Pivot in SQL server

PIVOT:

PIVOT rotates a table-valued expression by turning the unique values from one column in the expression into multiple columns in the output, and performs aggregations where they are required on any remaining column values that are wanted in the final output.

CREATE TABLE Sales (Make varchar(8), Year int, Sales int)
GO
Insert into Sales values ('Honda',2005,20000)
Insert into Sales values ('Honda',2005,60000)
Insert into Sales values ('Mahindra',2005,40000)
Insert into Sales values ('Honda',2006,30000)
Insert into Sales values ('Mahindra',2006,30000)
Insert into Sales values ('Mahindra',2006,60000)
Insert into Sales values ('Mahindra',2007,8000)
GO

Actual Group by, get Number of cars sold for each make in each year.

Select Make,YEAR,SUM(Sales) from Sales
group by Make,Year

Transposed of above query

Select * from Sales pivot(Sum(sales) for Year in ([2001],[2002],[2003],[2004],[2005],[2006],[2007])) t
go

Restricting the columns which we need to show.

Select * from Sales pivot(Sum(sales) for Year in ([2005],[2006],[2007])) t

List sum of each product sold for each year and product should have been sold all the years and not just 1 year.

Select * from Sales pivot(Sum(sales) for Year in ([2005],[2006],[2007])) t
where [2007] is not null

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Difference in Correlated Update

Difference in Correlated Update :

I came across the Correlated update statement with below format. It updates the address filed in the “Main” table and takes it from the “sub” table. The NVL function has given outside the sub query.

Outer NVL Query:

update main m
set address = nvl((select address from sub s where m.id=s.id),-1);

We can give the NVL function with the sub query (specifically for that column). Then what is the difference between “Outer NVL Query” and “Inner NVL Query”.
Inner NVL Query:
update main m
set address = (select nvl(address,-1) from sub s where m.id=s.id)
Initial setup:
This step creates the tables and populates it.
create table main ( id number, Address varchar2(10));

insert into main values (1,'NNNNNNN');
insert into main values (2,'NNNNNNN');
insert into main values (3,'NNNNNNN');
insert into main values (4,'NNNNNNN');

create table sub ( id number, Address varchar2(10));

insert into sub values (1,'A');
insert into sub values (2,null);
insert into sub values (4,'C');

select * from main;

select * from sub

Inner NVL Query:

update main m
set address = (select nvl(address,-1) from sub s where m.id=s.id);
It is applying the NVL function if NULL value comes from the “SUB” table. It is putting the NULL value in the “Main” table if it does not find the match in the “Sub” table.

We think that the outer NVL query works in the same way. But there is some difference.
Outer NVL Query:
update main m
set address = nvl((select address from sub s where m.id=s.id),-1)
We can see the difference in below result. It is applying the NVL function in the both the situation.

This is small Difference in the Correlated Update.

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Shell script to find the latest file

The below shell script will help to find the latest file and name starting with “sales”.

Example the below files are available in UNIX box.
sales_2007_02_23_0010.dat
sales_2007_02_24_0024.dat
sales_2007_02_24_0714.dat
sales_2007_02_25_0010.dat
sales_2007_02_25_0525.dat
sales_2007_02_25_0835.dat

There are 3 files on 25th with different time stamp. The aim is identify the latest file. So the script check the latest file based on the time when that file has created in the UNIX box.

Script : latest_file.sh

#!/usr/bin/ksh
rm -f /home/hb/ms/hb/karthi/sales.dat
val=`ls -lrt sales* tail -1 awk '{print $9}'`
echo $val
ls /home/hb/ms/hb/karthi/$val
cp /home/hb/ms/hb/karthi/$val /home/hb/ms/hb/karthi/sales.dat

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Cumulative Sum

Ways to get a Cumulative Sum:

Initial setup:
Creating the table and populating the table.
CREATE TABLE #Sales (DayCount smallint, Sales money, RunningTotal money)


INSERT INTO #Sales (DayCount, Sales, RunningTotal)
SELECT 1, 10, 0
union all
SELECT 2, 7, 0
union all
SELECT 3, 2, 0
union all
SELECT 4, 6, 0

GO

SELECT * FROM #Sales ORDER BY DayCount

GO




Method 1:
DECLARE @RunningTotal money
SET @RunningTotal = 0

UPDATE #Sales
SET @RunningTotal = RunningTotal = @RunningTotal + Sales

GO

SELECT * FROM #Sales ORDER BY DayCount

GO



Method 2:

SELECT
a.DayCount
, a.Sales
, SUM(b.Sales) AS 'Running Total'
FROM
#Sales a
JOIN #Sales b ON b.DayCount <= a.DayCount GROUP BY a.DayCount , a.Sales ORDER BY a.DayCount , a.Sales

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Partitioning in SQL server

Partitioning in SQL server with small sample:


Summary: Table partitioning can make very large tables and indexes easier to manage, and improve the performance of appropriately filtered queries.

Table Partitioning Components:

The partition function, partition scheme, and partitioned table or index form a dependency tree, with the partition function at the top, the partition scheme depending on the partition function, and then the partitioned table or index depending on being linked to a partition scheme.


Partition function:
The partition function defines the boundary values of the initial set of partitions and the data type of the partitioned column:
o It makes no reference to any tables or disk storage.
o It forms the basis for one or more partition schemes.

create PARTITION function PF_dept(int)
AS range right FOR values (10,20,30)
Partition scheme:
The partition scheme maps particular partitions to filegroups:
o A given partition scheme can be used for one or more partitioned tables, indexes, and indexed views.
create partition scheme ps_dept
as partition PF_dept
TO (FG1,FG2,FG3,FG4)
Partitioned table:
The partitioned table or index is tied to a particular partition scheme when it is created:
o The partition table has only an indirect relationship, through the partition scheme, to the partition function.
create table dept
(deptid int,
deptname varchar(10)
)
ON ps_dept(deptid)


CREATE TABLE EMP
(EMPID INT,
EMPNAME VARCHAR(10),
DEPT_ID INT)
ON PS_DEPT(DEPT_ID)

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WITH TIES

WITH TIES in SQL server

Specifies that additional rows be returned from the base result set with the same value in the ORDER BY columns appearing as the last of the TOP n (PERCENT) rows. TOP...WITH TIES can be specified only in SELECT statements, and only if an ORDER BY clause is specified.

create table sales( customer_name varchar(10), sales_amount int)

insert into sales values
('A',6000),
('B',6000),
('C',10000),
('D',12000),
('E',13000),
('F',15000),
('G',5000),
('H',4000)
Top 5:
The “Top 5” gives first 5 records based on the condition in query.
select top 5 * from sales
order by sales_amount desc

customer_name sales_amount
------------- ------------
F 15000
E 13000
D 12000
C 10000
B 6000

(5 row(s) affected)
Top 5 WITH TIES:
The “Top 5 WITH TIES” gives 6 records because the base result set with the same value in the ORDER BY columns appearing as the last of the TOP 5.
select top 5 WITH TIES * from sales
order by sales_amount desc

customer_name sales_amount
------------- ------------
F 15000
E 13000
D 12000
C 10000
A 6000
B 6000

(6 row(s) affected)

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Comparing the Contents of Two Tables in SQL Server

Comparing the Contents of Two Tables:

create table a
(
c1 int,
c2 varchar(3),
c3 varchar(3)
)


create table b
(
c1 int,
c2 varchar(3),
c3 varchar(3)
)

insert into a values(1,'x','y')
insert into a values(2,'xx','y')
insert into a values(3,'x','y')

insert into b values(1,'x','y')
insert into b values(2,'x','y')
insert into b values(3,'x','yy')

select * from a

C1 C2 C3
----- -- --
1 x y
2 xx y
3 x y

select * from b
C1 C2 C3
----- -- --
1 x y
2 x y
3 x yy

The first rows are the same, but the second and third rows differ. This is how we can find them:
select i.c1,i.c2, i.c3,
count(i.src1) CNT1,
count(i.src2) CNT2
from
( select a.*,
1 src1,
null as src2
from a
union all
select b.*,
null as src1,
2 src2
from b
)i
group by i.c1,i.c2,i.c3
having count(i.src1) <> count(i.src2)

C1 C2 C3 CNT1 CNT2
--- -- -- ---- ----
2 x y 0 1
2 xx y 1 0
3 x y 1 0
3 x yy 0 1

The below link having the script in Oracle:

http://karthikeyanbaskaran.blogspot.com/2010/02/comparing-contents-of-two-tables.html

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