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  • How do I keep a table in sync across multiple SQL Databases?

    - by Refracted Paladin
    I have a Win Form, Data Entry, application that uses 4 seperate Data Bases. This is an occasionally connected app that uses Merge Replication (SQL 2005) to stay in Sync. This is working just fine. The next hurdle I am trying to tackle is adding Filters to my Publications. Right now we are replicating 70mbs, compressed, to each of our 150 subscribers when, truthfully, they only need a tiny fraction of that. Using Filters I am able to accomplish this(see code below) but I had to make a mapping table in order to do so. This mapping table consists of 3 columns. A PrimaryID(Guid), WorkerName(varchar), and ClientID(int). The problem is I need this table present in all FOUR Databases in order to use it for the filter since, to my knowledge, views or cross-db query's are not allowed in a Filter Statement. What are my options? Seems like I would set it up to be maintained in 1 Database and then use Triggers to keep it updated in the other 3 Databases. In order to be a part of the Filter I have to include that table in the Replication Set so how do I flag it appropriately. Is there a better way, altogether? SELECT <published_columns> FROM [dbo].[tblPlan] WHERE [ClientID] IN (select ClientID from [dbo].[tblWorkerOwnership] where WorkerID = SUSER_SNAME()) Which allows you to chain together Filters, this next one is below the first one so it only pulls from the first's Filtered Set. SELECT <published_columns> FROM [dbo].[tblPlan] INNER JOIN [dbo].[tblHealthAssessmentReview] ON [tblPlan].[PlanID] = [tblHealthAssessmentReview].[PlanID] P.S. - I know how illogical the DB structure sounds. I didn't make it. I inherited it and was then told to make it a "disconnected app." Go figure!

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  • Data from a table in 1 DB needed for filter in different DB...

    - by Refracted Paladin
    I have a Win Form, Data Entry, application that uses 4 seperate Data Bases. This is an occasionally connected app that uses Merge Replication (SQL 2005) to stay in Sync. This is working just fine. The next hurdle I am trying to tackle is adding Filters to my Publications. Right now we are replicating 70mbs, compressed, to each of our 150 subscribers when, truthfully, they only need a tiny fraction of that. Using Filters I am able to accomplish this(see code below) but I had to make a mapping table in order to do so. This mapping table consists of 3 columns. A PrimaryID(Guid), WorkerName(varchar), and ClientID(int). The problem is I need this table present in all FOUR Databases in order to use it for the filter since, to my knowledge, views or cross-db query's are not allowed in a Filter Statement. What are my options? Seems like I would set it up to be maintained in 1 Database and then use Triggers to keep it updated in the other 3 Databases. In order to be a part of the Filter I have to include that table in the Replication Set so how do I flag it appropriately. Is there a better way, altogether? SELECT <published_columns> FROM [dbo].[tblPlan] WHERE [ClientID] IN (select ClientID from [dbo].[tblWorkerOwnership] where WorkerID = SUSER_SNAME()) Which allows you to chain together Filters, this next one is below the first one so it only pulls from the first's Filtered Set. SELECT <published_columns> FROM [dbo].[tblPlan] INNER JOIN [dbo].[tblHealthAssessmentReview] ON [tblPlan].[PlanID] = [tblHealthAssessmentReview].[PlanID]

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  • How do I keep a table in Sync across 4 db's to be used in SQL Replication Filtering?

    - by Refracted Paladin
    I have a Win Form, Data Entry, application that uses 4 seperate Data Bases. This is an occasionally connected app that uses Merge Replication (SQL 2005) to stay in Sync. This is working just fine. The next hurdle I am trying to tackle is adding Filters to my Publications. Right now we are replicating 70mbs, compressed, to each of our 150 subscribers when, truthfully, they only need a tiny fraction of that. Using Filters I am able to accomplish this(see code below) but I had to make a mapping table in order to do so. This mapping table consists of 3 columns. A PrimaryID(Guid), WorkerName(varchar), and ClientID(int). The problem is I need this table present in all FOUR Databases in order to use it for the filter since, to my knowledge, views or cross-db query's are not allowed in a Filter Statement. What are my options? Seems like I would set it up to be maintained in 1 Database and then use Triggers to keep it updated in the other 3 Databases. In order to be a part of the Filter I have to include that table in the Replication Set so how do I flag it appropriately. Is there a better way, altogether? SELECT <published_columns> FROM [dbo].[tblPlan] WHERE [ClientID] IN (select ClientID from [dbo].[tblWorkerOwnership] where WorkerID = SUSER_SNAME()) Which allows you to chain together Filters, this next one is below the first one so it only pulls from the first's Filtered Set. SELECT <published_columns> FROM [dbo].[tblPlan] INNER JOIN [dbo].[tblHealthAssessmentReview] ON [tblPlan].[PlanID] = [tblHealthAssessmentReview].[PlanID] P.S. - I know how illogical the DB structure sounds. I didn't make it. I inherited it and was then told to make it a "disconnected app."

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  • Excel or Access: how to group several lines in a table and insert contents in columns? ("split column")

    - by Martin
    I have a table containing data of sold products (shown in the example on the left): Columns: Number of the order Product Name Attribute - specifies what is given in the following field "value", e. g. Customer Name or Product Variant Value - is the value of the Attribute Count - is the number of products of this variant sold in the order That means: Product B has 2 variants "c" and "d" Note that in Order 1 Product B was sold in Variant d only, because the letter "N" in field "D4" means "none". Note, that in OrdnerNo 3 Product B was sold only in Variant c, because for Variant d field "D9" is "N"!! This is confusing, but it is the structure of the original data (which I can not change). I need a way to convert the table on the left in a table like that on the right: one line for each product type Order Number Product Name Customer Name Count (number of products sold in this order) Variant - this is the problem, as it has to be filled with the So all rows with the same OrderNo and same product have to be grouped in to one, and I hope it is clear what I need. I tried to do it with Pivot Tables, but that fails, as the Count is always in each line, no matter if it has Value "N" or not and for the products without variants there is only one line for each order, however for products with variants there are several... So how could I create the right table with a VBA macro in MS Excel or maybe there is a trick in MS Access to do it directly or with an SQL query?

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  • Mysql: create index on 1.4 billion records

    - by SiLent SoNG
    I have a table with 1.4 billion records. The table structure is as follows: CREATE TABLE text_page ( text VARCHAR(255), page_id INT UNSIGNED ) ENGINE=MYISAM DEFAULT CHARSET=ascii The requirement is to create an index over the column text. The table size is about 34G. I have tried to create the index by the following statement: ALTER TABLE text_page ADD KEY ix_text (text) After 10 hours' waiting I finally give up this approach. Is there any workable solution on this problem? UPDATE: the table is unlikely to be updated or inserted or deleted. The reason why to create index on the column text is because this kind of sql query would be frequently executed: SELECT page_id FROM text_page WHERE text = ? UPDATE: I have solved the problem by partitioning the table. The table is partitioned into 40 pieces on column text. Then creating index on the table takes about 1 hours to complete. It seems that MySQL index creation becomes very slow when the table size becomes very big. And partitioning reduces the table into smaller trunks.

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  • How to design data storage for partitioned tagging system?

    - by Morgan Cheng
    How to design data storage for huge tagging system (like digg or delicious)? There is already discussion about it, but it is about centralized database. Since the data is supposed to grow, we'll need to partition the data into multiple shards soon or later. So, the question turns to be: How to design data storage for partitioned tagging system? The tagging system basically has 3 tables: Item (item_id, item_content) Tag (tag_id, tag_title) TagMapping(map_id, tag_id, item_id) That works fine for finding all items for given tag and finding all tags for given item, if the table is stored in one database instance. If we need to partition the data into multiple database instances, it is not that easy. For table Item, we can partition its content with its key item_id. For table Tag, we can partition its content with its key tag_id. For example, we want to partition table Tag into K databases. We can simply choose number (tag_id % K) database to store given tag. But, how to partition table TagMapping? The TagMapping table represents the many-to-many relationship. I can only image to have duplication. That is, same content of TagMappping has two copies. One is partitioned with tag_id and the other is partitioned with item_id. In scenario to find tags for given item, we use partition with tag_id. If scenario to find items for given tag, we use partition with item_id. As a result, there is data redundancy. And, the application level should keep the consistency of all tables. It looks hard. Is there any better solution to solve this many-to-many partition problem?

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  • JQuery Tablesorter memorizeSortOrder widget

    - by echedey lorenzo
    Hi, I've found this code in the internet: $.tablesorter.addWidget({ id: "memorizeSortOrder", format: function(table) { if (!table.config.widgetMemorizeSortOrder.isBinded) { // only bind if not already binded table.config.widgetMemorizeSortOrder.isBinded = true; $("thead th:visible",table).click(function() { var i = $("thead th:visible",table).index(this); $.get(table.config.widgetMemorizeSortOrder.url+i+'|'+table.config.headerList[i].order); }); } // fi } }); Found in: http://www.adspeed.org/2008/10/jquery-extend-tablesorter-plugin.html I would like to memorize the sorting of my ajax tables so on each update (table changes completely so there is no append) it keeps sorted the as it was. Question is.. how can use this? $("#tablediv").load( "table.php", null, function (responseText, textStatus, req) { $("#table").trigger("update"); } ); What changes do I need?

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  • Booting Ubuntu EFI on Macbook Pro Apple Bootmanager

    - by user279771
    Following: http://www.rodsbooks.com/ubuntu-efi/ and https://help.ubuntu.com/community/UEFIBooting#Detect_.28U.29EFI_firmware_processor_architecture I have ubuntu booting on my macbook pro with both refined and the native apple loader working, however, I would like to get the "Improving the Boot Method" (i believe it is called the kernel efi stub loader) working with the Apple Boot manager (the article only explains for refined), but have found no articles explaining this. Can anyone help with this? Below is what I have: What I have is the following: /dev/sda apple partitions /boot ext3 (root) ext3 swap side notes: From what I understand, I should have had my boot partition as fat32/hfs+... I can always switch it or copy the kernel to the apple EFI partition. (I tried creating /boot as an hfs+ partition during installation but was unable. Even after installing hfsprogs, although I was able to create an hfs+ partition in gparted I couldn't use the partition as /boot in ubiquity).

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  • SQL SERVER – SmallDateTime and Precision – A Continuous Confusion

    - by pinaldave
    Some kinds of confusion never go away. Here is one of the ancient confusing things in SQL. The precision of the SmallDateTime is one concept that confuses a lot of people, proven by the many messages I receive everyday relating to this subject. Let me start with the question: What is the precision of the SMALLDATETIME datatypes? What is your answer? Write it down on your notepad. Now if you do not want to continue reading the blog post, head to my previous blog post over here: SQL SERVER – Precision of SMALLDATETIME. A Social Media Question Since the increase of social media conversations, I noticed that the amount of the comments I receive on this blog is a bit staggering. I receive lots of questions on facebook, twitter or Google+. One of the very interesting questions yesterday was asked on Facebook by Raghavendra. I am re-organizing his script and asking all of the questions he has asked me. Let us see if we could help him with his question: CREATE TABLE #temp (name VARCHAR(100),registered smalldatetime) GO DECLARE @test smalldatetime SET @test=GETDATE() INSERT INTO #temp VALUES ('Value1',@test) INSERT INTO #temp VALUES ('Value2',@test) GO SELECT * FROM #temp ORDER BY registered DESC GO DROP TABLE #temp GO Now when the above script is ran, we will get the following result: Well, the expectation of the query was to have the following result. The row which was inserted last was expected to return as first row in result set as the ORDER BY descending. Side note: Because the requirement is to get the latest data, we can’t use any  column other than smalldatetime column in order by. If we use name column in the order by, we will get an incorrect result as it can be any name. My Initial Reaction My initial reaction was as follows: 1) DataType DateTime2: If file precision of the column is expected from the column which store date and time, it should not be smalldatetime. The precision of the column smalldatetime is One Minute (Read Here) for finer precision use DateTime or DateTime2 data type. Here is the code which includes above suggestion: CREATE TABLE #temp (name VARCHAR(100), registered datetime2) GO DECLARE @test datetime2 SET @test=GETDATE() INSERT INTO #temp VALUES ('Value1',@test) INSERT INTO #temp VALUES ('Value2',@test) GO SELECT * FROM #temp ORDER BY registered DESC GO DROP TABLE #temp GO 2) Tie Breaker Identity: There are always possibilities that two rows were inserted at the same time. In that case, you may need a tie breaker. If you have an increasing identity column, you can use that as a tie breaker as well. CREATE TABLE #temp (ID INT IDENTITY(1,1), name VARCHAR(100),registered datetime2) GO DECLARE @test datetime2 SET @test=GETDATE() INSERT INTO #temp VALUES ('Value1',@test) INSERT INTO #temp VALUES ('Value2',@test) GO SELECT * FROM #temp ORDER BY ID DESC GO DROP TABLE #temp GO Those two were the quick suggestions I provided. It is not necessary that you should use both advices. It is possible that one can use only DATETIME datatype or Identity column can have datatype of BIGINT or have another tie breaker. An Alternate NO Solution In the facebook thread this was also discussed as one of the solutions: CREATE TABLE #temp (name VARCHAR(100),registered smalldatetime) GO DECLARE @test smalldatetime SET @test=GETDATE() INSERT INTO #temp VALUES ('Value1',@test) INSERT INTO #temp VALUES ('Value2',@test) GO SELECT name, registered, ROW_NUMBER() OVER(ORDER BY registered DESC) AS "Row Number" FROM #temp ORDER BY 3 DESC GO DROP TABLE #temp GO However, I believe it is not the solution and can be further misleading if used in a production server. Here is the example of why it is not a good solution: CREATE TABLE #temp (name VARCHAR(100) NOT NULL,registered smalldatetime) GO DECLARE @test smalldatetime SET @test=GETDATE() INSERT INTO #temp VALUES ('Value1',@test) INSERT INTO #temp VALUES ('Value2',@test) GO -- Before Index SELECT name, registered, ROW_NUMBER() OVER(ORDER BY registered DESC) AS "Row Number" FROM #temp ORDER BY 3 DESC GO -- Create Index ALTER TABLE #temp ADD CONSTRAINT [PK_#temp] PRIMARY KEY CLUSTERED (name DESC) GO -- After Index SELECT name, registered, ROW_NUMBER() OVER(ORDER BY registered DESC) AS "Row Number" FROM #temp ORDER BY 3 DESC GO DROP TABLE #temp GO Now let us examine the resultset. You will notice that an index which is created on the base table which is (indeed) schema change the table but can affect the resultset. As you can see, an index can change the resultset, so this method is not yet perfect to get the latest inserted resultset. No Schema Change Requirement After giving these two suggestions, I was waiting for the feedback of the asker. However, the requirement of the asker was there can’t be any schema change because the application was used by many other applications. I validated again, and of course, the requirement is no schema change at all. No addition of the column of change of datatypes of any other columns. There is no further help as well. This is indeed an interesting question. I personally can’t think of any solution which I could provide him given the requirement of no schema change. Can you think of any other solution to this? Need of Database Designer This question once again brings up another ancient question:  “Do we need a database designer?” I often come across databases which are facing major performance problems or have redundant data. Normalization is often ignored when a database is built fast under a very tight deadline. Often I come across a database which has table with unnecessary columns and performance problems. While working as Developer Lead in my earlier jobs, I have seen developers adding columns to tables without anybody’s consent and retrieving them as SELECT *.  There is a lot to discuss on this subject in detail, but for now, let’s discuss the question first. Do you have any suggestions for the above question? Reference: Pinal Dave (http://blog.sqlauthority.com) Filed under: CodeProject, Developer Training, PostADay, SQL, SQL Authority, SQL DateTime, SQL Query, SQL Server, SQL Tips and Tricks, SQLServer, T SQL, Technology

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  • Ubuntu confuses my partitions

    - by Diego
    I have 3 relevant partitions split between 2 disks, sda2: Windows 1 partition sda3: Ubuntu partition sdb1: Data partition I was using pysdm to add a label to my partitions and somehow I seem to have screwed up my installation. Now, every time I access the Data partition mounted in /media/Data I see the files in my Windows partition, and viceversa. I've tried unmounting and remounting correctly to no avail, it seems that wherever I mount sda2, if I access that folder I get the files in sdb1, and viceversa. Anyone know what may have happened and how to solve this? Update: This is the result of blkid: /dev/sda1: LABEL="System Reserved" UUID="C62603F02603E073" TYPE="ntfs" /dev/sda2: LABEL="Windows" UUID="00A6D498A6D49010" TYPE="ntfs" /dev/sda5: UUID="033cac3b-6f77-4f09-a629-495dc866866a" TYPE="ext4" /dev/sdb1: LABEL="Data" UUID="BCD83AE3D83A9B98" TYPE="ntfs" These are the contents of my ftsab file: UUID=033cac3b-6f77-4f09-a629-495dc866866a / ext4 errors=remount-ro,user_xattr 0 1 /dev/sda1 /media/Boot_old ntfs defaults 0 0 /dev/sda2 /media/Windows ntfs defaults 0 0 /dev/sdb1 /media/Data ntfs nls=iso8859-1,ro,users,umask=000 0 0

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  • How can I install Ubuntu Saucy Server with BTRFS?

    - by Walter Souto
    So, a week ago I installed the latest beta version of Ubuntu Saucy Server in a MacMini using BTRFS as only partition to mount "/" on it (I know it's not recommended to do this, but it's not the point here) with no problems et al. Everything went just "naturally"... Now with the released image of 13.10 server, I just can't get any BTRFS partition done from installer. I'm getting a "Can't create filesystem" error whenever I try to create any BTRFS partition, like the 13.10 server final installer can't handle format BTRFS partitions... Am I doing something wrong? Or it's a bug in the installer? Is there anything that I can do to workaround this and get my BTRFS partition set on installation, or I'll need to work this after the installation. I can just leave some space left and create a btrfs partition later on, with Saucy already installed and proceed to use to LXC (which is my solo purpose to have btrfs), but anyway, why installer can't do btrfs anymore?

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  • What is the easiest way to find a sql query returns a result or not?

    - by bala3569
    Consider the following sql server query , DECLARE @Table TABLE( Wages FLOAT ) INSERT INTO @Table SELECT 20000 INSERT INTO @Table SELECT 15000 INSERT INTO @Table SELECT 10000 INSERT INTO @Table SELECT 45000 INSERT INTO @Table SELECT 50000 SELECT * FROM ( SELECT *, ROW_NUMBER() OVER(ORDER BY Wages DESC) RowID FROM @Table ) sub WHERE RowID = 3 The result of the query would be 20000 ..... Thats fine as of now i have to find the result of this query, SELECT * FROM ( SELECT *, ROW_NUMBER() OVER(ORDER BY Wages DESC) RowID FROM @Table ) sub WHERE RowID = 6 It will not give any result because there are only 5 rows in the table..... so now my question is What is the easiest way to find a sql query returns a result or not?

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  • Disable Foreign key constraint on all tables didn't work ?

    - by Space Cracker
    i try a lot of commands to disable tables constraints in my database to make truncate to all tables but still now it give me the same error Cannot truncate table '' because it is being referenced by a FOREIGN KEY constraint. i try EXEC sp_msforeachtable "ALTER TABLE ? NOCHECK CONSTRAINT all" EXEC sp_MSforeachtable "TRUNCATE TABLE ?" and i tried this for each table ALTER TABLE [Table Name] NOCHECK CONSTRAINT ALL truncate table [Table Name] ALTER TABLE [Table Name] CHECK CONSTRAINT ALL and every time i have the previous error message .. could any please help me to solve sucha a problem ?

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  • Demantra Partitioning and the First PK Column

    - by user702295
      We have found that it is necessary in Demantra to have an index that matches the partition key, although it does not have to be the PK.  It is ok   to create a new index instead of changing the PK.   For example, if my PK on SALES_DATA is (ITEM_ID, LOCATION_ID, SALES_DATE) and I decide partition by SALES_DATE, then I should add an index starting   with the partition key like this: (SALES_DATE, ITEM_ID, LOCATION_ID).   * Note that the first column of the new index matches the partition key.   It might also be helpful to create a 2nd index with the other PK columns reversed (SALES_DATE, LOCATION_ID, ITEM_ID). Again, the first column   matches the partition key.

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  • Netbook partitioning scheme suggestions

    - by David B
    I got a new Asus EEE PC 1015PEM with 2GB RAM and a 250GB HD. After playing with the netbook edition a little, I would like to install the desktop edition I'm used to. In addition to ubunto partition(s), I would like to have one separate partition for data (documents, music, etc.), so I could try other OSs in the future without losing the data. What partition scheme would you recommend? I usually like to let the installation do it by itself, but when I try to that I can only use the entire disk, so I don't get the desired data partition. I wish there was a way to see the recommended default partitioning scheme, then just tweak it a bit to fit your needs (instead of building one from scratch). So, how would you recommend I partition my HD? Please be specific since I never manually partitioned before. Thanks!

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  • Can you add identity to existing column in sql server 2008?

    - by bmutch
    In all my searching I see that you essentially have to copy the existing table to a new table to chance to identity column for pre-2008, does this apply to 2008 also? thanks. most concise solution I have found so far: CREATE TABLE Test ( id int identity(1,1), somecolumn varchar(10) ); INSERT INTO Test VALUES ('Hello'); INSERT INTO Test VALUES ('World'); -- copy the table. use same schema, but no identity CREATE TABLE Test2 ( id int NOT NULL, somecolumn varchar(10) ); ALTER TABLE Test SWITCH TO Test2; -- drop the original (now empty) table DROP TABLE Test; -- rename new table to old table's name EXEC sp_rename 'Test2','Test'; -- see same records SELECT * FROM Test;

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  • Order of calls to set functions when invoking a flex component

    - by Jason
    I have a component called a TableDataViewer that contains the following pieces of data and their associated set functions: [Bindable] private var _dataSetLoader:DataSetLoader; public function get dataSetLoader():DataSetLoader {return _dataSetLoader;} public function set dataSetLoader(dataSetLoader:DataSetLoader):void { trace("setting dSL"); _dataSetLoader = dataSetLoader; } [Bindable] private var _table:Table = null; public function set table(table:Table):void { trace("setting table"); _table = table; _dataSetLoader.load(_table.definition.id, "viewData", _table.definition.id); } This component is nested in another component as follows: <ve:TableDataViewer width="100%" height="100%" paddingTop="10" dataSetLoader="{_openTable.dataSetLoader}" table="{_openTable.table}"/> Looking at the trace in the logs, the call to set table is coming before the call to set dataSetLoader. Which is a real shame because set table() needs dataSetLoader to already be set in order to call its load() function. So my question is, is there a way to enforce an order on the calls to the set functions when declaring a component?

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  • OBIA on Teradata - Part 1 Loader and Monitoring

    - by Mohan Ramanuja
    Normal 0 false false false EN-US X-NONE X-NONE /* Style Definitions */ table.MsoNormalTable {mso-style-name:"Table Normal"; mso-tstyle-rowband-size:0; mso-tstyle-colband-size:0; mso-style-noshow:yes; mso-style-priority:99; mso-style-qformat:yes; mso-style-parent:""; mso-padding-alt:0in 5.4pt 0in 5.4pt; mso-para-margin:0in; mso-para-margin-bottom:.0001pt; mso-pagination:widow-orphan; font-size:11.0pt; font-family:"Calibri","sans-serif"; mso-ascii-font-family:Calibri; mso-ascii-theme-font:minor-latin; mso-fareast-font-family:"Times New Roman"; mso-fareast-theme-font:minor-fareast; mso-hansi-font-family:Calibri; mso-hansi-theme-font:minor-latin; mso-bidi-font-family:"Times New Roman"; mso-bidi-theme-font:minor-bidi;} The out-of-the-box (OOB) OBIA Informatica mappings come with TPump loader.   TPUMP  FASTLOAD TPump does not lock the table. FastLoad applies exclusive lock on the table. The table that TPump is loading can have data. The table that FastLoad is loading needs to be empty. Normal 0 false false false EN-US X-NONE X-NONE /* Style Definitions */ table.MsoNormalTable {mso-style-name:"Table Normal"; mso-tstyle-rowband-size:0; mso-tstyle-colband-size:0; mso-style-noshow:yes; mso-style-priority:99; mso-style-qformat:yes; mso-style-parent:""; mso-padding-alt:0in 5.4pt 0in 5.4pt; mso-para-margin:0in; mso-para-margin-bottom:.0001pt; mso-pagination:widow-orphan; font-size:11.0pt; font-family:"Calibri","sans-serif"; mso-ascii-font-family:Calibri; mso-ascii-theme-font:minor-latin; mso-fareast-font-family:"Times New Roman"; mso-fareast-theme-font:minor-fareast; mso-hansi-font-family:Calibri; mso-hansi-theme-font:minor-latin; mso-bidi-font-family:"Times New Roman"; mso-bidi-theme-font:minor-bidi;} TPump is not efficient with lookups. FastLoad is more efficient in the absence of lookups. Normal 0 false false false EN-US X-NONE X-NONE /* Style Definitions */ table.MsoNormalTable {mso-style-name:"Table Normal"; mso-tstyle-rowband-size:0; mso-tstyle-colband-size:0; mso-style-noshow:yes; mso-style-priority:99; mso-style-qformat:yes; mso-style-parent:""; mso-padding-alt:0in 5.4pt 0in 5.4pt; mso-para-margin:0in; mso-para-margin-bottom:.0001pt; mso-pagination:widow-orphan; font-size:11.0pt; font-family:"Calibri","sans-serif"; mso-ascii-font-family:Calibri; mso-ascii-theme-font:minor-latin; mso-fareast-font-family:"Times New Roman"; mso-fareast-theme-font:minor-fareast; mso-hansi-font-family:Calibri; mso-hansi-theme-font:minor-latin; mso-bidi-font-family:"Times New Roman"; mso-bidi-theme-font:minor-bidi;} The out-of the box Informatica mappings come with TPump loader. There is chance for bottleneck in writer thread The out-of the box tables in Teradata supplied with OBAW features all Dimension and Fact tables using ROW_WID as the key for primary index. Also, all staging tables use integration_id as the key for primary index. This reduces skewing of data across Teradata AMPs.You can use an SQL statement similar to the following to determine if data for a given table is distributed evenly across all AMP vprocs. The SQL statement displays the AMP with the most used through the AMP with the least-used space, investigating data distribution in the Message table in database RST.SELECT vproc,CurrentPermFROM DBC.TableSizeWHERE Databasename = ‘PRJ_CRM_STGC’AND Tablename = ‘w_party_per_d’ORDER BY 2 descIf you suspect distribution problems (skewing) among AMPS, the following is a sample of what you might enter for a three-column PI:SELECT HASHAMP (HASHBUCKET (HASHROW (col_x, col_y, col_z))), count (*)FROM hash15GROUP BY 1ORDER BY 2 desc; ETL Error Monitoring Error Table – These are tables that start with ET. Location and name can be specified in Informatica session as well as the loader connection.Loader Log – Loader log is available in the Informatica server under the session log folder. These give feedback on the loader parameters such as Packing Factor to use. These however need to be monitored in the production environment. The recommendations made in one environment may not be used in another environment.Log Table – These are tables that start with TL. These are sparse on information.Bad File – This is the Informatica file generated in case there is data quality issues

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  • Is it Considered Good SQL practice to use GUID to link multiple tables to same Id field?

    - by Mallow
    I want to link several tables to a many-to-many(m2m) table. One table would be called location and this table would always be on one side of the m2m table. But I will have a list of several tables for example: Cards Photographs Illustrations Vectors Would using GUID's between these tables to link it to a single column in another table be considered 'Good Practice'? Will Mysql let me to have it automatically cascade updates and delete? If so, would multiple cascades lead to an issues? UPDATE I've read that GUID (a hex number) Generally takes up more space in a database and slows queries down. However I could still generate 'unique' ids by just having the table initial's as part of the id so that the table card's id would be c0001, and then Illustrations be I001. Regardless of this change, the questions still stands.

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  • Unable to start GRUB2 on Triple boot Macbook Pro with OS X Lion, Windows 7 and Ubuntu

    - by Shravan
    I installed Windows 7 using Bootcamp. Then I created a partition from the empty space in the Windows partition and another 4GB partition for the linux swap using GParted. I installed Ubuntu 12.04 LTS in the the newly created partition from the Windows partition. Now GRUB2 does not load and I can only see the blinking cursor on the top right when selecting "Windows HD" from the 'option' key at the start up. OS X works fine but nothing else. Could someone please help me fix this? I am attaching the boot-info from the boot-repair tool. http://paste.ubuntu.com/1040169/

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  • I can not install Ubuntu 14.04 due to some problem?

    - by user285643
    I have installed Window 8.1 on my HP laptop.Now I want to install Ubuntu 14.04,I try to use "Wubi" but when I use it, after installation on window, when my computer restart i have a message "No root file system is defined" I have read some thread here and i got some solution. One of them are "I must format my partition again, using ext4 format and mount on it". I did it by using Gparted in Try Ubuntu mode, but I got another message "/dev/sda contains GPT signatures,indicating that it has a GPT table. However, it does not have a valid fake msdos partition table,as it should. Perhaps it was corrupted--possibly by a program that does not understand GPT partition table. Or perhaps you deleted the GPT table, and are now using msdos partition table. Is this a GPT partition table?"

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  • Database design for a Fantasy league

    - by Samidh T
    Here's the basic schema for my database Table user{ userid numeber primary key, count number } Table player{ pid number primary key, } Table user-player{ userid number primary key foreign key(user), pid number primary key foreign key(player) } Table temp{ pid number primary key, points number } Here's what I intend to do... After every match the temp table is updated which holds the id of players that played the last match and the points they earned. Next run a procedure that will match the pid from temp table with every uid of user-player table having the same pid. add the points from temp table to the count of user table for every matching uid. empty temp table. My questions is considering 200 players and 10000 users,Will this method be efficient? I am going to be using mysql for this.

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  • Select records by comparing subsets

    - by devnull
    Given two tables (the rows in each table are distinct): 1) x | y z 2) x | y z ------- --- ------- --- 1 | a a 1 | a a 1 | b b 1 | b b 2 | a 1 | c 2 | b 2 | a 2 | c 2 | b 2 | c Is there a way to select the values in the x column of the first table for which all the values in the y column (for that x) are found in the z column of the second table? In case 1), expected result is 1. If c is added to the second table then the expected result is 2. In case 2), expected result is no record since neither of the subsets in the first table matches the subset in the second table. If c is added to the second table then the expected result is 1, 2. I've tried using except and intersect to compare subsets of first table with the second table, which works fine, but it takes too long on the intersect part and I can't figure out why (the first table has about 10.000 records and the second has around 10). EDIT: I've updated the question to provide an extra scenario.

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  • Sql (partial) search in a list and get matched fields

    - by qods
    I have two tables, I want to search TermID in Table-A through TermID in Table-B and If there is a termID like in Table-A and then want to get result table as shown below. TermIDs are in different length. There is no search pattern to search with "like %" TermIDs in Table-A are part of the TermIDs in Table-B Regards, Table-A ID TermID 101256666 126006230 101256586 126006231 101256810 126006233 101256841 126006238 101256818 126006239 101256734 1190226408 101256809 1190226409 101256585 1200096999 101256724 1200096997 101256748 1200097005 Table-B TermNo TermID 14 8990010901190226366F 16 8990010901190226374F 15 8990010901190226382F 18 8990010901190226408F 19 8990010901190226416F 11 8990010901200096981F 10 8990010901200096999F 12 8990010901200097005F 13 8990010901200097013F 17 8990010901260062337F As a result I want to get this table; Result Table -TableA.ID TableA.TermID TableB.TermNo A.ID A.TermID B.TermNo 101256734 1190226408 18 101256585 1200096999 10 101256748 1200097005 12

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