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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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  • 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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  • 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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  • TSQL Conditionally Select Specific Value

    - by Dzejms
    This is a follow-up to #1644748 where I successfully answered my own question, but Quassnoi helped me to realize that it was the wrong question. He gave me a solution that worked for my sample data, but I couldn't plug it back into the parent stored procedure because I fail at SQL 2005 syntax. So here is an attempt to paint the broader picture and ask what I actually need. This is part of a stored procedure that returns a list of items in a bug tracking application I've inherited. There are are over 100 fields and 26 joins so I'm pulling out only the mostly relevant bits. SELECT tickets.ticketid, tickets.tickettype, tickets_tickettype_lu.tickettypedesc, tickets.stage, tickets.position, tickets.sponsor, tickets.dev, tickets.qa, DATEDIFF(DAY, ticket_history_assignment.savedate, GETDATE()) as 'daysinqueue' FROM dbo.tickets WITH (NOLOCK) LEFT OUTER JOIN dbo.tickets_tickettype_lu WITH (NOLOCK) ON tickets.tickettype = tickets_tickettype_lu.tickettypeid LEFT OUTER JOIN dbo.tickets_history_assignment WITH (NOLOCK) ON tickets_history_assignment.ticketid = tickets.ticketid AND tickets_history_assignment.historyid = ( SELECT MAX(historyid) FROM dbo.tickets_history_assignment WITH (NOLOCK) WHERE tickets_history_assignment.ticketid = tickets.ticketid GROUP BY tickets_history_assignment.ticketid ) WHERE tickets.sponsor = @sponsor The area of interest is the daysinqueue subquery mess. The tickets_history_assignment table looks roughly as follows declare @tickets_history_assignment table ( historyid int, ticketid int, sponsor int, dev int, qa int, savedate datetime ) insert into @tickets_history_assignment values (1521402, 92774,20,14, 20, '2009-10-27 09:17:59.527') insert into @tickets_history_assignment values (1521399, 92774,20,14, 42, '2009-08-31 12:07:52.917') insert into @tickets_history_assignment values (1521311, 92774,100,14, 42, '2008-12-08 16:15:49.887') insert into @tickets_history_assignment values (1521336, 92774,100,14, 42, '2009-01-16 14:27:43.577') Whenever a ticket is saved, the current values for sponsor, dev and qa are stored in the tickets_history_assignment table with the ticketid and a timestamp. So it is possible for someone to change the value for qa, but leave sponsor alone. What I want to know, based on all of these conditions, is the historyid of the record in the tickets_history_assignment table where the sponsor value was last changed so that I can calculate the value for daysinqueue. If a record is inserted into the history table, and only the qa value has changed, I don't want that record. So simply relying on MAX(historyid) won't work for me. Quassnoi came up with the following which seemed to work with my sample data, but I can't plug it into the larger query, SQL Manager bitches about the WITH statement. ;WITH rows AS ( SELECT *, ROW_NUMBER() OVER (PARTITION BY ticketid ORDER BY savedate DESC) AS rn FROM @Table ) SELECT rl.sponsor, ro.savedate FROM rows rl CROSS APPLY ( SELECT TOP 1 rc.savedate FROM rows rc JOIN rows rn ON rn.ticketid = rc.ticketid AND rn.rn = rc.rn + 1 AND rn.sponsor <> rc.sponsor WHERE rc.ticketid = rl.ticketid ORDER BY rc.rn ) ro WHERE rl.rn = 1 I played with it yesterday afternoon and got nowhere because I don't fundamentally understand what is going on here and how it should fit into the larger context. So, any takers? UPDATE Ok, here's the whole thing. I've been switching some of the table and column names in an attempt to simplify things so here's the full unedited mess. snip - old bad code Here are the errors: Msg 102, Level 15, State 1, Procedure usp_GetProjectRecordsByAssignment, Line 159 Incorrect syntax near ';'. Msg 102, Level 15, State 1, Procedure usp_GetProjectRecordsByAssignment, Line 179 Incorrect syntax near ')'. Line numbers are of course not correct but refer to ;WITH rows AS And the ')' char after the WHERE rl.rn = 1 ) Respectively Is there a tag for extra super long question? UPDATE #2 Here is the finished query for anyone who may need this: CREATE PROCEDURE [dbo].[usp_GetProjectRecordsByAssignment] ( @assigned numeric(18,0), @assignedtype numeric(18,0) ) AS SET NOCOUNT ON WITH rows AS ( SELECT *, ROW_NUMBER() OVER (PARTITION BY recordid ORDER BY savedate DESC) AS rn FROM projects_history_assignment ) SELECT projects_records.recordid, projects_records.recordtype, projects_recordtype_lu.recordtypedesc, projects_records.stage, projects_stage_lu.stagedesc, projects_records.position, projects_position_lu.positiondesc, CASE projects_records.clientrequested WHEN '1' THEN 'Yes' WHEN '0' THEN 'No' END AS clientrequested, projects_records.reportingmethod, projects_reportingmethod_lu.reportingmethoddesc, projects_records.clientaccess, projects_clientaccess_lu.clientaccessdesc, projects_records.clientnumber, projects_records.project, projects_lu.projectdesc, projects_records.version, projects_version_lu.versiondesc, projects_records.projectedversion, projects_version_lu_projected.versiondesc AS projectedversiondesc, projects_records.sitetype, projects_sitetype_lu.sitetypedesc, projects_records.title, projects_records.module, projects_module_lu.moduledesc, projects_records.component, projects_component_lu.componentdesc, projects_records.loginusername, projects_records.loginpassword, projects_records.assistedusername, projects_records.browsername, projects_browsername_lu.browsernamedesc, projects_records.browserversion, projects_records.osname, projects_osname_lu.osnamedesc, projects_records.osversion, projects_records.errortype, projects_errortype_lu.errortypedesc, projects_records.gsipriority, projects_gsipriority_lu.gsiprioritydesc, projects_records.clientpriority, projects_clientpriority_lu.clientprioritydesc, projects_records.scheduledstartdate, projects_records.scheduledcompletiondate, projects_records.projectedhours, projects_records.actualstartdate, projects_records.actualcompletiondate, projects_records.actualhours, CASE projects_records.billclient WHEN '1' THEN 'Yes' WHEN '0' THEN 'No' END AS billclient, projects_records.billamount, projects_records.status, projects_status_lu.statusdesc, CASE CAST(projects_records.assigned AS VARCHAR(5)) WHEN '0' THEN 'N/A' WHEN '10000' THEN 'Unassigned' WHEN '20000' THEN 'Client' WHEN '30000' THEN 'Tech Support' WHEN '40000' THEN 'LMI Tech Support' WHEN '50000' THEN 'Upload' WHEN '60000' THEN 'Spider' WHEN '70000' THEN 'DB Admin' ELSE rtrim(users_assigned.nickname) + ' ' + rtrim(users_assigned.lastname) END AS assigned, CASE CAST(projects_records.assigneddev AS VARCHAR(5)) WHEN '0' THEN 'N/A' WHEN '10000' THEN 'Unassigned' ELSE rtrim(users_assigneddev.nickname) + ' ' + rtrim(users_assigneddev.lastname) END AS assigneddev, CASE CAST(projects_records.assignedqa AS VARCHAR(5)) WHEN '0' THEN 'N/A' WHEN '10000' THEN 'Unassigned' ELSE rtrim(users_assignedqa.nickname) + ' ' + rtrim(users_assignedqa.lastname) END AS assignedqa, CASE CAST(projects_records.assignedsponsor AS VARCHAR(5)) WHEN '0' THEN 'N/A' WHEN '10000' THEN 'Unassigned' ELSE rtrim(users_assignedsponsor.nickname) + ' ' + rtrim(users_assignedsponsor.lastname) END AS assignedsponsor, projects_records.clientcreated, CASE projects_records.clientcreated WHEN '1' THEN 'Yes' WHEN '0' THEN 'No' END AS clientcreateddesc, CASE projects_records.clientcreated WHEN '1' THEN rtrim(clientusers_createuser.firstname) + ' ' + rtrim(clientusers_createuser.lastname) + ' (Client)' ELSE rtrim(users_createuser.nickname) + ' ' + rtrim(users_createuser.lastname) END AS createuser, projects_records.createdate, projects_records.savedate, projects_resolution.sitesaffected, projects_sitesaffected_lu.sitesaffecteddesc, DATEDIFF(DAY, projects_history_assignment.savedate, GETDATE()) as 'daysinqueue', projects_records.iOnHitList, projects_records.changetype FROM dbo.projects_records WITH (NOLOCK) LEFT OUTER JOIN dbo.projects_recordtype_lu WITH (NOLOCK) ON projects_records.recordtype = projects_recordtype_lu.recordtypeid LEFT OUTER JOIN dbo.projects_stage_lu WITH (NOLOCK) ON projects_records.stage = projects_stage_lu.stageid LEFT OUTER JOIN dbo.projects_position_lu WITH (NOLOCK) ON projects_records.position = projects_position_lu.positionid LEFT OUTER JOIN dbo.projects_reportingmethod_lu WITH (NOLOCK) ON projects_records.reportingmethod = projects_reportingmethod_lu.reportingmethodid LEFT OUTER JOIN dbo.projects_lu WITH (NOLOCK) ON projects_records.project = projects_lu.projectid LEFT OUTER JOIN dbo.projects_version_lu WITH (NOLOCK) ON projects_records.version = projects_version_lu.versionid LEFT OUTER JOIN dbo.projects_version_lu projects_version_lu_projected WITH (NOLOCK) ON projects_records.projectedversion = projects_version_lu_projected.versionid LEFT OUTER JOIN dbo.projects_sitetype_lu WITH (NOLOCK) ON projects_records.sitetype = projects_sitetype_lu.sitetypeid LEFT OUTER JOIN dbo.projects_module_lu WITH (NOLOCK) ON projects_records.module = projects_module_lu.moduleid LEFT OUTER JOIN dbo.projects_component_lu WITH (NOLOCK) ON projects_records.component = projects_component_lu.componentid LEFT OUTER JOIN dbo.projects_browsername_lu WITH (NOLOCK) ON projects_records.browsername = projects_browsername_lu.browsernameid LEFT OUTER JOIN dbo.projects_osname_lu WITH (NOLOCK) ON projects_records.osname = projects_osname_lu.osnameid LEFT OUTER JOIN dbo.projects_errortype_lu WITH (NOLOCK) ON projects_records.errortype = projects_errortype_lu.errortypeid LEFT OUTER JOIN dbo.projects_resolution WITH (NOLOCK) ON projects_records.recordid = projects_resolution.recordid LEFT OUTER JOIN dbo.projects_sitesaffected_lu WITH (NOLOCK) ON projects_resolution.sitesaffected = projects_sitesaffected_lu.sitesaffectedid LEFT OUTER JOIN dbo.projects_gsipriority_lu WITH (NOLOCK) ON projects_records.gsipriority = projects_gsipriority_lu.gsipriorityid LEFT OUTER JOIN dbo.projects_clientpriority_lu WITH (NOLOCK) ON projects_records.clientpriority = projects_clientpriority_lu.clientpriorityid LEFT OUTER JOIN dbo.projects_status_lu WITH (NOLOCK) ON projects_records.status = projects_status_lu.statusid LEFT OUTER JOIN dbo.projects_clientaccess_lu WITH (NOLOCK) ON projects_records.clientaccess = projects_clientaccess_lu.clientaccessid LEFT OUTER JOIN dbo.users users_assigned WITH (NOLOCK) ON projects_records.assigned = users_assigned.userid LEFT OUTER JOIN dbo.users users_assigneddev WITH (NOLOCK) ON projects_records.assigneddev = users_assigneddev.userid LEFT OUTER JOIN dbo.users users_assignedqa WITH (NOLOCK) ON projects_records.assignedqa = users_assignedqa.userid LEFT OUTER JOIN dbo.users users_assignedsponsor WITH (NOLOCK) ON projects_records.assignedsponsor = users_assignedsponsor.userid LEFT OUTER JOIN dbo.users users_createuser WITH (NOLOCK) ON projects_records.createuser = users_createuser.userid LEFT OUTER JOIN dbo.clientusers clientusers_createuser WITH (NOLOCK) ON projects_records.createuser = clientusers_createuser.userid LEFT OUTER JOIN dbo.projects_history_assignment WITH (NOLOCK) ON projects_history_assignment.recordid = projects_records.recordid AND projects_history_assignment.historyid = ( SELECT ro.historyid FROM rows rl CROSS APPLY ( SELECT TOP 1 rc.historyid FROM rows rc JOIN rows rn ON rn.recordid = rc.recordid AND rn.rn = rc.rn + 1 AND rn.assigned <> rc.assigned WHERE rc.recordid = rl.recordid ORDER BY rc.rn ) ro WHERE rl.rn = 1 AND rl.recordid = projects_records.recordid ) WHERE (@assignedtype='0' and projects_records.assigned = @assigned) OR (@assignedtype='1' and projects_records.assigneddev = @assigned) OR (@assignedtype='2' and projects_records.assignedqa = @assigned) OR (@assignedtype='3' and projects_records.assignedsponsor = @assigned) OR (@assignedtype='4' and projects_records.createuser = @assigned)

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  • Cosmic Journeys – Supermassive Black Hole at the Center of the Galaxy

    - by Akemi Iwaya
    Even though the center of our galaxy is obscured by thick dust and blinding starlight, that has not stopped scientists from piecing together clues about what may lie there. Sit back and enjoy a ‘cosmic journey’ with this excellent half-hour video from YouTube channel SpaceRip discussing what scientists have learned about the supermassive black hole at the center of our galaxy, and their work on getting a ‘direct image’ of it. Cosmic Journeys: Supermassive Black Hole at the Center of the Galaxy [YouTube]     

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  • SQL different joins not making any difference to result

    - by Chrissi
    I'm trying to write a quick (ha!) program to organise some of my financial information. What I ideally want is a query that will return all records with financial information in them from TableA. There should be one row for each month, but in instances where there were no transactions for a month there will be no record. I get results like this: SELECT Period,Year,TotalValue FROM TableA WHERE Year='1997' Result: Period Year TotalValue 1 1997 298.16 2 1997 435.25 4 1997 338.37 8 1997 336.07 9 1997 578.97 11 1997 361.23 By joining on a table (well a View in this instance) which just contains a field Period with values from 1 to 12, I expect to get something like this: SELECT p.Period,a.Year,a.TotalValue FROM Periods AS p LEFT JOIN TableA AS a ON p.Period = a.Period WHERE Year='1997' Result: Period Year TotalValue 1 1997 298.16 2 1997 435.25 3 NULL NULL 4 1997 338.37 5 NULL NULL 6 NULL NULL 7 NULL NULL 8 1997 336.07 9 1997 578.97 10 NULL NULL 11 1997 361.23 12 NULL NULL What I'm actually getting though is the same result no matter how I join it (except CROSS JOIN which goes nuts, but it's really not what I wanted anyway, it was just to see if different joins are even doing anything). LEFT JOIN, RIGHT JOIN, INNER JOIN all fail to provide the NULL records I am expecting. Is there something obvious that I'm doing wrong in the JOIN? Does it matter that I'm joining onto a View?

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