Search Results

Search found 763 results on 31 pages for 'union'.

Page 9/31 | < Previous Page | 5 6 7 8 9 10 11 12 13 14 15 16  | Next Page >

• How does this Perl grep work to determine the union of several hashes?

  - by titaniumdecoy

  I don't understand the last line of this function from Programming Perl 3e. Here's how you might write a function that does a kind of set intersection by returning a list of keys occurring in all the hashes passed to it: @common = inter( \%foo, \%bar, \%joe ); sub inter { my %seen; for my $href (@_) { while (my $k = each %$href) { $seen{$k}++; } } return grep { $seen{$_} == @_ } keys %seen; } I understand that %seen is a hash which maps each key to the number of times it was encountered in any of the hashes provided to the function.

  Read the article

• Select all items of union but not the last one!

  - by 100r

  I have few texboxes, dropdown lists, etc.. They have their css classes. I want to select all elements of specific class(es) but NOT the last element from group of all classes. <asp:TextBox ID="TextBox1" runat="server" CssClass="class1" ></asp:TextBox> <asp:TextBox ID="TextBox2" runat="server" CssClass="class2" ></asp:TextBox> <asp:TextBox ID="TextBox3" runat="server" CssClass="class2" ></asp:TextBox> I want to select only TextBox1 and TextBox2, not TextBox3! selector should be something like this $("(.class1,.class1):not(:last)") or something like $(".class1,.class1").filter(":not(:last)") but of course none of it is working :) any sugestions? tnx in advance!

  Read the article

• How to find kth minimal element in the union of two sorted arrays?

  - by Michael

  This is a homework question. They say it takes O(logN + logM) where N and M are the arrays lengths. Let's name the arrays a and b. Obviously we can ignore all a[i] and b[i] where i k. First let's compare a[k/2] and b[k/2]. Let b[k/2] a[k/2]. Therefore we can discard also all b[i], where i k/2. Now we have all a[i], where i < k and all b[i], where i < k/2 to find the answer. What is the next step?

  Read the article

• Can you define values in a SQL statement that you can join/union, but are not stored in a table outs

  - by Mervyn

  I'm trying to create a query and need to join against something that I can define values in without creating a table. I'll attempt to describe what I'm trying to do: table1 is joined on field a with table2 (titles for FK in table 1) - Table1 has values outside of what exists in table2 - I want to add an additional 'table' to be unioned with table2 and then joined with table 1 Thanks

  Read the article

• SQL SERVER – Select and Delete Duplicate Records – SQL in Sixty Seconds #036 – Video

  - by pinaldave

  Developers often face situations when they find their column have duplicate records and they want to delete it. A good developer will never delete any data without observing it and making sure that what is being deleted is the absolutely fine to delete. Before deleting duplicate data, one should select it and see if the data is really duplicate. In this video we are demonstrating two scripts – 1) selects duplicate records 2) deletes duplicate records. We are assuming that the table has a unique incremental id. Additionally, we are assuming that in the case of the duplicate records we would like to keep the latest record. If there is really a business need to keep unique records, one should consider to create a unique index on the column. Unique index will prevent users entering duplicate data into the table from the beginning. This should be the best solution. However, deleting duplicate data is also a very valid request. If user realizes that they need to keep only unique records in the column and if they are willing to create unique constraint, the very first requirement of creating a unique constraint is to delete the duplicate records. Let us see how to connect the values in Sixty Seconds: Here is the script which is used in the video. USE tempdb GO CREATE TABLE TestTable (ID INT, NameCol VARCHAR(100)) GO INSERT INTO TestTable (ID, NameCol) SELECT 1, 'First' UNION ALL SELECT 2, 'Second' UNION ALL SELECT 3, 'Second' UNION ALL SELECT 4, 'Second' UNION ALL SELECT 5, 'Second' UNION ALL SELECT 6, 'Third' GO -- Selecting Data SELECT * FROM TestTable GO -- Detecting Duplicate SELECT NameCol, COUNT(*) TotalCount FROM TestTable GROUP BY NameCol HAVING COUNT(*) > 1 ORDER BY COUNT(*) DESC GO -- Deleting Duplicate DELETE FROM TestTable WHERE ID NOT IN ( SELECT MAX(ID) FROM TestTable GROUP BY NameCol) GO -- Selecting Data SELECT * FROM TestTable GO DROP TABLE TestTable GO Related Tips in SQL in Sixty Seconds: SQL SERVER – Delete Duplicate Records – Rows SQL SERVER – Count Duplicate Records – Rows SQL SERVER – 2005 – 2008 – Delete Duplicate Rows Delete Duplicate Records – Rows – Readers Contribution Unique Nonclustered Index Creation with IGNORE_DUP_KEY = ON – A Transactional Behavior What would you like to see in the next SQL in Sixty Seconds video? Reference: Pinal Dave (http://blog.sqlauthority.com) Filed under: Database, Pinal Dave, PostADay, SQL, SQL Authority, SQL in Sixty Seconds, SQL Query, SQL Scripts, SQL Server, SQL Server Management Studio, SQL Tips and Tricks, T SQL, Technology, Video Tagged: Excel

  Read the article

• Using set operation in LINQ

  - by vik20000in

  There are many set operation that are required to be performed while working with any kind of data. This can be done very easily with the help of LINQ methods available for this functionality. Below are some of the examples of the set operation with LINQ. Finding distinct values in the set of data. We can use the distinct method to find out distinct values in a given list.     int[] factorsOf300 = { 2, 2, 3, 5, 5 };     var uniqueFactors = factorsOf300.Distinct(); We can also use the set operation of UNION with the help of UNION method in the LINQ. The Union method takes another collection as a parameter and returns the distinct union values in  both the list. Below is an example.     int[] numbersA = { 0, 2, 4, 5, 6, 8, 9 };    int[] numbersB = { 1, 3, 5, 7, 8 };    var uniqueNumbers = numbersA.Union(numbersB); We can also get the set operation of INTERSECT with the help of the INTERSECT method. Below is an example.     int[] numbersA = { 0, 2, 4, 5, 6, 8, 9 };     int[] numbersB = { 1, 3, 5, 7, 8 };         var commonNumbers = numbersA.Intersect(numbersB);  We can also find the difference between the 2 sets of data with the help of except method.      int[] numbersA = { 0, 2, 4, 5, 6, 8, 9 };     int[] numbersB = { 1, 3, 5, 7, 8 };         IEnumerable<int> aOnlyNumbers = numbersA.Except(numbersB);  Vikram

  Read the article

• Programming logic to group a users activities like facebook. E.g. Chris is now friends with A, B and C

  - by Chris Dowdeswell

  So I am trying to develop an activity feed for my site, Basically If I UNION a bunch of activities into a feed I would end up with something like the following. Chris is now friends with Mark Chris is now friends with Dave What I want though is a neater way of grouping these similar posts so the feed doesn't give information overload... E.g. Chris is now friends with Mark, Dave and 4 Others Any ideas on how I can approach this logically? I am using Classic ASP on SQL server. Here is the UNION statement I have so far... SELECT U.UserID As UserID, L.UN As UN,Left(U.UID,13) As ProfilePic,U.Fname + ' ' + U.Sname As FullName, 'said ' + WP.Post AS Activity, WP.Ctime FROM Users AS U LEFT JOIN Logins L ON L.userID = U.UserID LEFT OUTER JOIN WallPosts AS WP ON WP.userID = U.userID WHERE WP.Ctime IS NOT NULL UNION SELECT U.UserID As UserID, L.UN As UN,Left(U.UID,13) As ProfilePic,U.Fname + ' ' + U.Sname As FullName, 'commented ' + C.Comment AS Activity, C.Ctime FROM Users AS U LEFT JOIN Logins L ON L.userID = U.UserID LEFT OUTER JOIN Comments AS C ON C.UserID = U.userID WHERE C.Ctime IS NOT NULL UNION SELECT U.UserID As UserID, L.UN As UN,Left(U.UID,13) As ProfilePic, U.Fname + ' ' + U.Sname As FullName, 'connected with <a href="/profile.asp?un='+(SELECT Logins.un FROM Logins WHERE Logins.userID = Cn.ToUserID)+'">' + (SELECT Users.Fname + ' ' + Users.Sname FROM Users WHERE userID = Cn.ToUserID) + '</a>' AS Activity, Cn.Ctime FROM Users AS U LEFT JOIN Logins L ON L.userID = U.UserID LEFT OUTER JOIN Connections AS Cn ON Cn.UserID = U.userID WHERE CN.Ctime IS NOT NULL

  Read the article

• SQL SERVER – Find First Non-Numeric Character from String

  - by pinaldave

  It is fun when you have to deal with simple problems and there are no out of the box solution. I am sure there are many cases when we needed the first non-numeric character from the string but there is no function available to identify that right away. Here is the quick script I wrote down using PATINDEX. The function PATINDEX exists for quite a long time in SQL Server but I hardly see it being used. Well, at least I use it and I am comfortable using it. Here is a simple script which I use when I have to identify first non-numeric character. -- How to find first non numberic character USE tempdb GO CREATE TABLE MyTable (ID INT, Col1 VARCHAR(100)) GO INSERT INTO MyTable (ID, Col1) SELECT 1, '1one' UNION ALL SELECT 2, '11eleven' UNION ALL SELECT 3, '2two' UNION ALL SELECT 4, '22twentytwo' UNION ALL SELECT 5, '111oneeleven' GO -- Use of PATINDEX SELECT PATINDEX('%[^0-9]%',Col1) 'Position of NonNumeric Character', SUBSTRING(Col1,PATINDEX('%[^0-9]%',Col1),1) 'NonNumeric Character', Col1 'Original Character' FROM MyTable GO DROP TABLE MyTable GO Here is the resultset: Where do I use in the real world – well there are lots of examples. In one of the future blog posts I will cover that as well. Meanwhile, do you have any better way to achieve the same. Do share it here. I will write a follow up blog post with due credit to you. Reference : Pinal Dave (http://blog.SQLAuthority.com) Filed under: PostADay, SQL, SQL Authority, SQL Function, SQL Query, SQL Server, SQL String, SQL Tips and Tricks, T SQL, Technology

  Read the article

• SQL SERVER – Three Methods to Insert Multiple Rows into Single Table – SQL in Sixty Seconds #024 – Video

  - by pinaldave

  One of the biggest ask I have always received from developers is that if there is any way to insert multiple rows into a single table in a single statement. Currently when developers have to insert any value into the table they have to write multiple insert statements. First of all this is not only boring it is also very much time consuming as well. Additionally, one has to repeat the same syntax so many times that the word boring becomes an understatement. In the following quick video we have demonstrated three different methods to insert multiple values into a single table. -- Insert Multiple Values into SQL Server CREATE TABLE #SQLAuthority (ID INT, Value VARCHAR(100)); Method 1: Traditional Method of INSERT…VALUE -- Method 1 - Traditional Insert INSERT INTO #SQLAuthority (ID, Value) VALUES (1, 'First'); INSERT INTO #SQLAuthority (ID, Value) VALUES (2, 'Second'); INSERT INTO #SQLAuthority (ID, Value) VALUES (3, 'Third'); Clean up -- Clean up TRUNCATE TABLE #SQLAuthority; Method 2: INSERT…SELECT -- Method 2 - Select Union Insert INSERT INTO #SQLAuthority (ID, Value) SELECT 1, 'First' UNION ALL SELECT 2, 'Second' UNION ALL SELECT 3, 'Third'; Clean up -- Clean up TRUNCATE TABLE #SQLAuthority; Method 3: SQL Server 2008+ Row Construction -- Method 3 - SQL Server 2008+ Row Construction INSERT INTO #SQLAuthority (ID, Value) VALUES (1, 'First'), (2, 'Second'), (3, 'Third'); Clean up -- Clean up DROP TABLE #SQLAuthority; Related Tips in SQL in Sixty Seconds: SQL SERVER – Insert Multiple Records Using One Insert Statement – Use of UNION ALL SQL SERVER – 2008 – Insert Multiple Records Using One Insert Statement – Use of Row Constructor I encourage you to submit your ideas for SQL in Sixty Seconds. We will try to accommodate as many as we can. If we like your idea we promise to share with you educational material. Reference: Pinal Dave (http://blog.sqlauthority.com) Filed under: Database, Pinal Dave, PostADay, SQL, SQL Authority, SQL in Sixty Seconds, SQL Query, SQL Scripts, SQL Server, SQL Server Management Studio, SQL Tips and Tricks, T SQL, Technology, Video

  Read the article

• Programming logic to group a users activities like Facebook

  - by Chris Dowdeswell

  So I am trying to develop an activity feed for my site. Basically If I UNION a bunch of activities into a feed I would end up with something like the following. Chris is now friends with Mark Chris is now friends with Dave What I want though is a neater way of grouping these similar posts so the feed doesn't give information overload... E.g. Chris is now friends with Mark, Dave and 4 Others Any ideas on how I can approach this logically? I am using Classic ASP on SQL server. Here is the UNION statement I have so far: SELECT U.UserID As UserID, L.UN As UN,Left(U.UID,13) As ProfilePic,U.Fname + ' ' + U.Sname As FullName, 'said ' + WP.Post AS Activity, WP.Ctime FROM Users AS U LEFT JOIN Logins L ON L.userID = U.UserID LEFT OUTER JOIN WallPosts AS WP ON WP.userID = U.userID WHERE WP.Ctime IS NOT NULL UNION SELECT U.UserID As UserID, L.UN As UN,Left(U.UID,13) As ProfilePic,U.Fname + ' ' + U.Sname As FullName, 'commented ' + C.Comment AS Activity, C.Ctime FROM Users AS U LEFT JOIN Logins L ON L.userID = U.UserID LEFT OUTER JOIN Comments AS C ON C.UserID = U.userID WHERE C.Ctime IS NOT NULL UNION SELECT U.UserID As UserID, L.UN As UN,Left(U.UID,13) As ProfilePic, U.Fname + ' ' + U.Sname As FullName, 'connected with <a href="/profile.asp?un='+(SELECT Logins.un FROM Logins WHERE Logins.userID = Cn.ToUserID)+'">' + (SELECT Users.Fname + ' ' + Users.Sname FROM Users WHERE userID = Cn.ToUserID) + '</a>' AS Activity, Cn.Ctime FROM Users AS U LEFT JOIN Logins L ON L.userID = U.UserID LEFT OUTER JOIN Connections AS Cn ON Cn.UserID = U.userID WHERE CN.Ctime IS NOT NULL

  Read the article

• Basics of Join Predicate Pushdown in Oracle

  - by Maria Colgan

  Happy New Year to all of our readers! We hope you all had a great holiday season. We start the new year by continuing our series on Optimizer transformations. This time it is the turn of Predicate Pushdown. I would like to thank Rafi Ahmed for the content of this blog.Normally, a view cannot be joined with an index-based nested loop (i.e., index access) join, since a view, in contrast with a base table, does not have an index defined on it. A view can only be joined with other tables using three methods: hash, nested loop, and sort-merge joins. Introduction The join predicate pushdown (JPPD) transformation allows a view to be joined with index-based nested-loop join method, which may provide a more optimal alternative. In the join predicate pushdown transformation, the view remains a separate query block, but it contains the join predicate, which is pushed down from its containing query block into the view. The view thus becomes correlated and must be evaluated for each row of the outer query block. These pushed-down join predicates, once inside the view, open up new index access paths on the base tables inside the view; this allows the view to be joined with index-based nested-loop join method, thereby enabling the optimizer to select an efficient execution plan. The join predicate pushdown transformation is not always optimal. The join predicate pushed-down view becomes correlated and it must be evaluated for each outer row; if there is a large number of outer rows, the cost of evaluating the view multiple times may make the nested-loop join suboptimal, and therefore joining the view with hash or sort-merge join method may be more efficient. The decision whether to push down join predicates into a view is determined by evaluating the costs of the outer query with and without the join predicate pushdown transformation under Oracle's cost-based query transformation framework. The join predicate pushdown transformation applies to both non-mergeable views and mergeable views and to pre-defined and inline views as well as to views generated internally by the optimizer during various transformations. The following shows the types of views on which join predicate pushdown is currently supported. UNION ALL/UNION view Outer-joined view Anti-joined view Semi-joined view DISTINCT view GROUP-BY view Examples Consider query A, which has an outer-joined view V. The view cannot be merged, as it contains two tables, and the join between these two tables must be performed before the join between the view and the outer table T4. A: SELECT T4.unique1, V.unique3 FROM T_4K T4,            (SELECT T10.unique3, T10.hundred, T10.ten             FROM T_5K T5, T_10K T10             WHERE T5.unique3 = T10.unique3) VWHERE T4.unique3 = V.hundred(+) AND       T4.ten = V.ten(+) AND       T4.thousand = 5; The following shows the non-default plan for query A generated by disabling join predicate pushdown. When query A undergoes join predicate pushdown, it yields query B. Note that query B is expressed in a non-standard SQL and shows an internal representation of the query. B: SELECT T4.unique1, V.unique3 FROM T_4K T4,           (SELECT T10.unique3, T10.hundred, T10.ten             FROM T_5K T5, T_10K T10             WHERE T5.unique3 = T10.unique3             AND T4.unique3 = V.hundred(+)             AND T4.ten = V.ten(+)) V WHERE T4.thousand = 5; The execution plan for query B is shown below. In the execution plan BX, note the keyword 'VIEW PUSHED PREDICATE' indicates that the view has undergone the join predicate pushdown transformation. The join predicates (shown here in red) have been moved into the view V; these join predicates open up index access paths thereby enabling index-based nested-loop join of the view. With join predicate pushdown, the cost of query A has come down from 62 to 32.  As mentioned earlier, the join predicate pushdown transformation is cost-based, and a join predicate pushed-down plan is selected only when it reduces the overall cost. Consider another example of a query C, which contains a view with the UNION ALL set operator.C: SELECT R.unique1, V.unique3 FROM T_5K R,            (SELECT T1.unique3, T2.unique1+T1.unique1             FROM T_5K T1, T_10K T2             WHERE T1.unique1 = T2.unique1             UNION ALL             SELECT T1.unique3, T2.unique2             FROM G_4K T1, T_10K T2             WHERE T1.unique1 = T2.unique1) V WHERE R.unique3 = V.unique3 and R.thousand < 1; The execution plan of query C is shown below. In the above, 'VIEW UNION ALL PUSHED PREDICATE' indicates that the UNION ALL view has undergone the join predicate pushdown transformation. As can be seen, here the join predicate has been replicated and pushed inside every branch of the UNION ALL view. The join predicates (shown here in red) open up index access paths thereby enabling index-based nested loop join of the view. Consider query D as an example of join predicate pushdown into a distinct view. We have the following cardinalities of the tables involved in query D: Sales (1,016,271), Customers (50,000), and Costs (787,766).  D: SELECT C.cust_last_name, C.cust_city FROM customers C,            (SELECT DISTINCT S.cust_id             FROM sales S, costs CT             WHERE S.prod_id = CT.prod_id and CT.unit_price > 70) V WHERE C.cust_state_province = 'CA' and C.cust_id = V.cust_id; The execution plan of query D is shown below. As shown in XD, when query D undergoes join predicate pushdown transformation, the expensive DISTINCT operator is removed and the join is converted into a semi-join; this is possible, since all the SELECT list items of the view participate in an equi-join with the outer tables. Under similar conditions, when a group-by view undergoes join predicate pushdown transformation, the expensive group-by operator can also be removed. With the join predicate pushdown transformation, the elapsed time of query D came down from 63 seconds to 5 seconds. Since distinct and group-by views are mergeable views, the cost-based transformation framework also compares the cost of merging the view with that of join predicate pushdown in selecting the most optimal execution plan. Summary We have tried to illustrate the basic ideas behind join predicate pushdown on different types of views by showing example queries that are quite simple. Oracle can handle far more complex queries and other types of views not shown here in the examples. Again many thanks to Rafi Ahmed for the content of this blog post.

  Read the article

• SQL SERVER – Disable Clustered Index and Data Insert

  - by pinaldave

  Earlier today I received following email. “Dear Pinal, [Removed unrelated content] We looked at your script and found out that in your script of disabling indexes, you have only included non-clustered index during the bulk insert and missed to disabled all the clustered index. Our DBA[name removed] has changed your script a bit and included all the clustered indexes. Since our application is not working. When DBA [name removed] tried to enable clustered indexes again he is facing error incorrect syntax error. We are in deep problem [word replaced] [Removed Identity of organization and few unrelated stuff ]“ I have replied to my client and helped them fixed the problem. What really came to my attention is the concept of disabling clustered index. Let us try to learn a lesson from this experience. In this case, there was no need to disable clustered index at all. I had done necessary work when I was called in to work on tuning project. I had removed unused indexes, created few optimal indexes and wrote a script to disable few selected high cost indexes when bulk insert (and similar) operations are performed. There was another script which rebuild all the indexes as well. The solution worked till they included clustered index in disabling the script. Clustered indexes are in fact original table (or heap) physically ordered (any more things – not scope of this article) according to one or more keys(columns). When clustered index is disabled data rows of the disabled clustered index cannot be accessed. This means there will be no insert possible. When non clustered indexes are disabled all the data related to physically deleted but the definition of the index is kept in the system. Due to the same reason even reorganization of the index is not possible till the clustered index (which was disabled) is rebuild. Now let us come to the second part of the question, regarding receiving the error when clustered index is ‘enabled’. This is very common question I receive on the blog. (The following statement is written keeping the syntax of T-SQL in mind) Clustered indexes can be disabled but can not be enabled, they have to rebuild. It is intuitive to think that something which we have ‘disabled’ can be ‘enabled’ but the syntax for the same is ‘rebuild’. This issue has been explained here: SQL SERVER – How to Enable Index – How to Disable Index – Incorrect syntax near ‘ENABLE’. Let us go over this example where inserting the data is not possible when clustered index is disabled. USE AdventureWorks GO -- Create Table CREATE TABLE [dbo].[TableName]( [ID] [int] NOT NULL, [FirstCol] [varchar](50) NULL, CONSTRAINT [PK_TableName] PRIMARY KEY CLUSTERED ([ID] ASC) ) GO -- Create Nonclustered Index CREATE UNIQUE NONCLUSTERED INDEX [IX_NonClustered_TableName] ON [dbo].[TableName] ([FirstCol] ASC) GO -- Populate Table INSERT INTO [dbo].[TableName] SELECT 1, 'First' UNION ALL SELECT 2, 'Second' UNION ALL SELECT 3, 'Third' GO -- Disable Nonclustered Index ALTER INDEX [IX_NonClustered_TableName] ON [dbo].[TableName] DISABLE GO -- Insert Data should work fine INSERT INTO [dbo].[TableName] SELECT 4, 'Fourth' UNION ALL SELECT 5, 'Fifth' GO -- Disable Clustered Index ALTER INDEX [PK_TableName] ON [dbo].[TableName] DISABLE GO -- Insert Data will fail INSERT INTO [dbo].[TableName] SELECT 6, 'Sixth' UNION ALL SELECT 7, 'Seventh' GO /* Error: Msg 8655, Level 16, State 1, Line 1 The query processor is unable to produce a plan because the index 'PK_TableName' on table or view 'TableName' is disabled. */ -- Reorganizing Index will also throw an error ALTER INDEX [PK_TableName] ON [dbo].[TableName] REORGANIZE GO /* Error: Msg 1973, Level 16, State 1, Line 1 Cannot perform the specified operation on disabled index 'PK_TableName' on table 'dbo.TableName'. */ -- Rebuliding should work fine ALTER INDEX [PK_TableName] ON [dbo].[TableName] REBUILD GO -- Insert Data should work fine INSERT INTO [dbo].[TableName] SELECT 6, 'Sixth' UNION ALL SELECT 7, 'Seventh' GO -- Clean Up DROP TABLE [dbo].[TableName] GO I hope this example is clear enough. There were few additional posts I had written years ago, I am listing them here. SQL SERVER – Enable and Disable Index Non Clustered Indexes Using T-SQL SQL SERVER – Enabling Clustered and Non-Clustered Indexes – Interesting Fact Reference : Pinal Dave (http://blog.SQLAuthority.com) Filed under: Pinal Dave, SQL, SQL Authority, SQL Constraint and Keys, SQL Query, SQL Server, SQL Tips and Tricks, T SQL, Technology

  Read the article

• SQL SERVER – Quiz and Video – Introduction to Hierarchical Query using a Recursive CTE

  - by pinaldave

  This blog post is inspired from SQL Queries Joes 2 Pros: SQL Query Techniques For Microsoft SQL Server 2008 – SQL Exam Prep Series 70-433 – Volume 2.[Amazon] | [Flipkart] | [Kindle] | [IndiaPlaza] This is follow up blog post of my earlier blog post on the same subject - SQL SERVER – Introduction to Hierarchical Query using a Recursive CTE – A Primer. In the article we discussed various basics terminology of the CTE. The article further covers following important concepts of common table expression. What is a Common Table Expression (CTE) Building a Recursive CTE Identify the Anchor and Recursive Query Add the Anchor and Recursive query to a CTE Add an expression to track hierarchical level Add a self-referencing INNER JOIN statement Above six are the most important concepts related to CTE and SQL Server.  There are many more things one has to learn but without beginners fundamentals one can’t learn the advanced  concepts. Let us have small quiz and check how many of you get the fundamentals right. Quiz 1) You have an employee table with the following data. EmpID FirstName LastName MgrID 1 David Kennson 11 2 Eric Bender 11 3 Lisa Kendall 4 4 David Lonning 11 5 John Marshbank 4 6 James Newton 3 7 Sally Smith NULL You need to write a recursive CTE that shows the EmpID, FirstName, LastName, MgrID, and employee level. The CEO should be listed at Level 1. All people who work for the CEO will be listed at Level 2. All of the people who work for those people will be listed at Level 3. Which CTE code will achieve this result? WITH EmpList AS (SELECT Boss.EmpID, Boss.FName, Boss.LName, Boss.MgrID, 1 AS Lvl FROM Employee AS Boss WHERE Boss.MgrID IS NULL UNION ALL SELECT E.EmpID, E.FirstName, E.LastName, E.MgrID, EmpList.Lvl + 1 FROM Employee AS E INNER JOIN EmpList ON E.MgrID = EmpList.EmpID) SELECT * FROM EmpList WITH EmpListAS (SELECT EmpID, FirstName, LastName, MgrID, 1 as Lvl FROM Employee WHERE MgrID IS NULL UNION ALL SELECT EmpID, FirstName, LastName, MgrID, 2 as Lvl ) SELECT * FROM BossList WITH EmpList AS (SELECT EmpID, FirstName, LastName, MgrID, 1 as Lvl FROM Employee WHERE MgrID is NOT NULL UNION SELECT EmpID, FirstName, LastName, MgrID, BossList.Lvl + 1 FROM Employee INNER JOIN EmpList BossList ON Employee.MgrID = BossList.EmpID) SELECT * FROM EmpList 2) You have a table named Employee. The EmployeeID of each employee’s manager is in the ManagerID column. You need to write a recursive query that produces a list of employees and their manager. The query must also include the employee’s level in the hierarchy. You write the following code segment: WITH EmployeeList (EmployeeID, FullName, ManagerName, Level) AS ( –PICK ANSWER CODE HERE ) SELECT EmployeeID, FullName, ” AS [ManagerID], 1 AS [Level] FROM Employee WHERE ManagerID IS NULL UNION ALL SELECT emp.EmployeeID, emp.FullName mgr.FullName, 1 + 1 AS [Level] FROM Employee emp JOIN Employee mgr ON emp.ManagerID = mgr.EmployeeId SELECT EmployeeID, FullName, ” AS [ManagerID], 1 AS [Level] FROM Employee WHERE ManagerID IS NULL UNION ALL SELECT emp.EmployeeID, emp.FullName, mgr.FullName, mgr.Level + 1 FROM EmployeeList mgr JOIN Employee emp ON emp.ManagerID = mgr.EmployeeId Now make sure that you write down all the answers on the piece of paper. Watch following video and read earlier article over here. If you want to change the answer you still have chance. Solution 1) 1 2) 2 Now compare let us check the answers and compare your answers to following answers. I am very confident you will get them correct. Available at USA: Amazon India: Flipkart | IndiaPlaza Volume: 1, 2, 3, 4, 5 Please leave your feedback in the comment area for the quiz and video. Did you know all the answers of the quiz? Reference: Pinal Dave (http://blog.sqlauthority.com) Filed under: Joes 2 Pros, PostADay, SQL, SQL Authority, SQL Query, SQL Server, SQL Tips and Tricks, T SQL, Technology

  Read the article

• Handy SQL Server Functions Series (HSSFS) Part 2.0 - Prelude to Parsing Patterns Properly

  - by Most Valuable Yak (Rob Volk)

  In Part 1 of the series I wrote about 2 lesser-known and somewhat undocumented functions. In this part, I'm going to cover some familiar string functions like Substring(), Parsename(), Patindex(), and Charindex() and delve into their strengths and weaknesses. I'm also splitting this part up into sub-parts to help focus on a particular technique and/or problem with the technique, hence the Part 2.0. Consider this a composite post, or com-post, if you will. (It may just turn out to be a pile of sh_t after all) I'll be using a contrived example, perhaps the most frustratingly useful, or usefully frustrating, function in SQL Server: @@VERSION. Contrived, because there are better ways to get the information (which I'll cover later); frustrating, because of the way Microsoft formatted the value; and useful because it does have 1 or 2 bits of information not found elsewhere. First let's take a look at the output of @@VERSION: Microsoft SQL Server 2008 R2 (RTM) - 10.50.1600.1 (Intel X86) Apr 2 2010 15:53:02 Copyright (c) Microsoft Corporation Developer Edition on Windows NT 5.1 <X86> (Build 2600: Service Pack 3) There are 4 lines, with lines 2-4 indented with a tab character.  In case your browser (or this blog software) doesn't show it correctly, I gave each line a different color.  While this PRINTs nicely, if you SELECT @@VERSION in grid mode it all runs together because it ignores carriage return/line feed (CR/LF) characters.  Not fatal, but annoying. Note that @@VERSION's output will vary depending on edition and version of SQL Server, and also the OS it's installed on.  Despite the differences, the output is laid out the same way and the relevant pieces are in the same order. I'll be using the following view for Parts 2.1 onward, so we have a nice collection of @@VERSION information: create view version(SQLVersion,VersionString) AS ( select 2000, 'Microsoft SQL Server 2000 - 8.00.2055 (Intel X86) Dec 16 2008 19:46:53 Copyright (c) 1988-2003 Microsoft Corporation Developer Edition on Windows NT 5.1 (Build 2600: Service Pack 3)' union all select 2005, 'Microsoft SQL Server 2005 - 9.00.4053.00 (Intel X86) May 26 2009 14:24:20 Copyright (c) 1988-2005 Microsoft Corporation Developer Edition on Windows NT 5.1 (Build 2600: Service Pack 3)' union all select 2008, 'Microsoft SQL Server 2008 R2 (RTM) - 10.50.1600.1 (Intel X86) Apr 2 2010 15:53:02 Copyright (c) Microsoft Corporation Developer Edition on Windows NT 5.1 <X86> (Build 2600: Service Pack 3)' union all select 2005, 'Microsoft SQL Server 2005 - 9.00.3080.00 (Intel X86) Sep 6 2009 01:43:32 Copyright (c) 1988-2005 Microsoft Corporation Standard Edition on Windows NT 5.2 (Build 3790: Service Pack 2)' union all select 2008, 'Microsoft SQL Server 2008 R2 (RTM) - 10.50.1600.1 (X64) Apr 2 2010 15:48:46 Copyright (c) Microsoft Corporation Developer Edition (64-bit) on Windows NT 6.1 <X64> (Build 7600: ) (Hypervisor)' union all select 2008, 'Microsoft SQL Server 2008 R2 (RTM) - 10.50.1600.1 (X64) Apr 2 2010 15:48:46 Copyright (c) Microsoft Corporation Express Edition with Advanced Services (64-bit) on Windows NT 6.1 <X64> (Build 7600: ) (Hypervisor)' ) Feel free to add your own @@VERSION info if it's not already there. In Part 2.1 I'll focus on extracting the SQL Server version number (10.50.1600.1 in first example) and the Edition (Developer), but will have a solution that works with all versions.  Stay tuned!

  Read the article

• Building Tag Cloud Declarative ADF Component

  - by Arunkumar Ramamoorthy

  When building a website, there could a requirement to add a tag cloud to let the users know the popular tags (or terms) used in the site. In this blog, we would build a simple declarative component to be used as tag cloud in the page. To start with, we would first create the declarative component, which could display the tag cloud. We will do that by creating a new custom application from the new gallery. Give a name for the app and the project and from the new gallery, let us create a new ADF Declarative Component We need to specify the name for the declarative component, attributes in it etc. as follows For displaying the tags as cloud, we need to pass the content to this component. So, we will create an attribute to hold the values for the tag. Let us name it as "value" and make it as java.lang.String  type. Once after this, to hold the component, we need to create a tag library. This can be done by clicking on the Add Tag Library button. Clicking on OK buttons in all the open dialogs would create a declarative component for us. Now, we need to display the tag cloud based on the value passed to the component. To do that, we assume that the value is a Tree Binding and has two attributes in it, say "Name" and "Weight". To make a tag cloud, we would put together the "Name" in a loop and set it's font size based on the "Weight". After putting our logic to work, here is how the source look Attributes added to the declarative components can be retrieved by using #{attrs.<attribute_name>}. Now, we need to deploy this project as ADF Library Jar file, so that this can be distributed to the consuming applications. We'll select ADF Library Jar as type and create the profile. We would be getting the jar file after deployment. To test the functionality, we could create a simple Fusion Web Application. To add our custom component to the consuming application, we can create a file system connection pointing to the location where the jar file is and add it or, add through the project properties of the ViewController project. Now, our custom component has been added to the consuming application. We could test that by creating a VO in the model project with a query like, select 'Faces' as Name,25 as Weight from dual union all select 'ADF', 15 from dual  union all select 'ADFdi', 30 from dual union all select 'BC4J', 20 from dual union all select 'EJB', 40 from dual union all select 'WS', 35 from dual Add this VO to the AppModule, so that it would be exposed to the data control. Then, we could create a jspx page, and add a tree binding to the VO created. We can now see our Tag Cloud declarative component is available in the component palette.  It can be inserted from the component palette to our page and set it's value property to CollectionModel of the tree binding created. Now that we've created the Declarative component and added that to our page successfully, we can run the page to see how it looks. As per the query, the Tags are displayed in different fonts, based on their weight.

  Read the article

• SQL Server 2008 Splitting string variable number of token per line

  - by josephj1989

  I have a fairly simple requirement -I have a table with the following (relevant) structure. with cte as( select 1 id,'AA,AB,AC,AD' names union all select 2,'BA,BB' union all select 3,'CA,CB,CC,CD,CE' union all select 4,'DA,DB,DC' ) i would like to create a select statement which will split each "names" column into multiple rows. For example the first row should produce 1,'AA' 1,'AB' 1,'AC' 1,'AD' Can we do it using only SQL. This is failry easy to do in Oracle.

  Read the article

• SQL Server - CAST AND DIVIDE

  - by rs

  DECLARE @table table(XYZ VARCHAR(8) , id int) INSERT INTO @table SELECT '4000', 1 UNION ALL SELECT '3.123', 2 UNION ALL SELECT '7.0', 3 UNION ALL SELECT '80000', 4 UNION ALL SELECT NULL, 5 SELECT CASE WHEN PATINDEX('^[0-9]{1,5}[\.][0-9]{1,3}$', XYZ) = 0 THEN XYZ WHEN PATINDEX('^[0-9]{1,8}$',XYZ) = 0 THEN CAST(XYZ AS decimal(18,3))/1000 ELSE NULL END FROM @table This part - CAST(XYZ AS decimal(18,3))/1000 doesn't divide value it gives me more number of zeros after decimal instead of dividing it. (I even enclosed that in brackets and tried but same result) Am i doing something wrong here?

  Read the article

• Searching a table MySQL & PHP.

  - by S1syphus

  I want to be able to search through a MySQL table using values from a search string, from the url and display the results as an XML output. I think I have got the formatting and declaring the variables from the search string down. The issue I have is searching the entire table, I've looked over SO for previous answers, and they all seem to have to declare each column in the table to search through. So for example my database layout is as follows: **filesindex** -filename -creation -length -wall -playlocation First of all would the following be appropriate: $query = "SELECT * FROM filesindex WHERE filename LIKE '".$searchterm."%' UNION SELECT * FROM filesindex WHERE creation LIKE '".$searchterm."%' UNION SELECT * FROM filesindex WHERE length LIKE '".$searchterm."%' UNION SELECT * FROM filesindex WHERE wall LIKE '".$searchterm."%' UNION SELECT * FROM filesindex WHERE location LIKE '".$searchterm."%'"; Or ideally, is there an easier way that involves less hardcoding to search a table. Any ideas? Thanks

  Read the article

• Unexpected behaviour of Order by clause

  - by Newbie

  I have a table which looks like Col1 col2 col3 col4 col5 1 5 1 4 6 1 4 0 3 7 0 1 5 6 3 1 8 2 1 5 4 3 2 1 4 The script is declare @t table(col1 int, col2 int, col3 int,col4 int,col5 int) insert into @t select 1,5,1,4,6 union all select 1,4,0,3,7 union all select 0,1,5,6,3 union all select 1,8,2,1,5 union all select 4,3,2,1,4 If I do a sorting (ascending), the output is Col1 col2 col3 col4 col5 0 1 5 6 3 1 4 0 3 7 1 5 1 4 6 1 8 2 1 5 4 3 2 1 4 The query is Select * from @t order by col1,col2,col3,col4,col5 But as can be seen that the sorting output is wrong (col2 to col5). I want the output to be every column being sorted in ascending order i.e. Col1 col2 col3 col4 col5 0 1 0 1 3 1 3 1 1 4 1 4 2 3 5 1 5 2 4 6 4 8 5 6 7 Why so and how to overcome this? Thanks in advance

  Read the article

• Random select is not always returning a single row.

  - by Lieven

  The intention of following (simplified) code fragment is to return one random row. Unfortunatly, when we run this fragment in the query analyzer, it returns between zero and three results. As our input table consists of exactly 5 rows with unique ID's and as we perform a select on this table where ID equals a random number, we are stumped that there would ever be more than one row returned. Note: among other things, we already tried casting the checksum result to an integer with no avail. DECLARE @Table TABLE ( ID INTEGER IDENTITY (1, 1) , FK1 INTEGER ) INSERT INTO @Table SELECT 1 UNION ALL SELECT 2 UNION ALL SELECT 3 UNION ALL SELECT 4 UNION ALL SELECT 5 SELECT * FROM @Table WHERE ID = ABS(CHECKSUM(NEWID())) % 5 + 1

  Read the article

• MySQL search for user and their roles

  - by Jenkz

  I am re-writing the SQL which lets a user search for any other user on our site and also shows their roles. An an example, roles can be "Writer", "Editor", "Publisher". Each role links a User to a Publication. Users can take multiple roles within multiple publications. Example table setup: "users" : user_id, firstname, lastname "publications" : publication_id, name "link_writers" : user_id, publication_id "link_editors" : user_id, publication_id Current psuedo SQL: SELECT * FROM ( (SELECT user_id FROM users WHERE firstname LIKE '%Jenkz%') UNION (SELECT user_id FROM users WHERE lastname LIKE '%Jenkz%') ) AS dt JOIN (ROLES STATEMENT) AS roles ON roles.user_id = dt.user_id At the moment my roles statement is: SELECT dt2.user_id, dt2.publication_id, dt.role FROM ( (SELECT 'writer' AS role, link_writers.user_id, link_writers.publication_id FROM link_writers) UNION (SELECT 'editor' AS role, link_editors.user_id, link_editors.publication_id FROM link_editors) ) AS dt2 The reason for wrapping the roles statement in UNION clauses is that some roles are more complex and require a table join to find the publication_id and user_id. As an example "publishers" might be linked accross two tables "link_publishers": user_id, publisher_group_id "link_publisher_groups": publisher_group_id, publication_id So in that instance, the query forming part of my UNION would be: SELECT 'publisher' AS role, link_publishers.user_id, link_publisher_groups.publication_id FROM link_publishers JOIN link_publisher_groups ON lpg.group_id = lp.group_id I'm pretty confident that my table setup is good (I was warned off the one-table-for-all system when researching the layout). My problem is that there are now 100,000 rows in the users table and upto 70,000 rows in each of the link tables. Initial lookup in the users table is fast, but the joining really slows things down. How can I only join on the relevant roles? -------------------------- EDIT ---------------------------------- Explain above (open in a new window to see full resolution). The bottom bit in red, is the "WHERE firstname LIKE '%Jenkz%'" the third row searches WHERE CONCAT(firstname, ' ', lastname) LIKE '%Jenkz%'. Hence the large row count, but I think this is unavoidable, unless there is a way to put an index accross concatenated fields? The green bit at the top just shows the total rows scanned from the ROLES STATEMENT. You can then see each individual UNION clause (#6 - #12) which all show a large number of rows. Some of the indexes are normal, some are unique. It seems that MySQL isn't optimizing to use the dt.user_id as a comparison for the internal of the UNION statements. Is there any way to force this behaviour? Please note that my real setup is not publications and writers but "webmasters", "players", "teams" etc.

  Read the article

• Unexpected behaviour of Order by clause(SQL SERVER 2005)

  - by Newbie

  I have a table which looks like Col1 col2 col3 col4 col5 1 5 1 4 6 1 4 0 3 7 0 1 5 6 3 1 8 2 1 5 4 3 2 1 4 The script is declare @t table(col1 int, col2 int, col3 int,col4 int,col5 int) insert into @t select 1,5,1,4,6 union all select 1,4,0,3,7 union all select 0,1,5,6,3 union all select 1,8,2,1,5 union all select 4,3,2,1,4 If I do a sorting (ascending), the output is Col1 col2 col3 col4 col5 0 1 5 6 3 1 4 0 3 7 1 5 1 4 6 1 8 2 1 5 4 3 2 1 4 The query is Select * from @t order by col1,col2,col3,col4,col5 But as can be seen that the sorting output is wrong (col2 to col5). Why so and how to overcome this? Thanks in advance

  Read the article

• Help needed for writing a Set Based query for finding the highest marks obtained by the students

  - by priyanka.sarkar_2

  I have the below table declare @t table (id int identity, name varchar(50),sub1 int,sub2 int,sub3 int,sub4 int) insert into @t select 'name1',20,30,40,50 union all select 'name2',10,30,40,50 union all select 'name3',40,60,100,50 union all select 'name4',80,30,40,80 union all select 'name5',80,70,40,50 union all select 'name6',10,30,40,80 The desired output should be Id Name Sub1 Sub2 Sub3 Sub4 3 Name3 100 4 Name4 80 80 5 Name5 80 70 6 Name6 80 What I have done so far is ;with cteSub1 as ( select rn1 = dense_rank() over(order by sub1 desc),t.id,t.name,t.sub1 from @t t ) ,cteSub2 as ( select rn2 = dense_rank() over(order by sub2 desc),t.id,t.name,t.sub2 from @t t ) ,cteSub3 as ( select rn3 = dense_rank() over(order by sub3 desc),t.id,t.name,t.sub3 from @t t ) ,cteSub4 as ( select rn4 = dense_rank() over(order by sub4 desc),t.id,t.name,t.sub4 from @t t ) select x1.id,x2.id,x3.id,x4.id ,x1.sub1,x2.sub2,x3.sub3,x4.sub4 from (select c1.id,c1.sub1 from cteSub1 c1 where rn1 =1) as x1 full join (select c2.id,c2.sub2 from cteSub2 c2 where rn2 =1)x2 on x1.id = x2.id full join (select c3.id,c3.sub3 from cteSub3 c3 where rn3 =1)x3 on x1.id = x3.id full join (select c4.id,c4.sub4 from cteSub4 c4 where rn4 =1)x4 on x1.id = x4.id which is giving me the output as id id id id sub1 sub2 sub3 sub4 5 5 NULL NULL 80 70 NULL NULL 4 NULL NULL 4 80 NULL NULL 80 NULL NULL 3 NULL NULL NULL 100 NULL NULL NULL NULL 6 NULL NULL NULL 80 Help needed. Also how can I reduce the number of CTE's?

  Read the article

• SQL SERVER – Spatial Database Queries – What About BLOB – T-SQL Tuesday #006

  - by pinaldave

  Michael Coles is one of the most interesting book authors I have ever met. He has a flair of writing complex stuff in a simple language. There are a very few people like that.  I really enjoyed reading his recent book, Expert SQL Server 2008 Encryption. I strongly suggest taking a look at it. This blog is written in response to T-SQL Tuesday #006: “What About BLOB? by Michael Coles. Spatial Database is my favorite subject. Since I did my TechEd India 2010 presentation, I have enjoyed this subject a lot. Before I continue this blog post, there are a few other blog posts, so I suggest you read them.  To help build the environment run the queries, I am going to present them in this single blog post. SQL SERVER – What is Spatial Database? – Developing with SQL Server Spatial and Deep Dive into Spatial Indexing This blog post explains the basics of Spatial Database and also provides a good introduction to Indexing concept. SQL SERVER – World Shapefile Download and Upload to Database – Spatial Database This blog post will enable you with how to load the shape file into database. SQL SERVER – Spatial Database Definition and Research Documents This blog post links to the white paper about Spatial Database written by Microsoft experts. SQL SERVER – Introduction to Spatial Coordinate Systems: Flat Maps for a Round Planet This blog post links to the white paper explaining coordinate system, as written by Microsoft experts. After reading the above listed blog posts, I am very confident that you are ready to run the following script. Once you create a database using the World Shapefile, as mentioned in the second link above,you can display the image of India just like the following. Please note that this is not an accurate political map. The boundary of this map has many errors and it is just a representation. You can run the following query to generate the map of India from the database spatial which you have created after following the instructions here. USE Spatial GO -- India Map SELECT [CountryName] ,[BorderAsGeometry] ,[Border] FROM [Spatial].[dbo].[Countries] WHERE Countryname = 'India' GO Now, let us find the longitude and latitude of the two major IT cities of India, Hyderabad and Bangalore. I find their values as the following: the values of longitude-latitude for Bangalore is 77.5833300000 13.0000000000; for Hyderabad, longitude-latitude is 78.4675900000 17.4531200000. Now, let us try to put these values on the India Map and see their location. -- Bangalore DECLARE @GeoLocation GEOGRAPHY SET @GeoLocation = GEOGRAPHY::STPointFromText('POINT(77.5833300000 13.0000000000)',4326).STBuffer(20000); -- Hyderabad DECLARE @GeoLocation1 GEOGRAPHY SET @GeoLocation1 = GEOGRAPHY::STPointFromText('POINT(78.4675900000 17.4531200000)',4326).STBuffer(20000); -- Bangalore and Hyderabad on Map of India SELECT name, [GeoLocation] FROM [IndiaGeoNames] I WHERE I.[GeoLocation].STDistance(@GeoLocation) <= 0 UNION ALL SELECT name, [GeoLocation] FROM [IndiaGeoNames] I WHERE I.[GeoLocation].STDistance(@GeoLocation1) <= 0 UNION ALL SELECT '',[Border] FROM [Spatial].[dbo].[Countries] WHERE Countryname = 'India' GO Now let us quickly draw a straight line between them. DECLARE @GeoLocation GEOGRAPHY SET @GeoLocation = GEOGRAPHY::STPointFromText('POINT(78.4675900000 17.4531200000)',4326).STBuffer(10000); DECLARE @GeoLocation1 GEOGRAPHY SET @GeoLocation1 = GEOGRAPHY::STPointFromText('POINT(77.5833300000 13.0000000000)',4326).STBuffer(10000); DECLARE @GeoLocation2 GEOGRAPHY SET @GeoLocation2 = GEOGRAPHY::STGeomFromText('LINESTRING(78.4675900000 17.4531200000, 77.5833300000 13.0000000000)',4326) SELECT name, [GeoLocation] FROM [IndiaGeoNames] I WHERE I.[GeoLocation].STDistance(@GeoLocation) <= 0 UNION ALL SELECT name, [GeoLocation] FROM [IndiaGeoNames] I1 WHERE I1.[GeoLocation].STDistance(@GeoLocation1) <= 0 UNION ALL SELECT '' name, @GeoLocation2 UNION ALL SELECT '',[Border] FROM [Spatial].[dbo].[Countries] WHERE Countryname = 'India' GO Let us use the distance function of the spatial database and find the straight line distance between this two cities. -- Distance Between Hyderabad and Bangalore DECLARE @GeoLocation GEOGRAPHY SET @GeoLocation = GEOGRAPHY::STPointFromText('POINT(78.4675900000 17.4531200000)',4326) DECLARE @GeoLocation1 GEOGRAPHY SET @GeoLocation1 = GEOGRAPHY::STPointFromText('POINT(77.5833300000 13.0000000000)',4326) SELECT @GeoLocation.STDistance(@GeoLocation1)/1000 'KM'; GO The result of above query is as displayed in following image. As per SQL Server, the distance between these two cities is 501 KM, but according to what I know, the distance between those two cities is around 562 KM by road. However, please note that roads are not straight and they have lots of turns, whereas this is a straight-line distance. What would be more accurate is the distance between these two cities by air travel. When we look at the air travel distance between Bangalore and Hyderabad, the total distance covered is 495 KM, which is very close to what SQL Server has estimated, which is 501 KM. Bravo! SQL Server has accurately provided the distance between two of the cities. SQL Server Spatial Database can be very useful simply because it is very easy to use, as demonstrated above. I appreciate your comments, so let me know what your thoughts and opinions about this are. Reference: Pinal Dave (http://blog.SQLAuthority.com) Filed under: Pinal Dave, SQL, SQL Authority, SQL Query, SQL Scripts, SQL Server, SQL Tips and Tricks, T SQL, Technology Tagged: Spatial Database

  Read the article

• SQL SERVER – Subquery or Join – Various Options – SQL Server Engine Knows the Best – Part 2

  - by pinaldave

  This blog post is part 2 of the earlier written article SQL SERVER – Subquery or Join – Various Options – SQL Server Engine knows the Best by Paulo R. Pereira. Paulo has left excellent comment to earlier article once again proving the point that SQL Server Engine is smart enough to figure out the best plan itself and uses the same for the query. Let us go over his comment as he has posted. “I think IN or EXISTS is the best choice, because there is a little difference between ‘Merge Join’ of query with JOIN (Inner Join) and the others options (Left Semi Join), and JOIN can give more results than IN or EXISTS if the relationship is 1:0..N and not 1:0..1. And if I try use NOT IN and NOT EXISTS the query plan is different from LEFT JOIN too (Left Anti Semi Join vs. Left Outer Join + Filter). So, I found a case where EXISTS has a different query plan than IN or ANY/SOME:” USE AdventureWorks GO -- use of SOME SELECT * FROM HumanResources.Employee E WHERE E.EmployeeID = SOME ( SELECT EA.EmployeeID FROM HumanResources.EmployeeAddress EA UNION ALL SELECT EA.EmployeeID FROM HumanResources.EmployeeDepartmentHistory EA ) -- use of IN SELECT * FROM HumanResources.Employee E WHERE E.EmployeeID IN ( SELECT EA.EmployeeID FROM HumanResources.EmployeeAddress EA UNION ALL SELECT EA.EmployeeID FROM HumanResources.EmployeeDepartmentHistory EA ) -- use of EXISTS SELECT * FROM HumanResources.Employee E WHERE EXISTS ( SELECT EA.EmployeeID FROM HumanResources.EmployeeAddress EA UNION ALL SELECT EA.EmployeeID FROM HumanResources.EmployeeDepartmentHistory EA ) When looked into execution plan of the queries listed above indeed we do get different plans for queries and SQL Server Engines creates the best (least cost) plan for each query. Click on image to see larger images. Thanks Paulo for your wonderful contribution. Reference : Pinal Dave (http://blog.SQLAuthority.com) Filed under: Pinal Dave, Readers Contribution, SQL, SQL Authority, SQL Joins, SQL Optimization, SQL Performance, SQL Query, SQL Scripts, SQL Server, SQL Tips and Tricks, T SQL, Technology

  Read the article

< Previous Page | 5 6 7 8 9 10 11 12 13 14 15 16  | Next Page >