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  • Do you want to learn about developing Web, Mobile and beyond Oracle based applications? Join our online virtual event on November 26th

    - by JuergenKress
    Learn about the latest innovations in Oracle ADF. Our virtual event provides sessions that range from introductory to deep dive, covering Oracle’s strategic framework for developing multi-channel enterprise applications for the Oracle platforms. Multiple tracks cover every interest and every level and include live online Q&A chats with Oracle’s technical staff. For details please visit our registration page. WebLogic Partner Community For regular information become a member in the WebLogic Partner Community please visit: http://www.oracle.com/partners/goto/wls-emea ( OPN account required). If you need support with your account please contact the Oracle Partner Business Center. Blog Twitter LinkedIn Mix Forum Wiki Technorati Tags: ADF,ADF mobile,education,training,Oracle OpenWorld,WebLogic,WebLogic Community,Oracle,OPN,Jürgen Kress

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  • An invitation to join a JDeveloper and ADF productivity clinic (and more!) at KScope

    - by Chris Muir
    Would you like a chance to influence Oracle's decisions on tool usability and productivity? If you're attending ODTUG's Kaleidoscope conference this year in San Antonio, Oracle would like to invite you to participate in our Usability Activity Research and separately our JDeveloper and ADF Productivity Clinics with our experienced user experience teams.  The teams are keen to hear what you have to say about your experiences with our tools in general and specifically JDeveloper and ADF.  The details of each event are described below. Invitation to Usability Activity - Sunday June 24th to Wednesday June 27th Oracle is constantly working on new tools and new features for developers, and invites YOU to become a key part of the process!  As a special addition to Kscope 12, Oracle will be conducting onsite usability research in the Alyssum room, from Sunday June 24 to Wednesday June 27. Usability activities are scheduled ahead of time for participants' convenience.  If you would like to take part, please fill out this form to let us know of the session(s) that you would like to attend and your development experience. You will be emailed with your scheduled session before the start of the conference. JDeveloper and ADF Productivity Clinic - Thursday June 28th Are you concerned that Java, Oracle ADF or JDeveloper is difficult? Is JDeveloper making you jump through hoops?  Do you hate a particular dialog or feature of JDeveloper? Well, come and get things off your chest! Oracle is hosting a product management and user experience clinic where we want to hear about your issues and concerns. What's difficult to use?  What doesn't work the way you want, and how would you want it to work?  What isn't behaving like your current favorite tool?  If we can't help on you the spot, we'll take your feedback and use it to improve the product experience.  A great opportunity to get answers, or get improvements. Drop by the Alyssum room, anytime from 8:30 to 10:30 on Thursday, June 28. We look forward to seeing you at KScope soon! 

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  • How can I be prepared to join a company?

    - by Aerovistae
    There's more to it than that, but this title was the best way I could think of to sum it up. I'm a senior in a good computer science program, and I'm graduating early. About to start interviews and all whatnot. I'm not a super-experienced programmer, not one of those people who started in middle school. I'm decent at this, but I'm not among the best, not nearly. I have to do an awful lot of googling. So today I'm meeting some fellow for lunch at a campus cafe to discuss some front-end details when this tall, good-looking guy begs pardon, says he's new to campus, says he's wondering if we know where he can go to sign up for recruiting developers. Quickly evolves into long conversation: he's the CEO of a seems-to-be-doing-well start-up. Hiring passionate interns and full-times. Sounds great! I take one look at his site on my own computer later, immediately spot a major bug. No idea how to fix it, but I see it. I go over to the page code, and good god. It's the standard amount of code you would expect from a full-scale web application, a couple dozen pages of HTML and scripts. I don't even know where to start reading it. I've built sites from scratch, but obviously never on that scale, nor have I ever worked on one of that scale. I have no idea which bit might generate the bug. But that sets me thinking: How could someone like me possibly settle into an environment like that? A start-up is a very high-pressure working environment. I don't know if I can work at that pace under those constraints-- I would hate to let people down. And with only 10 employees, it's not like anyone has much time to help you get your bearings. Somewhere in there is a question. Can you see it? I'm asking for general advice here. Maybe even anecdotal advice. Is joining a start-up right out of college a scary process? Am I overestimating what it would take to figure out the mass of code behind this site? What's the likelihood a decent but only moderately-experienced coder could earn his pay at such a place? For instance, I know nothing of server-side/back-end programming. Never touched it. That scares me.

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  • Angry Birds and Star Wars Join Forces for an Awesome New Edition [Plus Wallpaper!]

    - by Asian Angel
    Are you ready for a new version of Angry Birds? Then rejoice, you are less than a month away from an awesome new release of everyone’s favorite bird-slinging, pig smashing game! Prepare for a journey to a galaxy far, far away… From the blog post: From the deserts of Tatooine to the depths of the Death Star – the game and merchandise will feature the Angry Birds characters starring as the iconic heroes of the beloved Saga. In the coming weeks, fans can expect additional new videos, characters, and much more exciting content to be revealed. The game will be available on iOS, Android, Amazon Kindle Fire, Mac, PC, Windows Phone and Windows 8. Here is the first of the promo videos for the new version. Also, make sure to download the first official wallpaper (linked to below)! How To Get a Better Wireless Signal and Reduce Wireless Network Interference How To Troubleshoot Internet Connection Problems 7 Ways To Free Up Hard Disk Space On Windows

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  • And if I ask the job interviewer for reasons to join the company?

    - by Oscar
    In job interviews I am frequently asked if I know the company, explain why do I think I would be the best choice for this company, etc and I have never liked this kinds of questions. Using your experiences in job interviewing, what do you think it would happen if I ask the interviewer to explain me why he think the company is the best company for me and why I should accept their answer? Do you think it would be a good think or bad thing to do? Edit: the idea of the question would not be to "challenge" the interviewer. The idea of the question would be to see what he thinks about the company, what values he thinks are given importance inside the company and what are the strong points of the company.

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  • If you are in Brussels next week... come join a policy dialog on the European Accessibility Act

    - by Peter Korn
    Next week I will be in Brussels attending a policy dialog held at the European Policy Centre on Thursday October 11th. It is titled "The Accessibility Act – ensuring access to goods and services across the EU", and I will be part of a distinguished panel exploring some of the issues the upcoming European Accessibility Act may address - with a particular emphasis in my case on the role of ICT accessibility. This morning policy dialog will be followed by a more focused workshop in the afternoon looking at specific challenges and potential solutions to those challenges. Oracle is sponsoring this policy dialog and workshop, alongside the European Disability Forum. If you are in Brussels, you are invited to attend. Registration by e-mail is now open.

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  • Convert SQL to LINQ in MVC3 with Ninject

    - by Jeff
    I'm using MVC3 and still learning LINQ. I'm having some trouble trying to convert a query to LINQ to Entities. I want to return an employee object. SELECT E.EmployeeID, E.FirstName, E.LastName, MAX(EO.EmployeeOperationDate) AS "Last Operation" FROM Employees E INNER JOIN EmployeeStatus ES ON E.EmployeeID = ES.EmployeeID INNER JOIN EmployeeOperations EO ON ES.EmployeeStatusID = EO.EmployeeStatusID INNER JOIN Teams T ON T.TeamID = ES.TeamID WHERE T.TeamName = 'MyTeam' GROUP BY E.EmployeeID, E.FirstName, E.LastName ORDER BY E.FirstName, E.LastName What I have is a few tables, but I need to get only the newest status based on the EmployeeOpertionDate. This seems to work fine in SQL. I'm also using Ninject and set my query to return Ienumerable. I played around with the group by option but it then returns IGroupable. Any guidance on converting and returning the property object type would be appreciated. Edit: I started writing this out in LINQ but I'm not sure how to properly return the correct type or cast this. public IQueryable<Employee> GetEmployeesByTeam(int teamID) { var q = from E in context.Employees join ES in context.EmployeeStatuses on E.EmployeeID equals ES.EmployeeID join EO in context.EmployeeOperations on ES.EmployeeStatusID equals EO.EmployeeStatusID join T in context.Teams on ES.TeamID equals T.TeamID where T.TeamName == "MyTeam" group E by E.EmployeeID into G select G; return q; } Edit2: This seems to work for me public IQueryable<Employee> GetEmployeesByTeam(int teamID) { var q = from E in context.Employees join ES in context.EmployeeStatuses on E.EmployeeID equals ES.EmployeeID join EO in context.EmployeeOperations.OrderByDescending(eo => eo.EmployeeOperationDate) on ES.EmployeeStatusID equals EO.EmployeeStatusID join T in context.Teams on ES.TeamID equals T.TeamID where T.TeamID == teamID group E by E.EmployeeID into G select G.FirstOrDefault(); return q; }

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  • NHibernate Native SQL multiple joins

    - by Chris
    Hi all, I"m having some problems with Nhibernate and native sql. I've got an entity with alot of collections and I am doing an SQL Fulltext search on it. So when returning 100 or so entities, I dont want all collections be lazy loaded. For this I changed my SQL query: SELECT Query.* FROM (SELECT {spr.*}, {adr.*}, {adrt.*}, {cty.*}, {com.*}, {comt.*}, spft.[Rank] AS [Rak], Row_number() OVER(ORDER BY spft.[Rank] DESC) AS rownum FROM customer spr INNER JOIN CONTAINSTABLE ( customerfulltext , computedfulltextindex , '" + parsedSearchTerm + @"' ) AS spft ON spr.customerid = spft.[Key] LEFT JOIN [Address] adr ON adr.customerid = spr.customerid INNER JOIN [AddressType] adrt ON adrt.addresstypeid = adr.addresstypeid INNER JOIN [City] cty ON cty.cityid = adr.cityid LEFT JOIN [Communication] com ON com.customerid = spr.customerid INNER JOIN [CommunicationType] comt ON comt.communicationtypeid = com.communicationtypeid) as Query ORDER BY Query.[Rank] DESC This is how I setup the query: var items = GetCurrentSession() .CreateSQLQuery(query) .AddEntity("spr", typeof(Customer)) .AddJoin("adr", "spr.addresses") .AddJoin("adrt", "adr.Type") .AddJoin("cty", "adr.City") .AddJoin("com", "spr.communicationItems") .AddJoin("comt", "com.Type") .List<Customer>(); What happens now is, that the query returns customers twice (or more), I assume this is because of the joins since for each customer address, communicationItem (e.g. phone, email), a new sql row is returned. In this case I thought I could use the DistinctRootEntityResultTransformer. var items = GetCurrentSession() .CreateSQLQuery(query) .AddEntity("spr", typeof(Customer)) .AddJoin("adr", "spr.addresses") .AddJoin("adrt", "adr.Type") .AddJoin("cty", "adr.City") .AddJoin("com", "spr.communicationItems") .AddJoin("comt", "com.Type") .SetResultTransformer(new DistinctRootEntityResultTransformer()) .List<Customer>(); Doing so an exception is thrown. This is because I try to list customers .List<Customer>() but the transformer returns only entities of the last join added. E.g. in the case above, the entity with alias "comt" is returned when doing .List() instead of .List(). If I would switch last join with the join alias "cty", then the transformer returns a list of cities only... Anyone knows how I can return a clean list of customers in this case?

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  • Help optimizing a query with 16 subqueries

    - by Webnet
    I have indexes/primaries on all appropriate ID fields for each type. I'm wondering though how I could make this more efficient. It takes a while to load the page with only 15,000 rows and that'll quickly grow to 500k. The $whereSql variable simply has a few more parameters for the main ebay_archive_listing table. NOTE: This is all done in a single query because I have ASC/DESC sorting for each subquery value. NOTE: I've converted some of the sub queries to INNER JOIN's SELECT product_master.product_id, ( SELECT COUNT(listing_id) FROM ebay_archive_product_listing_assoc '.$listingCountJoin.' WHERE ebay_archive_product_listing_assoc.product_id = product_master.product_id) as listing_count, sku, type_id, ( SELECT AVG(ebay_archive_listing.current_price) FROM ebay_archive_listing INNER JOIN ebay_archive_product_listing_assoc ON ( ebay_archive_product_listing_assoc.listing_id = ebay_archive_listing.id AND ebay_archive_product_listing_assoc.product_id = product_master.product_id ) WHERE '.$whereSql.' AND ebay_archive_listing.current_price > 0 ) as average_bid_price, ( SELECT AVG(ebay_archive_listing.buy_it_now_price) FROM ebay_archive_listing INNER JOIN ebay_archive_product_listing_assoc ON ( ebay_archive_product_listing_assoc.listing_id = ebay_archive_listing.id AND ebay_archive_product_listing_assoc.product_id = product_master.product_id ) WHERE '.$whereSql.' AND ebay_archive_listing.buy_it_now_price > 0 ) as average_buyout_price, ( SELECT MIN(ebay_archive_listing.current_price) FROM ebay_archive_listing INNER JOIN ebay_archive_product_listing_assoc ON ( ebay_archive_product_listing_assoc.listing_id = ebay_archive_listing.id AND ebay_archive_product_listing_assoc.product_id = product_master.product_id ) WHERE '.$whereSql.' AND ebay_archive_listing.current_price > 0 ) as lowest_bid_price, ( SELECT MAX(ebay_archive_listing.current_price) FROM ebay_archive_listing INNER JOIN ebay_archive_product_listing_assoc ON ( ebay_archive_product_listing_assoc.listing_id = ebay_archive_listing.id AND ebay_archive_product_listing_assoc.product_id = product_master.product_id ) WHERE '.$whereSql.' AND ebay_archive_listing.current_price > 0 ) as highest_bid_price, ( SELECT MIN(ebay_archive_listing.buy_it_now_price) FROM ebay_archive_listing INNER JOIN ebay_archive_product_listing_assoc ON ( ebay_archive_product_listing_assoc.listing_id = ebay_archive_listing.id AND ebay_archive_product_listing_assoc.product_id = product_master.product_id ) WHERE '.$whereSql.' AND ebay_archive_listing.current_price > 0 ) as lowest_buyout_price, ( SELECT MAX(ebay_archive_listing.buy_it_now_price) FROM ebay_archive_listing INNER JOIN ebay_archive_product_listing_assoc ON ( ebay_archive_product_listing_assoc.listing_id = ebay_archive_listing.id AND ebay_archive_product_listing_assoc.product_id = product_master.product_id ) WHERE '.$whereSql.' AND ebay_archive_listing.current_price > 0 ) as highest_buyout_price, round((( SELECT COUNT(ebay_archive_listing.id) FROM ebay_archive_listing INNER JOIN ebay_archive_product_listing_assoc ON ( ebay_archive_product_listing_assoc.listing_id = ebay_archive_listing.id AND ebay_archive_product_listing_assoc.product_id = product_master.product_id ) WHERE '.$whereSql.' AND ebay_archive_listing.status_id = 2 ) / ( SELECT COUNT(listing_id) FROM ebay_archive_product_listing_assoc '.$listingCountJoin.' WHERE ebay_archive_product_listing_assoc.product_id = product_master.product_id ) * 100), 1) as sold_percent FROM product_master '.$joinSql.' WHERE product_master.product_id IN ( SELECT product_id FROM ebay_archive_product_listing_assoc INNER JOIN ebay_archive_listing ON ( ebay_archive_listing.id = ebay_archive_product_listing_assoc.listing_id AND '.$whereSql.' ) )

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  • Calculating estimated data loss with Always on

    - by blakmk
    Ever wondered how calculate estimated data loss (time) for always on. The metric in the always on dashboard shows the metric quite nicely but there does seem to be a lack of documentation about where the metrics ---come from. Heres a script that calculates the data loss ( lag ) so you can set up alerts based on your DR SLA's:       WITH DR_CTE ( replica_server_name, database_name, last_commit_time) AS                 (                                 select ar.replica_server_name, database_name, rs.last_commit_time                                 from master.sys.dm_hadr_database_replica_states  rs                                 inner join master.sys.availability_replicas ar on rs.replica_id = ar.replica_id                                 inner join sys.dm_hadr_database_replica_cluster_states dcs on dcs.group_database_id = rs.group_database_id and rs.replica_id = dcs.replica_id                                 where replica_server_name != @@servername                 ) select ar.replica_server_name, dcs.database_name, rs.last_commit_time, DR_CTE.last_commit_time 'DR_commit_time', datediff(ss,  DR_CTE.last_commit_time, rs.last_commit_time) 'lag_in_seconds' from master.sys.dm_hadr_database_replica_states  rs inner join master.sys.availability_replicas ar on rs.replica_id = ar.replica_id inner join sys.dm_hadr_database_replica_cluster_states dcs on dcs.group_database_id = rs.group_database_id and rs.replica_id = dcs.replica_id inner join DR_CTE on DR_CTE.database_name = dcs.database_name where ar.replica_server_name = @@servername order by lag_in_seconds desc

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  • Using LINQ to XML, how can I join two sets of data based on ordinal position?

    - by Donald Hughes
    Using LINQ to XML, how can I join two sets of data based on ordinal position? <document> <set1> <value>A</value> <value>B</value> <value>C</value> </set1> <set2> <value>1</value> <value>2</value> <value>3</value> </set2> </document> Based on the above fragment, I would like to join the two sets together such that "A" and "1" are in the same record, "B" and "2" are in the same record, and "C" and "3" are in the same record.

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  • How to get query result even if JOIN hasn't found any results?

    - by user1734651
    I want select data for user, and join another info from other table that related to the user. The problem is that this extra data not always exist for any user, just for few. How can I write a query that will return NULL for not found data, instead just return null for the whole query? SELECT a.*, b.* FROM user AS a LEFT JOIN extra AS b ON (a.userid = b.userid) WHERE a.userid = {$userid} LIMIT 1 When extra data found for the user, I get the resource as expected. If not, I get NULL for the whole query. Bottom line, I don't care if "extra" exist for the user or not, if yes - select it as well, if not - ignore that.

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  • Ruby and Forking

    - by Cory
    Quick question about Ruby forking - I ran across a bit of forking code in Resque earlier that was sexy as hell but tripped me up for a few. I'm hoping for someone to give me a little more detail about what's going on here. Specifically - it would appear that forking spawns a child (expected) and kicks it straight into the 'else' side of my condition (less expected. Is that expected behavior? A Ruby idiom? My IRB hack here: def fork return true if @cant_fork begin if Kernel.respond_to?(:fork) Kernel.fork else raise NotImplementedError end rescue NotImplementedError @cant_fork = true nil end end def do_something puts "Starting do_something" if foo = fork puts "we are forking from #{Process.pid}" Process.wait else puts "no need to fork, let's get to work: #{Process.pid} under #{Process.ppid}" puts "doing it" end end do_something

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  • Implement Tree/Details With Taskflow Regions Using EJB

    - by Deepak Siddappa
    This article describes on Display Tree/Details using taskflow regions.Use Case DescriptionLet us take scenario where we need to display Tree/Details, left region contains category hierarchy with items listed in a tree structure (ex:- Region-Countries-Locations-Departments in tree format) and right region contains the Employees list.In detail, Here User may drills down through categories using a tree until Employees are listed. Clicking the tree node name displays Employee list in the adjacent pane related to particular tree node. Implementation StepsThe script for creating the tables and inserting the data required for this application CreateSchema.sql Lets create a Java EE Web Application with Entities based on Regions, Countries, Locations, Departments and Employees table. Create a Stateless Session Bean and data control for the Stateless Session Bean. Add the below code to the session bean and expose the method in local/remote interface and generate a data control for that.Note:- Here in the below code "em" is a EntityManager. public List<Employees> empFilteredByTreeNode(String treeNodeType, String paramValue) { String queryString = null; try { if (treeNodeType == "null") { queryString = "select * from Employees emp ORDER BY emp.employee_id ASC"; } else if (Pattern.matches("[a-zA-Z]+[_]+[a-zA-Z]+[_]+[[0-9]+]+", treeNodeType)) { queryString = "select * from employees emp INNER JOIN departments dept\n" + "ON emp.department_id = dept.department_id JOIN locations loc\n" + "ON dept.location_id = loc.location_id JOIN countries cont\n" + "ON loc.country_id = cont.country_id JOIN regions reg\n" + "ON cont.region_id = reg.region_id and reg.region_name = '" + paramValue + "' ORDER BY emp.employee_id ASC"; } else if (treeNodeType.contains("regionsFindAll_bc_countriesList_1")) { queryString = "select * from employees emp INNER JOIN departments dept \n" + "ON emp.department_id = dept.department_id JOIN locations loc \n" + "ON dept.location_id = loc.location_id JOIN countries cont \n" + "ON loc.country_id = cont.country_id and cont.country_name = '" + paramValue + "' ORDER BY emp.employee_id ASC"; } else if (treeNodeType.contains("regionsFindAll_bc_locationsList_1")) { queryString = "select * from employees emp INNER JOIN departments dept ON emp.department_id = dept.department_id JOIN locations loc ON dept.location_id = loc.location_id and loc.city = '" + paramValue + "' ORDER BY emp.employee_id ASC"; } else if (treeNodeType.trim().contains("regionsFindAll_bc_departmentsList_1")) { queryString = "select * from Employees emp INNER JOIN Departments dept ON emp.DEPARTMENT_ID = dept.DEPARTMENT_ID and dept.DEPARTMENT_NAME = '" + paramValue + "'"; } } catch (NullPointerException e) { System.out.println(e.getMessage()); } return em.createNativeQuery(queryString, Employees.class).getResultList(); } In the ViewController project, create two ADF taskflow with page Fragments and name them as FirstTaskflow and SecondTaskflow respectively. Open FirstTaskflow,from component palette drop view(Page Fragment) name it as TreeList.jsff. Open SeconfTaskflow, from component palette drop view(Page Fragment) name it as EmpList.jsff and create two paramters in its overview parameters tab as shown in below image. Open TreeList.jsff , from data control palette drop regionsFindAll->Tree as ADF Tree. In Edit Tree Binding dialog, for Tree Level Rules select the display attributes as follows:-model.Regions - regionNamemodel.Countries - countryNamemodel.Locations - citymodel.Departments - departmentName In structure panel, click on af:Tree - t1 and select selectionListener with edit property. Create a "TreeBean" managed bean with scope as "session" as shown in below Image. Create new method as getTreeNodeSelectedValue and click ok. Open TreeBean managed bean and add the below code: private String treeNodeType; private String paramValue; public void getTreeNodeSelectedValue(SelectionEvent selectionEvent) { RichTree tree = (RichTree)selectionEvent.getSource(); RowKeySet addedSet = selectionEvent.getAddedSet(); Iterator i = addedSet.iterator(); TreeModel model = (TreeModel)tree.getValue(); model.setRowKey(i.next()); JUCtrlHierNodeBinding node = (JUCtrlHierNodeBinding)tree.getRowData(); //oracle.jbo.Row Row rw = node.getRow(); Object selectedTreeNode = node.getAttribute(0); Object treeListType = node.getBindings(); String treeNodeType = treeListType.toString(); this.setParamValue(selectedTreeNode.toString()); this.setTreeNodeType(treeNodeType); } public void setTreeNodeType(String treeNodeType) { this.treeNodeType = treeNodeType; } public String getTreeNodeType() { return treeNodeType; } public void setParamValue(String paramValue) { this.paramValue = paramValue; } public String getParamValue() { return paramValue; }<br /> Open EmpList.jsff , from data control palette drop empFilteredByTreeNode->Employees->Table as ADF Read-only Table. After selecting the  Employees result set, in Edit Action Binding dialog window pass the pageFlowScope parameters as shown in below Image. In empList.jsff page, click Binding tab and click on Create Executable binding and select Invoke action and follow as shown in below image. Edit executeEmpFiltered invoke action properties and set the Refresh to ifNeeded, So when ever the page needs the method will be executed. Create Main.jspx page with page template as Oracle Three Column Layout. Drop FirstTaskflow as Region in start facet and drop SecondTaskflow as Region in center facet, Edit task Flow Binding dialog window pass the Input Paramters as shown in below Image. Run the Main.jspx, tree will be displayed in left region and emp details will displyaed on the right region. Click on the Americas in tree node, all emp related to the Americas related will be displayed. Click on Americas->United States of America->South San Francisco->Accounting, only employee belongs to the Accounting department will be displayed.

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  • SQL query: how to translate IN() into a JOIN?

    - by tangens
    I have a lot of SQL queries like this: SELECT o.Id, o.attrib1, o.attrib2 FROM table1 o WHERE o.Id IN ( SELECT DISTINCT Id FROM table1, table2, table3 WHERE ... ) These queries have to run on different database engines (MySql, Oracle, DB2, MS-Sql, Hypersonic), so I can only use common SQL syntax. Here I read, that with MySql the IN statement isn't optimized and it's really slow, so I want to switch this into a JOIN. I tried: SELECT o.Id, o.attrib1, o.attrib2 FROM table1 o, table2, table3 WHERE ... But this does not take into account the DISTINCT keyword. Question: How do I get rid of the duplicate rows using the JOIN approach?

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  • Does CakePHP treat all INT fields as ID's for join tables?

    - by Jonnie
    I am trying to save a User, their Profile, and some tags and my join table that links the profile and the tags keeps getting messed up. The profile model is called Instructor, the tag model is called Subject. The Instructor has a phone number and a zip code and for some reason CakePHP thinks these are the fields it should use when creating entries in my join table. My Join table always comes out as: id | instructor_id | subject_id | 1 | 90210 | 1 | // thinks that the zip code is an instructor_id 2 | 1112223333 | 1 | // thinks that the phone number is an instructor_id 3 | 1 | 1 | // thinks that user_id is an instructor_id 4 | 1 | 1 | // the actual instructor_id, this one's correct 5 | 90210 | 2 | 6 | 1112223333 | 2 | 3 | 1 | 2 | 4 | 1 | 2 | My Models: class Instructor extends AppModel { var $name = 'Instructor'; var $belongsTo = array('User', 'State'); var $hasAndBelongsToMany = array( 'Subject' = array( 'className' = 'Subject', 'joinTable' = 'instructors_subjects', 'foreignKey' = 'instructor_id', 'associationForeignKey' = 'subject_id', 'unique' = true, 'conditions' = '', 'fields' = '', 'order' = '', 'limit' = '', 'offset' = '', 'finderQuery' = '', 'deleteQuery' = '', 'insertQuery' = '' ) ); } class Subject extends AppModel { var $name = 'Subject'; var $hasAndBelongsToMany = array( 'Instructor' = array( 'className' = 'Instructor', 'joinTable' = 'instructors_subjects', 'foreignKey' = 'subject_id', 'associationForeignKey' = 'instructor_id', 'unique' = true, 'conditions' = '', 'fields' = '', 'order' = '', 'limit' = '', 'offset' = '', 'finderQuery' = '', 'deleteQuery' = '', 'insertQuery' = '' ) ); } My Model Associations: User hasOne Instructor Instructor belongsTo User Instructor hasAndBelongsToMany Subject Subject hasAndBelongsToMany Instructor My form data looks like: Array ( [User] = Array ( [username] = MrInstructor [password] = cddb06c93c72f34eb9408610529a34645c29c55d [group_id] = 2 ) [Instructor] = Array ( [name] = Jimmy Bob [email] = [email protected] [phone] = 1112223333 [city] = Beverly Hills [zip_code] = 90210 [states] = 5 [website] = www.jimmybobbaseballschool.com [description] = Jimmy Bob is an instructor. [user_id] = 1 [id] = 1 ) [Subject] = Array ( [name] = hitting, pitching ) ) My function for processing the form looks like: function instructor_register() { $this-set('groups', $this-User-Group-find('list')); $this-set('states', $this-User-Instructor-State-find('list')); if (!empty($this-data)) { // Set the group to Instructor $this-data['User']['group_id'] = 2; // Save the user data $user = $this-User-save($this-data, true, array( 'username', 'password', 'group_id' )); // If the user was saved, save the instructor's info if (!empty($user)) { $this-data['Instructor']['user_id'] = $this-User-id; $instructor = $this-User-Instructor-save($this-data, true, array( 'user_id', 'name', 'email', 'phone', 'city', 'zip_code', 'state_id', 'website', 'description' )); // If the instructor was saved, save the rest if(!empty($instructor)) { $instructorId = $this-User-Instructor-id; $this-data['Instructor']['id'] = $instructorId; // Save each subject seperately $subjects = explode(",", $this-data['Subject']['name']); foreach ($subjects as $_subject) { // Get the correct subject format $_subject = strtolower(trim($_subject)); $this-User-Instructor-Subject-create($this-data); $this-User-Instructor-Subject-set(array( 'name' = $_subject )); $this-User-Instructor-Subject-save(); echo ''; print_r($this-data); echo ''; } } } } }

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  • how to add column in SQL Query that incl. LEFT OUTER JOIN

    - by radbyx
    I have this Query: SELECT p.ProductName, dt.MaxTimeStamp, p.Responsible FROM Product p LEFT JOIN (SELECT ProductID, MAX(TimeStamp) AS MaxTimeStamp FROM StateLog WHERE State = 0 GROUP BY ProductID, Status) dt ON p.ProductID = dt.ProductID ORDER BY p.ProductName; It works like it should, but now I need to SELECT "State" out too. The tricky part is, that I only want the lastest "TimeStamp" where "State" was false. But now I also need the "State" for the lastest "TimeStamp". I tried this: SELECT p.ProductName, dt.State, dt.MaxTimeStamp, p.Responsible FROM Product p LEFT JOIN (SELECT ProductID, MAX(TimeStamp) AS MaxTimeStamp, State FROM StateLog WHERE State = 0 GROUP BY ProductID, Status) dt ON p.ProductID =dt.ProductID ORDER BY p.ProductName; But it didn't work, because it gave me the "State" for the lastest "TimeStamp". So I hope there is some clever heads out there that can help me. I'm guessing that this is either very simple or very hard to solve.

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  • [Seamless Handover of WiFi] Can we program to automatically join the WiFi with the most strongest si

    - by Dobby
    Hi all, I am making a iphone program, I wish that: it can sense and display the current signals of wifi, with strength, - as I checked, this cannot be done by official SDK, but there were some private library that can handle. 2. i wish it can always automatically join the best wifi with the most highest strength ( assume that all wifi APs have no password, and all are not remembered into the used-list inside iphone), which is something like "seamless handover"... I just guess that, there should be some APIs to let the system join one designated wifi signal from the list, and wish someone could help me. I would like to appreciate a lot : ) Thanks a lot!

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  • Does INNER JOIN performance depends on order of tables?

    - by Kartic
    A question suddenly came to my mind while I was tuning one stored procedure. Let me ask it - I have two tables, table1 and table2. table1 contains huge data and table2 contains less data. Is there performance-wise any difference between these two queries(I am changing order of the tables)? Query1: SELECT t1.col1, t2.col2 FROM table1 t1 INNER JOIN table2 t2 ON t1.col1=t2.col2 Query2: SELECT t1.col1, t2.col2 FROM table2 t2 INNER JOIN table1 t1 ON t1.col1=t2.col2 We are using Microsoft SQL server 2005.

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  • Split a string and join it back together in a different order?

    - by Xaisoft
    What is the most concise, yet readable way to split a string and put join it back together in a different order. For example, I want to split the following string: 10-20-30-4000-50000 and I would do this via: string[] tokens = original.Split('-'); and now I want to put it back together in this order: 30-20-10-4000-50000 I know I can use Join to put it back together in it's original form, but I don't want that. The only thing I can think of right now is: string modified = string.Format("{0}{1}{2}{3}{4}",tokens[2],tokens[1],tokens[0],tokens[3], tokens[4]); I realized that if I do: string modified = string.Format("{2}{1}{0}{3}{4}", tokens); it does not keep the dashes which is what I want so is to do that, should I just do: string modified = string.Format("{2}-{1}-{0}-{3}-{4}", tokens);

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  • how to join on varchar(32) and binary(16) columns in sybase?

    - by Paul Sanwald
    I want to join two tables on a UUID. table A's UUID is represented as varchar(32). table B's UUID is represented as binary(16). what's the best way to join a varchar to a binary column? I've tried using some sybase functions for this, but I'm getting different results and unsure of why: select hextobigint('0x000036ca4c4c11d88b8dcd1344cdb512') 3948051912944290701 select convert(bigint,0x000036ca4c4c11d88b8dcd1344cdb512) -2877434794219274240 what am I missing about convert and hextobigint? I must be misundstanding at least one of these functions. thanks for your help!

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  • Is there a registry key change that will by-pass the Windows Domain Join Welcome page?

    - by user1256194
    I'm scripting some Windows Server 2008 R2 builds using Power Shell. Some software needs to be installed after the server has joined the domain. Since I want to automate everything, I'm looking to by-pass the domain controllers Welcome page using a registry hack script. I work for a large company and the Active Directory people are unwilling to change group policy. I figure if it's a registry key I can script the change, install the software, replace the key and reboot as the final step. Is there a registry key change that will by-pass the Domain Join Welcome page?

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  • Can not join comp to the domain... greyed out

    - by Logman
    I have an old WinXP Pro SP3 computer I need to join to the domain, simple right? not really. When I go to control panel - system - computer name and click on CHANGE ("rename this computer") everything is greyed out. I can not set it from workgroup to a domain. I am logged on locally as an admin. (Builtin account and one I created) I have checked local policy (gpedit.msc) on the comp, but it feels like a needle in the haystack. I could probably reload an image faster than trying to fix this...but I am curious so I post here to see if anyone knows of it/fix. I tried reseting the policy to defaults, but no luck: secedit /configure /cfg %windir%\repair\secsetup.inf /db secsetup.sdb /verbose EDIT:

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  • Is there any trick to join and use Windows 8/8.1 with Samba 4 (4.1.6)?

    - by tenshimsm
    It seems that Samba doesn't like at all. I've followed various tutorials and I can't get Windows 8 to work properly with a Ubuntu Server as domain controller. This week i've downloaded ubuntu 14.04 lts and set a fast domain configuration. As usual all other Windows version (XP and 7) work but the newest M$ nightmare doesn't. In this try it doesn't even join the domain, keeps saying the my username or password are wrong. My /etc/samba/smb.conf # Global parameters [global] workgroup = DOMAIN realm = DOMAIN.LAN netbios name = DOM server role = active directory domain controller dns forwarder = 8.8.8.8 idmap_ldb:use rfc2307 = yes [netlogon] path = /var/lib/samba/sysvol/domain.lan/scripts read only = No [sysvol] path = /var/lib/samba/sysvol read only = No [test] directory mode = 0750 path = /SHARES/test read only = no Does anyone have a tutorial that really works? Because I've tried many, each one with different configurations that works only with the people that made them. And is there a way to import my old AD users, computers and ID in a way that I won't need to rejoin all computers?

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  • LINQ count query returns a 1 instead of a 0

    - by user335810
    I have the following view:- CREATE VIEW tbl_adjudicator_result_view AS SELECT a.adjudicator_id, sar.section_adjudicator_role_id, s.section_id, sdr.section_dance_role_id, d.dance_id, c.contact_id, ro.round_id, r.result_id, c.title, c.first_name, c.last_name, d.name, r.value, ro.type FROM tbl_adjudicator a INNER JOIN tbl_section_adjudicator_role sar on sar.section_adjudicator_role2adjudicator = a.adjudicator_id INNER JOIN tbl_section s on sar.section_adjudicator_role2section = s.section_id INNER JOIN tbl_section_dance_role sdr on sdr.section_dance_role2section = s.section_id INNER JOIN tbl_dance d on sdr.section_dance_role2dance = d.dance_id INNER JOIN tbl_contact c on a.adjudicator2contact = c.contact_id INNER JOIN tbl_round ro on ro.round2section = s.section_id LEFT OUTER JOIN tbl_result r on r.result2adjudicator = a.adjudicator_id AND r.result2dance = d.dance_id When I run the following query directly against the db I get 0 in the count column where there is no result select adjudicator_id, first_name, COUNT(result_id) from tbl_adjudicator_result_view arv where arv.round_id = 16 group by adjudicator_id, first_name However when I use LINQ query I always get 1 in the Count Column var query = from arv in db.AdjudicatorResultViews where arv.round_id == id group arv by new { arv.adjudicator_id, arv.first_name} into grp select new AdjudicatorResultViewGroupedByDance { AdjudicatorId = grp.Key.adjudicator_id, FirstName = grp.Key.first_name, Count = grp.Select(p => p.result_id).Distinct().Count() }; What do I need to change in the View / Linq query.

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