cross-join | 易学教程

SQL Server 2008 using SUM() OVER(ORDER BY…)

阅读更多关于 SQL Server 2008 using SUM() OVER(ORDER BY…)

问题 I am trying to use a CTE and CROSS JOIN the result set. I want to sum up the 4 rows leading up to the current row. The example online I found does not use a CTE, only a newly created table (http://sqlandme.com/2011/08/17/sql-server-denali-over-rows-range/). The syntax should work, but I get an error saying "Incorrect syntax near 'ROWS'". An Example output would be this using the following statement: SUM(y) OVER(ORDER BY x ROWS 4 PRECEDING) sum X Y SUM 1 7 0 No prev rows, so sum is 0 2 1 7 Sum

How to generate a matrix of combinations

阅读更多关于 How to generate a matrix of combinations

I have 5 items each of which can take on the value of 1 or -1. I want to generate a matrix that consists of rows of the possible combinations. The order of the items does not matter and the order of the combinations does not matter. I know I could do this mechanically, but I thought that someone must know a shortcut to generating this matrix. I apologize if this is similar to other questions but none of the solutions I have found can be applied to this particular problem with my programming skills. expand.grid(c(-1,1), c(-1,1), c(-1,1), c(-1,1), c(-1,1)) To generalize Greg's answer: N <- 5 vec

How do you perform a CROSS JOIN with LINQ to SQL?

阅读更多关于 How do you perform a CROSS JOIN with LINQ to SQL?

How do you perform a CROSS JOIN with LINQ to SQL? A cross-join is simply the Cartesian product of two sets. There's no explicit join operator for it. var combo = from p in people from c in cars select new { p.Name, c.Make, c.Model, c.Colour }; Rzv.im The same thing with linq extension methods: var names = new string[] { "Ana", "Raz", "John" }; var numbers = new int[] { 1, 2, 3 }; var newList=names.SelectMany( x => numbers, (y, z) => { return y + z + " test "; }); foreach (var item in newList) { Console.WriteLine(item); } Based on Steve's answer, the simplest expression would be this: var combo

combinations (not permutations) from cross join in sql

阅读更多关于 combinations (not permutations) from cross join in sql

问题 If I have a table that I'd like to cross join to itself, how can I remove the duplicate rows? Or to put it another way, how can I do a "order doesn't matter" cross join? So for example, if I have a table T: field | ------- A | B | C | and I cross join to itself so that i don't get the A | A rows T as t1 cross join T as t2 on t1.field != t2.field I would get the following: field | field ------+------- A | B A | C B | A B | C C | A C | B However, to me A, B is the same as B, A. Is there a good

How to do cross join in R?

阅读更多关于 How to do cross join in R?

How can I achieve a cross join in R ? I know that "merge" can do inner join, outer join. But I do not know how to achieve a cross join in R. Thanks Is it just all=TRUE ? x<-data.frame(id1=c("a","b","c"),vals1=1:3) y<-data.frame(id2=c("d","e","f"),vals2=4:6) merge(x,y,all=TRUE) From documentation of merge : If by or both by.x and by.y are of length 0 (a length zero vector or NULL), the result, r, is the Cartesian product of x and y, i.e., dim(r) = c(nrow(x)*nrow(y), ncol(x) + ncol(y)). If speed is an issue, I suggest checking out the excellent data.table package. In the example at the end it's

Mixing implicit and explicit JOINs

阅读更多关于 Mixing implicit and explicit JOINs

问题 I am having a problem with Hibernate generating invalid SQL. Specifically, mixing and matching implicit and explicit joins. This seems to be an open bug. However, I\'m not sure why this is invalid SQL. I have come up with a small toy example that generates the same syntax exception. Schema CREATE TABLE Employee ( employeeID INT, name VARCHAR(255), managerEmployeeID INT ) Data INSERT INTO Employee (employeeID, name) VALUES (1, \'Gary\') INSERT INTO Employee (employeeID, name, managerEmployeeID

SQL Server: What is the difference between CROSS JOIN and FULL OUTER JOIN?

阅读更多关于 SQL Server: What is the difference between CROSS JOIN and FULL OUTER JOIN?

问题 What is the difference between CROSS JOIN and FULL OUTER JOIN in SQL Server? Are they the same, or not? Please explain. When would one use either of these? 回答1: A cross join produces a cartesian product between the two tables, returning all possible combinations of all rows. It has no on clause because you're just joining everything to everything. A full outer join is a combination of a left outer and right outer join. It returns all rows in both tables that match the query's where clause,

How do you perform a CROSS JOIN with LINQ to SQL?

阅读更多关于 How do you perform a CROSS JOIN with LINQ to SQL?

问题 How do you perform a CROSS JOIN with LINQ to SQL? 回答1: A cross-join is simply the Cartesian product of two sets. There's no explicit join operator for it. var combo = from p in people from c in cars select new { p.Name, c.Make, c.Model, c.Colour }; 回答2: The same thing with linq extension methods: var names = new string[] { "Ana", "Raz", "John" }; var numbers = new int[] { 1, 2, 3 }; var newList=names.SelectMany( x => numbers, (y, z) => { return y + z + " test "; }); foreach (var item in

What is the expected behaviour for multiple set-returning functions in SELECT clause?

阅读更多关于 What is the expected behaviour for multiple set-returning functions in SELECT clause?

问题 I\'m trying to get a \"cross join\" with the result of two set-returning functions, but in some cases I don\'t get the \"cross join\", see example Behaviour 1 : When set lenghts are the same, it matches item by item from each set postgres=# SELECT generate_series(1,3), generate_series(5,7) order by 1,2; generate_series | generate_series -----------------+----------------- 1 | 5 2 | 6 3 | 7 (3 rows) Behaviour 2 : When set lenghts are different, it \"cross join\"s the sets postgres=# SELECT

How to do cross join in R?

阅读更多关于 How to do cross join in R?

问题 How can I achieve a cross join in R ? I know that \"merge\" can do inner join, outer join. But I do not know how to achieve a cross join in R. Thanks 回答1: Is it just all=TRUE ? x<-data.frame(id1=c("a","b","c"),vals1=1:3) y<-data.frame(id2=c("d","e","f"),vals2=4:6) merge(x,y,all=TRUE) From documentation of merge : If by or both by.x and by.y are of length 0 (a length zero vector or NULL), the result, r, is the Cartesian product of x and y, i.e., dim(r) = c(nrow(x)*nrow(y), ncol(x) + ncol(y)).