Cross Apply And Outer Apply in SQL Server

Introduction

The APPLY operator allows us to invoke a table-valued function for each row returned by an outer table expression of a query. The APPLY operator allows us to join two table expressions; the right table expression is processed every time for each row from the left table expression. Left table expression is evaluated first and then right table expression is evaluated against each row of the left table expression for final result-set. The list of columns produced by the APPLY operator is the set of columns in the left input followed by the list of columns returned by the right input.

Now we create two tables, first is Employee and second Project table.

Employee Table

  1. CREATE TABLE Employee  
  2. (  
  3.    Emp_IdintPRIMARYKEY,  
  4.    Emp_Name [nvarchar](max),  
  5.    Manager_Idint,  
  6.    Project_Idint  
  7. )  
Now insert some data into Employee table.

 Employee table

Project Table
  1. CREATETABLE Project  
  2. (  
  3.    Project_Idint,  
  4.    Project_Name [nvarchar](max),  
  5.    Department [nvarchar](max)  
  6. )  
Insert data into project table.

Project table

Forms of Apply

SQL Server contain two forms of Apply: CROSS APPLY and OUTER APPLY.

CROSS APPLY

CROSS APPLY returns only rows from the outer table that produce a result set from the table-valued function. It other words, result of CROSS APPLY doesn’t contain any row of left side table expression for which no result is obtained from right side table expression. CROSS APPLY work as a row by row INNER JOIN.

Example

INNER JOIN Query:
  1. SELECT FROM ProjectAS PRO   
  2. INNERJOIN  
  3. Employee AS EMP  
  4. ON  
  5. PRO.Project_Id=EMP.Project_Id  
CROSS APPLY Query:
  1. SELECT FROM ProjectAS PRO   
  2. CROSSAPPLY  
  3. (SELECT FROM Employee AS EMP WHERE PRO.Project_Id=EMP.Project_Id)Tab  
Above both query produce same result.

query output

OUTER APPLY

OUTER APPLY returns both rows that produce a result set, and rows that do not, with NULL values in the columns produced by the table-valued function. OUTER APPLY work as LEFT OUTER JOIN.

Example

LEFT OUTER JOIN Query:
  1. SELECT*FROM ProjectAS PRO   
  2. LEFTOUTERJOIN  
  3. Employee AS EMP  
  4. ON  
  5. PRO.Project_Id=EMP.Project_Id  
OUTER APPLY Query:
  1. SELECT*FROM ProjectAS PRO   
  2. OUTERAPPLY  
  3. (SELECT*FROM Employee AS EMP WHERE PRO.Project_Id=EMP.Project_Id)Tab  
Above both query produce same result.

query

APPLY with User Define function:

We can perform APPLY operation with a function that may be scalar or table valued function. This function will invoke each row and return result that will be associated with the outer table.

Example 1:

Firstly, create a function.
  1. CREATEFUNCTIONReturn_Info(@Project_IDint)  
  2. RETURNS[nvarchar](max)  
  3. AS  
  4. BEGIN  
  5. DECLARE @Info[nvarchar](max);  
  6. SET @Info = (SELECT 'Project Name is= ' + Project.Project_Name + 'Deratment is= ' + Project.DepartmentFROM Project WHEREProject.Project_Id = @Project_ID)  
  7. RETURN @Info  
  8. END  
Above function will return a scalar value that is the combine result of Department column and Project column.

Now we perform APPLY on this function.
  1. SELECT * FROM dbo.EmployeeAS EMP  
  2. CROSSAPPLY(select [dbo].[Return_Info](EMP.Project_Id))Tab(Project_Id)  
Output

function

Example 2:

Firstly, create a table valued function.
  1. CREATE FUNCTION [dbo].[fun_Return_Info](@Project_IDint)  
  2. RETURNS @Tab TABLE  
  3. (  
  4.    Project_Idint,  
  5.    Project_Namenvarchar(max),  
  6.    Department nvarchar(max)  
  7. )  
  8. AS  
  9. BEGIN  
  10. INSERT INTO @Tab  
  11. (  
  12.    Project_Id,  
  13.    Project_Name,  
  14.    Department  
  15. )  
  16. SELECT*FROM dbo.ProjectAS PRO WHERE PRO.Project_Id>@Project_ID  
  17. RETURN  
  18. END  
  19.   
  20. Above function return information of all project that that’s Project_Id is greater than given Project_id.  
  21.   
  22. SELECT EMP.Project_Id,COUNT(EMP.Project_Id)AS [Total] FROM dbo.Employee AS EMP  
  23. OUTERAPPLY  
  24. [fun_Return_Info](EMP.Project_Id)  
  25. GROUP BY EMP.Project_Id  
Output:

id

APPLY with TOP Command
  1. SELECT FROM  
  2. (SELECT PRO.*,EMP.Emp_Name,EMP.Emp_Id,EMP.Manager_Id,ROW_NUMBER()OVER(PartitionByPRO.Project_IdOrderByPRO.Project_Id)as RankFROMdbo.ProjectAS PRO  
  3. LEFTOUTERJOIN  
  4. dbo.EmployeeAS EMP  
  5. ON  
  6. PRO.Project_Id=EMP.Project_Id)Tab1  
  7. WHERE Tab1.Rank<=2  
Output:

run

Above query return top Employee details for each project. We can perform same operation using APPLY.
  1. SELECT PRO.*,Tab.*FROMdbo.ProjectAS PRO  
  2. OUTERAPPLY(SELECTTOP 2 EMP.Emp_Name,EMP.Emp_Id,EMP.Manager_IdFromdbo.EmployeeAS EMP WHERE EMP.Project_Id=PRO.Project_Id)Tab  
Above query produce same result as previous query.
 
Example 4:
  1. DECLARE @Tab TABLE  
  2. (  
  3.    [State] [nvarchar](max),  
  4.    City [nvarchar](max)  
  5. )  
  6.   
  7. INSERTINTO @Tab  
  8. SELECT'Rajasthan','Alwar,Laipur,Ajmer,Kota'UNIONALL  
  9. SELECT'Haryana','Hisar,Jhajjar,Rohtak'UNIONALL  
  10. SELECT'Maharaster','Mumbai,Pune'  
  11.   
  12. SELECT*FROM @Tab AS [@TA]  
Output:

table

Suppose we have a table that contain State name and City names but city names are stored in comma separated manner, now we want to split each city name. For this we can use APPLY Method as below.

For this we create a function that split city name and return a table that contain the list of city names.
  1. ALTER FUNCTION SplitString(@Input NVARCHAR(MAX))  
  2. RETURNS @Output TABLE(City NVARCHAR(1000))  
  3. AS  
  4. BEGIN  
  5. DECLARE @Index int;  
  6. SET @Input = @Input + ',';  
  7. WHILE(LEN(@Input) > 0)  
  8. BEGIN  
  9. SET @Index = CHARINDEX(',', @Input);  
  10. INSERT INTO @Output(City)  
  11. VALUES(SUBSTRING(@Input, 0, @Index))  
  12. SET @Input = SUBSTRING(@Input, @Index + 1, LEN(@Input));  
  13. END  
  14. RETURN  
  15. END  
Now we use this function to split the city name.
  1. DECLARE @Tab TABLE  
  2. (  
  3. [State] [nvarchar](max),  
  4. City [nvarchar](max)  
  5. )  
  6.   
  7. INSERT INTO @Tab  
  8. SELECT'Rajasthan','Alwar,Laipur,Ajmer,Kota'UNIONALL  
  9. SELECT'Haryana','Hisar,Jhajjar,Rohtak'UNIONALL  
  10. SELECT'Maharaster','Mumbai,Pune'  
  11.   
  12. SELECT T.State,Tab.*FROM @Tab AS T  
  13. OUTERAPPLY dbo.SplitString(T.City)Tab  
Output:

Output

Thanks for reading the article.

 

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