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  • JPA : Add and remove operations on lazily initialized collection behaviour ?

    - by Albert Kam
    Hello, im currently trying out JPA 2 and using Hibernate 3.6.x as the engine. I have an entity of ReceivingGood that contains a List of ReceivingGoodDetail, and has a bidirectional relation. Some related codes for each entity follows : ReceivingGood.java @OneToMany(mappedBy="receivingGood", targetEntity=ReceivingGoodDetail.class, fetch=FetchType.LAZY, cascade = CascadeType.ALL) private List<ReceivingGoodDetail> details = new ArrayList<ReceivingGoodDetail>(); public void addReceivingGoodDetail(ReceivingGoodDetail receivingGoodDetail) { receivingGoodDetail.setReceivingGood(this); } void internalAddReceivingGoodDetail(ReceivingGoodDetail receivingGoodDetail) { this.details.add(receivingGoodDetail); } public void removeReceivingGoodDetail(ReceivingGoodDetail receivingGoodDetail) { receivingGoodDetail.setReceivingGood(null); } void internalRemoveReceivingGoodDetail(ReceivingGoodDetail receivingGoodDetail) { this.details.remove(receivingGoodDetail); } @ManyToOne @JoinColumn(name = "receivinggood_id") private ReceivingGood receivingGood; ReceivingGoodDetail.java : public void setReceivingGood(ReceivingGood receivingGood) { if (this.receivingGood != null) { this.receivingGood.internalRemoveReceivingGoodDetail(this); } this.receivingGood = receivingGood; if (receivingGood != null) { receivingGood.internalAddReceivingGoodDetail(this); } } In my experiements with both of these entities, both adding the detail to the receivingGood's collection, and even removing the detail from the receivingGood's collection, will trigger a query to fill the collection before doing the add or remove. This assumption is based on my experiments that i will paste below. My concern is that : is it ok to do changes on only a little bit of records on the collection, and the engine has to query all of the details belonging to the collection ? What if the collection would have to be filled with 1000 records when i just want to edit a single record ? Here are my experiments with the output as the comment above each method : /* Hibernate: select receivingg0_.id as id9_14_, receivingg0_.creationDate as creation2_9_14_, ... too long Hibernate: select receivingg0_.id as id10_20_, receivingg0_.creationDate as creation2_10_20_, ... too long removing existing detail from lazy collection Hibernate: select details0_.receivinggood_id as receivi13_9_8_, details0_.id as id8_, details0_.id as id10_7_, details0_.creationDate as creation2_10_7_, details0_.modificationDate as modifica3_10_7_, details0_.usercreate_id as usercreate10_10_7_, details0_.usermodify_id as usermodify11_10_7_, details0_.version as version10_7_, details0_.buyQuantity as buyQuant5_10_7_, details0_.buyUnit as buyUnit10_7_, details0_.internalQuantity as internal7_10_7_, details0_.internalUnit as internal8_10_7_, details0_.product_id as product12_10_7_, details0_.receivinggood_id as receivi13_10_7_, details0_.supplierLotNumber as supplier9_10_7_, user1_.id as id2_0_, user1_.creationDate as creation2_2_0_, user1_.modificationDate as modifica3_2_0_, user1_.usercreate_id as usercreate6_2_0_, user1_.usermodify_id as usermodify7_2_0_, user1_.version as version2_0_, user1_.name as name2_0_, user2_.id as id2_1_, user2_.creationDate as creation2_2_1_, user2_.modificationDate as modifica3_2_1_, user2_.usercreate_id as usercreate6_2_1_, user2_.usermodify_id as usermodify7_2_1_, user2_.version as version2_1_, user2_.name as name2_1_, user3_.id as id2_2_, user3_.creationDate as creation2_2_2_, user3_.modificationDate as modifica3_2_2_, user3_.usercreate_id as usercreate6_2_2_, user3_.usermodify_id as usermodify7_2_2_, user3_.version as version2_2_, user3_.name as name2_2_, user4_.id as id2_3_, user4_.creationDate as creation2_2_3_, user4_.modificationDate as modifica3_2_3_, user4_.usercreate_id as usercreate6_2_3_, user4_.usermodify_id as usermodify7_2_3_, user4_.version as version2_3_, user4_.name as name2_3_, product5_.id as id0_4_, product5_.creationDate as creation2_0_4_, product5_.modificationDate as modifica3_0_4_, product5_.usercreate_id as usercreate7_0_4_, product5_.usermodify_id as usermodify8_0_4_, product5_.version as version0_4_, product5_.code as code0_4_, product5_.name as name0_4_, user6_.id as id2_5_, user6_.creationDate as creation2_2_5_, user6_.modificationDate as modifica3_2_5_, user6_.usercreate_id as usercreate6_2_5_, user6_.usermodify_id as usermodify7_2_5_, user6_.version as version2_5_, user6_.name as name2_5_, user7_.id as id2_6_, user7_.creationDate as creation2_2_6_, user7_.modificationDate as modifica3_2_6_, user7_.usercreate_id as usercreate6_2_6_, user7_.usermodify_id as usermodify7_2_6_, user7_.version as version2_6_, user7_.name as name2_6_ from ReceivingGoodDetail details0_ left outer join COMMON_USER user1_ on details0_.usercreate_id=user1_.id left outer join COMMON_USER user2_ on user1_.usercreate_id=user2_.id left outer join COMMON_USER user3_ on user2_.usermodify_id=user3_.id left outer join COMMON_USER user4_ on details0_.usermodify_id=user4_.id left outer join Product product5_ on details0_.product_id=product5_.id left outer join COMMON_USER user6_ on product5_.usercreate_id=user6_.id left outer join COMMON_USER user7_ on product5_.usermodify_id=user7_.id where details0_.receivinggood_id=? after removing try selecting the size : 4 after removing, now flushing Hibernate: update ReceivingGood set creationDate=?, modificationDate=?, usercreate_id=?, usermodify_id=?, version=?, purchaseorder_id=?, supplier_id=?, transactionDate=?, transactionNumber=?, transactionType=?, transactionYearMonth=?, warehouse_id=? where id=? and version=? Hibernate: update ReceivingGoodDetail set creationDate=?, modificationDate=?, usercreate_id=?, usermodify_id=?, version=?, buyQuantity=?, buyUnit=?, internalQuantity=?, internalUnit=?, product_id=?, receivinggood_id=?, supplierLotNumber=? where id=? and version=? detail size : 4 */ public void removeFromLazyCollection() { String headerId = "3b373f6a-9cd1-4c9c-9d46-240de37f6b0f"; ReceivingGood receivingGood = em.find(ReceivingGood.class, headerId); // get existing detail ReceivingGoodDetail detail = em.find(ReceivingGoodDetail.class, "323fb0e7-9bb2-48dc-bc07-5ff32f30e131"); detail.setInternalUnit("MCB"); System.out.println("removing existing detail from lazy collection"); receivingGood.removeReceivingGoodDetail(detail); System.out.println("after removing try selecting the size : " + receivingGood.getDetails().size()); System.out.println("after removing, now flushing"); em.flush(); System.out.println("detail size : " + receivingGood.getDetails().size()); } /* Hibernate: select receivingg0_.id as id9_14_, receivingg0_.creationDate as creation2_9_14_, ... too long Hibernate: select receivingg0_.id as id10_20_, receivingg0_.creationDate as creation2_10_20_, ... too long adding existing detail into lazy collection Hibernate: select details0_.receivinggood_id as receivi13_9_8_, details0_.id as id8_, details0_.id as id10_7_, details0_.creationDate as creation2_10_7_, details0_.modificationDate as modifica3_10_7_, details0_.usercreate_id as usercreate10_10_7_, details0_.usermodify_id as usermodify11_10_7_, details0_.version as version10_7_, details0_.buyQuantity as buyQuant5_10_7_, details0_.buyUnit as buyUnit10_7_, details0_.internalQuantity as internal7_10_7_, details0_.internalUnit as internal8_10_7_, details0_.product_id as product12_10_7_, details0_.receivinggood_id as receivi13_10_7_, details0_.supplierLotNumber as supplier9_10_7_, user1_.id as id2_0_, user1_.creationDate as creation2_2_0_, user1_.modificationDate as modifica3_2_0_, user1_.usercreate_id as usercreate6_2_0_, user1_.usermodify_id as usermodify7_2_0_, user1_.version as version2_0_, user1_.name as name2_0_, user2_.id as id2_1_, user2_.creationDate as creation2_2_1_, user2_.modificationDate as modifica3_2_1_, user2_.usercreate_id as usercreate6_2_1_, user2_.usermodify_id as usermodify7_2_1_, user2_.version as version2_1_, user2_.name as name2_1_, user3_.id as id2_2_, user3_.creationDate as creation2_2_2_, user3_.modificationDate as modifica3_2_2_, user3_.usercreate_id as usercreate6_2_2_, user3_.usermodify_id as usermodify7_2_2_, user3_.version as version2_2_, user3_.name as name2_2_, user4_.id as id2_3_, user4_.creationDate as creation2_2_3_, user4_.modificationDate as modifica3_2_3_, user4_.usercreate_id as usercreate6_2_3_, user4_.usermodify_id as usermodify7_2_3_, user4_.version as version2_3_, user4_.name as name2_3_, product5_.id as id0_4_, product5_.creationDate as creation2_0_4_, product5_.modificationDate as modifica3_0_4_, product5_.usercreate_id as usercreate7_0_4_, product5_.usermodify_id as usermodify8_0_4_, product5_.version as version0_4_, product5_.code as code0_4_, product5_.name as name0_4_, user6_.id as id2_5_, user6_.creationDate as creation2_2_5_, user6_.modificationDate as modifica3_2_5_, user6_.usercreate_id as usercreate6_2_5_, user6_.usermodify_id as usermodify7_2_5_, user6_.version as version2_5_, user6_.name as name2_5_, user7_.id as id2_6_, user7_.creationDate as creation2_2_6_, user7_.modificationDate as modifica3_2_6_, user7_.usercreate_id as usercreate6_2_6_, user7_.usermodify_id as usermodify7_2_6_, user7_.version as version2_6_, user7_.name as name2_6_ from ReceivingGoodDetail details0_ left outer join COMMON_USER user1_ on details0_.usercreate_id=user1_.id left outer join COMMON_USER user2_ on user1_.usercreate_id=user2_.id left outer join COMMON_USER user3_ on user2_.usermodify_id=user3_.id left outer join COMMON_USER user4_ on details0_.usermodify_id=user4_.id left outer join Product product5_ on details0_.product_id=product5_.id left outer join COMMON_USER user6_ on product5_.usercreate_id=user6_.id left outer join COMMON_USER user7_ on product5_.usermodify_id=user7_.id where details0_.receivinggood_id=? after adding try selecting the size : 5 after adding, now flushing Hibernate: update ReceivingGood set creationDate=?, modificationDate=?, usercreate_id=?, usermodify_id=?, version=?, purchaseorder_id=?, supplier_id=?, transactionDate=?, transactionNumber=?, transactionType=?, transactionYearMonth=?, warehouse_id=? where id=? and version=? detail size : 5 */ public void editLazyCollection() { String headerId = "3b373f6a-9cd1-4c9c-9d46-240de37f6b0f"; ReceivingGood receivingGood = em.find(ReceivingGood.class, headerId); // get existing detail ReceivingGoodDetail detail = em.find(ReceivingGoodDetail.class, "323fb0e7-9bb2-48dc-bc07-5ff32f30e131"); detail.setInternalUnit("MCB"); System.out.println("adding existing detail into lazy collection"); receivingGood.addReceivingGoodDetail(detail); System.out.println("after adding try selecting the size : " + receivingGood.getDetails().size()); System.out.println("after adding, now flushing"); em.flush(); System.out.println("detail size : " + receivingGood.getDetails().size()); } Please share your experience on this matter ! Thank you !

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  • Mac has IP address, can connect to router but can't connect outside

    - by partition
    Weird problem, my MacBook can't connect anywhere right now! The router works, it gets an IP, it can log into to the router but it can't resolve anything! The router works as I connected another device to it and it connected to the net. The MacBook doesn't have any strange DNS configurations either, just 192.168.1.1 for the router I even tried tethering it to my phone, and it still would not connect to the net... help?

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  • Can PetaPoco populate an object using a stored procedure with a join clause?

    - by Mark Kadlec
    I have a stored procedure that does something similar to: SELECT a.TaskId, b.CompanyCode FROM task a JOIN company b ON b.CompanyId = a.CompanyId; I have an object called TaskItem that has the TaskId and CompanyCode properties, but when I execute the following (which I would have assumed worked): var masterDatabase = new Database("MasterConnectionString"); var s = PetaPoco.Sql.Builder.Append("EXEC spGetTasks @@numberOfTasks = @0", numberOfTasks); var tasks = masterDatabase.Query<Task>(s); The problem is that the CompanyCode column does not exist in the task table, I did a trace and it seems that PetaPoco is trying to select all the properties from the task table and populating using the stored procedure. How can I use PetaPoco to simply populate the list of task objects with the results of the stored procedure?

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  • how to join a set of XElements to the values of a struct?

    - by jcollum
    Let's say I have a struct that contains local environments: public struct Environments { public const string Dev = "DEV"; public const string Qa1 = "SQA"; public const string Prod1 = "PROD"; public const string Prod2 = "PROD_SA"; public const string Uat = "UAT"; } And I'd like to pull a set of XElements out of an xml doc, but only those elements that have a key that matches a value in a struct. this.environments =(from e in settings.Element("Settings").Element("Environments") .Elements("Environment") .Where( x => x.HasAttribute("name") ) join f in [struct?] on e.Attribute("name") equals [struct value?]).ToDictionary(...) How would I go about doing this? Do I need reflection to get the values of the constants in the struct?

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  • Rails - How do i update a records value in a join model ?

    - by ChrisWesAllen
    I have a join model in a HABTM relationship with a through association (details below). I 'm trying to find a record....find the value of that records attribute...change the value and update the record but am having a hard time doing it. The model setup is this User.rb has_many :choices has_many :interests, :through => :choices Interest.rb has_many :choices has_many :users, :through => :choices Choice.rb belongs_to :user belongs_to :interest and Choice has the user_id, interest_id, score as fields. And I find the ?object? like so @choice = Choice.where(:user_id => @user.id, :interest_id => interest.id) So the model Choice has an attribute called :score. How do I find the value of the score column....and +1/-1 it and then resave? I tried @choice.score = @choice.score + 1 @choice.update_attributes(params[:choice]) flash[:notice] = "Successfully updated choices value." but I get "undefined method score"......What did i miss?

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  • Join DELETE in MySql? How to?

    - by Camran
    I have this: $query="DELETE FROM classified, $sql_table WHERE classified.ad_id = '$id' AND classified.classified_id = $sql_table.classified_id AND classified.poster_password='$pass'"; I get this error: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'WHERE classified.ad_id = 'Bmw_M3_E46_Full_29920' AND classified.cla' at line 1 Any help? As you can see the $sql_table is linked to the classifieds table with the fields classified_id I need to JOIN DELETE somehow. Basically classified table is the main table, then every category has its own tables with vehicle data. classified table has a field called classified_id which is the same as the Here is the full query echoed: DELETE FROM classified, vehicles WHERE classified.ad_id = 'Bmw_M3_E46_410811305' AND classified.classified_id = vehicles.classified_id AND classified.poster_password='some_password' Why isn't this working, Should it be so hard to delete from multiple tables? Thanks

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  • Linux - How do i know the block map of the given file and/or the free space map of the partition.

    - by Inso Reiges
    Hello, I am on Linux and need to know either of the two things: 1) If i have a regular file on some file system on a partition under Linux is there a way to know the set of the physical blocks that this file occupies on the drive from user space? Or at least the set of the file system's clusters? 2) Is there a way to get the same information about the whole free space of the given file system? In both cases i understand that if there is any possible way to extract this info it will probably be totally unsafe and racy (anything could happen to these set of blocks between the time i see them and act on them somehow). I also really don't want an implementation that will have to know a lot about every filesystem.

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  • How to perform a join with CodeIgniter's Active Record class on a multicolumn key?

    - by Scott Southworth
    I've been able to make this code work using CodeIgniter's db->query as follows: $sql = 'SELECT mapping_code,zone_name,installation_name FROM installations,appearances,zones WHERE installations.installation_id = zones.installation_fk_id AND appearances.installation_fk_id = installations.installation_id AND appearances.zone_fk_id = zones.zone_id AND appearances.barcode = ? '; return $this->db->query($sql,array($barcode)); The 'appearances' table throws a 'not unique table' error if I try this using the Active Record class. I need to join appearances on both the zone and installations tables. How can I do this?

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  • Can I join between two MySQL tables stores on separate machines?

    - by CuriousCoder
    I have a relatively light query that needs information from a local MySQL table along with another MySQL table which is stored on a physically separate machine (on the same network). I'm keen to avoid setting up replication just to facilitate this light query that only needs executed once a day. Is there any way that I can join with a table on a remote machine using one query? Or run a SELECT INTO into a local table. Notes I'm using C# & .NET 4.

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  • In which layer should I join 2 entities together?

    - by William
    I use Spring MVC and a regular JDBC. I've just learned that I should separate business process into layers which are presentation layer, controller layer, service layer, and repository/DAO layer. Now suppose that I have an Entity called Person that can have multiple Jobs. Job itself is another entity which have its own properties. From what I gathered, the repository layer only manages one entity. Now I have one entity that contains another entity. Where do I "join" them? The service layer? Suppose I want to get a person whose job isn't known yet (lazy loading). But the system might ask what the job of that particular person is later on. What is the role of each layer in this case? Please let me know if I need to add any detail into this question.

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  • Can I join 2+ styles together into a superstyle?

    - by Kurru
    Hi I was looking to join 2 styles together to make a super style for easy use and customisation of my page. Is it possible to define something like this? (if so how) .bold { font-weight: bold;} .color1 {color: white;} .boldColor {.bold; .color1;} where .boldColor is effectively .boldColor {font-weight:bold; color:white;} I want this so that I can have styles thoughout the page and be able to easily change the colors in many places in 1 place. I'm currently using <p class="bold color"> but some of my class defs are becoming long so I'd like to be able to use <p class="boldColor"> Thanks

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  • Can anyone help me with a complex sum, 3 table join mysql query?

    - by Scarface
    Hey guys I have a query and it works fine, but I want to add another table to the mix. The invite table I want to add has two fields: username and user_invite. Much like this site, I am using a point system to encourage diligent users. The current query which is displayed below adds the up votes and down votes based on the user in question: $creator. I want to count the number of entries for that same user from the invite table, and add 50 for each row it finds to the current output/sum of my query. Is this possible with one query, or do I need two? "SELECT *, SUM(IF(points_id = \"1\", 1,0))-SUM(IF(points_id = \"2\", 1,0)) AS 'total' FROM points LEFT JOIN post ON post.post_id=points.points_id WHERE post.creator='$creator'"

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  • How to deal with the Hibernate hql multi-join query result in an Object-Oriented Way?

    - by EugeneP
    How to deal with the Hibernate hql multi-join query result in an Object-Oriented Way? As I see it returns a list of Objects. yes, it is tricky and only you who write the query know what should the query return (what objects). But are there ways to simplify things, so that it returned specific objects with no need in casting Object to a specific class according to its position in the query ? Maybe Spring can simplify things here? It has the similar functionality for JDBC, but I don't see if it can help in a similar way with Hibernate.

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  • Any Way to Use a Join in a Lambda Where() on a Table<>?

    - by lush
    I'm in my first couple of days using Linq in C#, and I'm curious to know if there is a more concise way of writing the following. MyEntities db = new MyEntities(ConnString); var q = from a in db.TableA join b in db.TableB on a.SomeFieldID equals b.SomeFieldID where (a.UserID == CurrentUser && b.MyField == Convert.ToInt32(MyDropDownList.SelectedValue)) select new { a, b }; if(q.Any()) { //snip } I know that if I were to want to check the existence of a value in the field of a single table, I could just use the following: if(db.TableA.Where(u => u.UserID == CurrentUser).Any()) { //snip } But I'm curious to know if there is a way to do the lambda technique, but where it would satisfy the first technique's conditions across those two tables. Sorry for any mistakes or clarity, I'll edit as necessary. Thanks in advance.

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  • Booting Fedora guest VBox on /dev/mapper/vg0-fc17-root

    - by NevilleDNZ
    I already have the following logical volumes: host:/dev/mapper/vg0-fc17-boot (guestOS:/dev/hdb) formatted as ext4 (no partition table) host:/dev/mapper/vg0-fc17-root (guestOS:/dev/hdc) formatted as ext4 (no partition table) Do I have to create the following grub partition to boot a guest VM under VirtualBox? host:/dev/mapper/vg-fc17-mbr (guestOS:/dev/hda) with a partition table and install grub MBR here? Or is there a better way? (Maybe grub on vg0-fc17-boot?)

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  • Exchange 2007 Standard Edition

    - by Phrontiste
    We Have : Exchange 2007 Standard Edition IBM System X3650 2 x Intel Xeon 5430 2.66 GHz Version 8.1 Build 240.6 Mailbox, Hub Transport, Client Access Role Installed on One Box Total Number of Mailboxes : 110 - 130 6 Physical Disks Disk 0,1 (68 GB) = Raid-1, OS Partition ( C: Partition) Disk 2,3 (279GB) = Raid-1, Exchange Database (First and Second Storage Groups) ( D: Partition ) Disk 4,5 (68 GB) = Raid-1, Exchange Transaction Logs ( E: Partition ) Setup: Storage Groups : D:\First Storage group\Mailbox database.edb Storage Groups : D:\Second Storage Group\Public Folder Database.edb Transaction Logs : E Partition Problem 1: On our D Partition (Mailbox Database Partition), total size is 279 GB, free space remaining is 64.7 GB, when I select the first storage group and second storage group folders and right click properties they report a size of 165 GB. Mailbox database reports a size of 157GB when right clicked Properties. where as the size displayed in the folder is 164,893,456 KB So, we are missing around 50-54 GB, there is nothing else on these drives, no page file, nothing at all. The partition housing the Transaction logs is reporting the sizes accurately. Any suggestions / fixes on the above ? Problem 2: As you may have already read in Problem 1, the size of the mailbox database is 157GB or 164GB reported; which is not recommended, a) What would you suggest we should do to divide mailboxes in storage groups on this same server ? b) How would we move mailboxes into different storage groups ? c) This is the information store size ? (Am I right in thinking that this is not recommended) d) Having multiple storage groups with one Mailbox DB in each, would that reduce the size of the Information Store? e) Any suggestions / how-to reduce the size of information store ? We didn't install this, we have inherited this - what other recommendations you can make in order to keep ourselves better prepared for any server disaster? We are backing up with Yosemite Backup on RD1000 (320GB) at the moment, which is backing up successfully, flushing the logs daily. We haven't done a test restore YET. I have tried to provide as much info as possible, please let me know if you need further info. Also, we haven't yet faced any problems in mailflow, access speeds, everything is working fine, we have two to five people accessing OWA or Outlook via vpn only. Thanks for your time to read the above - will look forward to your expert suggestions.

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  • Windows 7 Home Premium Disk Partitioning

    - by Tamir
    Hi all, I'm having new Dell studio 1749 laptop with one partition (C). there is another backup partition - hidden. How can I create new partition for all the files and the other stuff to be seperated from the C partition? I'm looking for a clean and simple way to do it, thanks!

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  • Windows 7 Home Premium x64 Disk Partitioning

    - by Tamir
    Hi all, I'm having new Dell studio 1749 laptop with one partition (C). there is another backup partition - hidden. How can I create new partition for all the files and the other stuff to be seperated from the C partition? I'm looking for a clean and simple way to do it, thanks!

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  • Convert MBR to GPT - Without any OS

    - by Sourabh
    I just got a new laptop without any OS (it had FreeDOS, but not anymore). When I go to Windows Installer and try to create a new partition from un-allocated space, I don't get any Error message but the installer is unable to create the partition. At the bottom of the installer window, there's a warning which says something like, Windows cannot install on MBR *partition. On EFI systems, Windows can only install on GPT *partition How can I convert MBR to GPT without any OS?

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  • rdisk value in boot.ini maps to which disk?

    - by MA1
    Hi All Following are the contents of a sample boot.ini: [boot loader] timeout=30 default=multi(0)disk(0)rdisk(0)partition(1)\WINDOWS [operating systems] multi(0)disk(0)rdisk(0)partition(1)\WINDOWS="Microsoft Windows XP Professional" /NOEXECUTE=OPTIN /FASTDETECT multi(0)disk(0)rdisk(0)partition(2)\WINNT="Windows 2000 Professional" /fastdetect multi(0)disk(0)rdisk(1)partition(1)\WINDOWS="Microsoft Windows XP Home Edition" /NOEXECUTE=OPTIN /FASTDETECT rdisk value tells the physical disk number. so, if i have three hard disks say: /dev/sda /dev/sdb /dev/sdc than how to know which disk(/dev/sda or /dev/sdb or /dev/sdc) is rdisk(0) and which disk is rdisk(1) etc Regards,

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  • Not your usual "The multi-part identifier could not be bound" error

    - by Eugene Niemand
    I have the following query, now the strange thing is if I run this query on my development and pre-prod server it runs fine. If I run it on production it fails. I have figured out that if I run just the Select statement its happy but as soon as I try insert into the table variable it complains. DECLARE @RESULTS TABLE ( [Parent] VARCHAR(255) ,[client] VARCHAR(255) ,[ComponentName] VARCHAR(255) ,[DealName] VARCHAR(255) ,[Purchase Date] DATETIME ,[Start Date] DATETIME ,[End Date] DATETIME ,[Value] INT ,[Currency] VARCHAR(255) ,[Brand] VARCHAR(255) ,[Business Unit] VARCHAR(255) ,[Region] VARCHAR(255) ,[DealID] INT ) INSERT INTO @RESULTS SELECT DISTINCT ClientName 'Parent' ,F.ClientID 'client' ,ComponentName ,A.DealName ,CONVERT(SMALLDATETIME , ISNULL(PurchaseDate , '1900-01-01')) 'Purchase Date' ,CONVERT(SMALLDATETIME , ISNULL(StartDate , '1900-01-01')) 'Start Date' ,CONVERT(SMALLDATETIME , ISNULL(EndDate , '1900-01-01')) 'End Date' ,DealValue 'Value' ,D.Currency 'Currency' ,ShortBrand 'Brand' ,G.BU 'Business Unit' ,C.DMRegion 'Region' ,DealID FROM LTCDB_admin_tbl_Deals A INNER JOIN dbo_DM_Brand B ON A.BrandID = B.ID INNER JOIN LTCDB_admin_tbl_DM_Region C ON A.Region = C.ID INNER JOIN LTCDB_admin_tbl_Currency D ON A.Currency = D.ID INNER JOIN LTCDB_admin_tbl_Deal_Clients E ON A.DealID = E.Deal_ID INNER JOIN LTCDB_admin_tbl_Clients F ON E.Client_ID = F.ClientID INNER JOIN LTCDB_admin_tbl_DM_BU G ON G.ID = A.BU INNER JOIN LTCDB_admin_tbl_Deal_Components H ON A.DealID = H.Deal_ID INNER JOIN LTCDB_admin_tbl_Components I ON I.ComponentID = H.Component_ID WHERE EndDate != '1899-12-30T00:00:00.000' AND StartDate < EndDate AND B.ID IN ( 1 , 2 , 5 , 6 , 7 , 8 , 10 , 12 ) AND C.SalesRegionID IN ( 1 , 3 , 4 , 11 , 16 ) AND A.BU IN ( 1 , 2 , 3 , 4 , 5 , 6 , 8 , 9 , 11 , 12 , 15 , 16 , 19 , 20 , 22 , 23 , 24 , 26 , 28 , 30 ) AND ClientID = 16128 SELECT ... FROM @Results I get the following error Msg 8180, Level 16, State 1, Line 1 Statement(s) could not be prepared. Msg 4104, Level 16, State 1, Line 1 The multi-part identifier "Tbl1021.ComponentName" could not be bound. Msg 4104, Level 16, State 1, Line 1 The multi-part identifier "Tbl1011.Currency" could not be bound. Msg 207, Level 16, State 1, Line 1 Invalid column name 'Col2454'. Msg 207, Level 16, State 1, Line 1 Invalid column name 'Col2461'. Msg 207, Level 16, State 1, Line 1 Invalid column name 'Col2491'. Msg 207, Level 16, State 1, Line 1 Invalid column name 'Col2490'. Msg 207, Level 16, State 1, Line 1 Invalid column name 'Col2482'. Msg 207, Level 16, State 1, Line 1 Invalid column name 'Col2478'. Msg 207, Level 16, State 1, Line 1 Invalid column name 'Col2477'. Msg 207, Level 16, State 1, Line 1 Invalid column name 'Col2475'. EDIT - EDIT - EDIT - EDIT - EDIT - EDIT through a process of elimination I have found that following and wondered if anyone can shed some light on this. If I remove only the DISTINCT the query runs fine, add the DISTINCT and I get strange errors. Also I have found that if I comment the following lines then the query runs with the DISTINCT what is strange is that none of the columns in the table LTCDB_admin_tbl_Deal_Components is referenced so I don't see how the distinct will affect that. INNER JOIN LTCDB_admin_tbl_Deal_Components H ON A.DealID = H.Deal_ID

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  • TSQL -- Make it better

    - by user319353
    Hi: -- Very Narrow (all IDs are passed in) IF(@EmpID IS NOT NULL AND @DeptID IS NOT NULL AND @CityID IS NOT NULL) BEGIN SELECT e.EmpName ,d.DeptName ,c.CityName FROM Employee e WITH (NOLOCK) JOIN Department d WITH (NOLOCK) ON e.deptid = d.deptid JOIN City c WITH (NOLOCK) ON e.CityID = c.CityID WHERE e.EmpID = @EmpID END -- Just 2 IDs passed in ELSE IF(@DeptID IS NOT NULL AND @CityID IS NOT NULL) BEGIN SELECT e.EmpName ,d.DeptName ,NULL AS [CityName] FROM Employee e WITH (NOLOCK) JOIN Department d WITH (NOLOCK) ON e.deptid = d.deptid JOIN City c WITH (NOLOCK) ON e.CityID = c.CityID WHERE d.deptID = @DeptID END -- Very Broad (just 1 ID passed in) ELSE IF(@CityID IS NOT NULL) BEGIN SELECT e.EmpName ,NULL AS [DeptName] ,NULL AS [CityName] FROM Employee e WITH (NOLOCK) JOIN Department d WITH (NOLOCK) ON e.deptid = d.deptid JOIN City c WITH (NOLOCK) ON e.CityID = c.CityID WHERE c.CityID = @CityID END -- None (Nothing passed in) ELSE BEGIN SELECT NULL AS [EmpName] ,NULL AS [DeptName] ,NULL AS [CityName] END Question: Is there any better way (OR specifically can I do anything without IF...ELSE condition?

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  • SQL Loop over a family tree

    - by simon831
    Using SQL server 2008. I have a family tree of animals stored in a table, and want to give some information on how 'genetically diverse' (or not) the offspring is. In SQL how can I produce sensible metrics to show how closely related the parents are? Perhaps some sort of percentage of shared blood, or a number of generations to go back before there is a shared ancestor? AnimalTable Id Name mumId dadId select * from AnimalTable child inner join AnimalTable mum on child.[mumId] = mum.[Id] inner join AnimalTable dad on child.[dadId] = dad.[Id] inner join AnimalTable mums_mum on mum.[mumId] = mums_mum.[Id] inner join AnimalTable mums_dad on mum.[dadId] = mums_dad.[Id] inner join AnimalTable dads_mum on dad.[mumId] = dads_mum.[Id] inner join AnimalTable dads_dad on dad.[dadId] = dads_dad.[Id]

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  • DB function failed with error number 1 in joomla admin panel

    - by sabuj
    When i access joomla article manager or module manager then i had faced the bellow output: 500 - An error has occurred! DB function failed with error number 1 Can't create/write to file '/tmp/#sql_57c0_0.MYD' (Errcode: 17) SQL=SELECT c.*, g.name AS groupname, cc.title AS name, u.name AS editor, f.content_id AS frontpage, s.title AS section_name, v.name AS author FROM jos_content AS c LEFT JOIN jos_categories AS cc ON cc.id = c.catid LEFT JOIN jos_sections AS s ON s.id = c.sectionid LEFT JOIN jos_groups AS g ON g.id = c.access LEFT JOIN jos_users AS u ON u.id = c.checked_out LEFT JOIN jos_users AS v ON v.id = c.created_by LEFT JOIN jos_content_frontpage AS f ON f.content_id = c.id WHERE c.state != -2 ORDER BY section_name , section_name, cc.title, c.ordering LIMIT 0, 20

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  • Access is refusing to run an query with linked table?

    - by Mahmoud
    Hey all i have 3 tables each as follow cash_credit Bank_Name-------in_date-------Com_Id---Amount America Bank 15/05/2010 1 200 HSBC 17/05/2010 3 500 Cheque_credit Bank_Name-----Cheque_Number-----in_date-------Com_Id---Amount America Bank 74835435-5435 15/05/2010 2 600 HSBC 41415454-2851 17/05/2010 5 100 Companies com_id----Com_Name 1 Ebay 2 Google 3 Facebook 4 Amazon Companies table is a linked table when i tried to create an query as follow SELECT cash_credit.Amount, Companies.Com_Name, cheque_credit.Amount FROM cheque_credit INNER JOIN (cash_credit INNER JOIN Companies ON cash_credit.com_id = Companies.com_id) ON cheque_credit.com_id = Companies.com_id; I get an error saying that my inner Join is incorrectly, this query was created using Access 2007 query design the error is Type mismatch in expression then i thought it might be the inner join so i tried Left Join and i get an error that this method is not used JOIN expression is not supported I am confused on where is the problem that is causing all this

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