10/1/12, 4:30 PM - 5:15 PM 101 Ways to Improve Java: Why Developer Participation Matters Session ID: BOF6283 Location: Hilton San Francisco - Continental Ballroom 4 Date and Time: 10/1/12, 5:30 PM - 6:15 PM

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  • Right-Time Retail Part 1

    - by David Dorf
    This is the first in a three-part series. Normal 0 false false false EN-US X-NONE X-NONE /* 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; mso-bidi-font-family:"Times New Roman"; mso-bidi-theme-font:minor-bidi;} Right-Time Revolution Technology enables some amazing feats in retail. I can order flowers for my wife while flying 30,000 feet in the air. I can order my groceries in the subway and have them delivered later that day. I can even see how clothes look on me without setting foot in a store. Who knew that a TV, diamond necklace, or even a car would someday be as easy to purchase as a candy bar? Can technology make a mattress an impulse item? Wake-up and your back is hurting, so you rollover and grab your iPad, then a new mattress is delivered the next day. Behind the scenes the many processes are being choreographed to make the sale happen. This includes moving data between systems with the least amount for friction, which in some cases is near real-time. But real-time isn’t appropriate for all the integrations. Think about what a completely real-time retailer would look like. A consumer grabs toothpaste off the shelf, and all systems are immediately notified so that the backroom clerk comes running out and pushes the consumer aside so he can replace the toothpaste on the shelf. Such a system is not only cost prohibitive, but it’s also very inefficient and ineffectual. Retailers must balance the realities of people, processes, and systems to find the right speed of execution. That’ what “right-time retail” means. Retailers used to sell during the day and count the money and restock at night, but global expansion and the Web have complicated that simplistic viewpoint. Our 24hr society demands not only access but also speed, which constantly pushes the boundaries of our IT systems. In the last twenty years, there have been three major technology advancements that have moved us closer to real-time systems. Networking is the first technology that drove the real-time trend. As systems became connected, it became easier to move data between them. In retail we no longer had to mail the daily business report back to corporate each day as the dial-up modem could transfer the data. That was soon replaced with trickle-polling, when sale transactions were occasionally sent from stores to corporate throughout the day, often through VSAT. Then we got terrestrial networks like DSL and Ethernet that allowed the constant stream of data between stores and corporate. When corporate could see the sales transactions coming from stores, it could better plan for replenishment and promotions. That drove the need for speed into the supply chain and merchandising, but for many years those systems were stymied by the huge volumes of data. Nordstrom has 150 million SKU/Store combinations when planning (RPAS); The Gap generates 110 million price changes during end-of-season (RPM); Argos does 1.78 billion calculations executed each day for replenishment planning (AIP). These areas are now being alleviated by the second technology, storage. The typical laptop disk drive runs at 5,400rpm with PCs stepping up to 7,200rpm and servers hitting 15,000rpm. But the platters can only spin so fast, so to squeeze more performance we’ve had to rely on things like disk striping. Then solid state drives (SSDs) were introduced and prices continue to drop. (Augmenting your harddrive with a SSD is the single best PC upgrade these days.) RAM continues to be expensive, but compressing data in memory has allowed more efficient use. So a few years back, Oracle decided to build a box that incorporated all these advancements to move us closer to real-time. This family of products, often categorized as engineered systems, combines the hardware and software so that they work together to provide better performance. How much better? If Exadata powered a 747, you’d go from New York to Paris in 42 minutes, and it would carry 5,000 passengers. If Exadata powered baseball, games would last only 18 minutes and Boston’s Fenway would hold 370,000 fans. The Exa-family enables processing more data in less time. So with faster networks and storage, that brings us to the third and final ingredient. If we continue to process data in traditional ways, we won’t be able to take advantage of the faster networks and storage. Enter what Harvard calls “The Sexiest Job of the 21st Century” – the data scientist. New technologies like the Hadoop-powered Oracle Big Data Appliance, Oracle Advanced Analytics, and Oracle Endeca Information Discovery change the way in which we organize data. These technologies allow us to extract actionable information from raw data at incredible speeds, often ad-hoc. So the foundation to support the real-time enterprise exists, but how does a retailer begin to take advantage? The most visible way is through real-time marketing, but I’ll save that for part 3 and instead begin with improved integrations for the assets you already have in part 2.

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  • Join the CodePlex community on Geeklist

    Community is very important to us at CodePlex. And we love partnering with other like-minded organizations. Geeklist is one of the new kids on the block, building a great place for geeks to share what they've done, who they did it with and connect with great companies and communities.     There are some exciting new experiences coming on-line soon that you won’t want to miss out on. Geeklist is currently in private beta, so if you don't already have an account, use the CodePlex invite code to create your own account. Then, join the CodePlex community and follow the CodePlex team on Geeklist. Once you’ve joined, be proud, tell the world what you have worked on, and who you did it with. And don’t be shy to give out a few high fives to the amazing work others in the community have created.

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  • Advise some swing based (open source) project to join

    - by user592704
    I am looking for some open source Swing based projects which wanted volunteer Java developers to join and the projects which show their products authors' names. I watched many links but most projects for some reason hide their authors names (showing some nick names or something...) and all developing process relative information... For example this one project it seems fine but still I couldn't find any information concerning some current project task(s), its developers group, some chronicles (tips, milestones, feedbacks etc) :( I googled a lot but found less :S So I wondering maybe you know some? I dearly hope you can give me a piece of advice Any useful comment is much appreciated

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