Generating series, now it’s very simple using recursive CTE

A interesting question asked by one of my Community friend to generate series of 4,8,16…. Without using any kind of looping structure.

This is very simple with the recursive CTE, lets see how we are implementing the solution

; with sam_series (myrow) as (

select 4 as myrow

union all

select myrow+ myrow from sam_series

where myrow<1000

)

select * from sam_series;

 myrow

-----------

4

8

16

32

64

128

256

512

1024

If we dissect the above CTE we can see

The anchor member is:  select 4

Recursive invocation member is: select myrow + myrow which is tied with Union All.

Termination Point is: myRow < 1000.

Conclusion: CTE provides the significant advantage of being able to reference itself, thereby creating a recursive CTE.

Get your cup filled with Tea … Enjoy coding  :)

Post Reference: Vikram Aristocratic Elfin Share

First insertion will always takes the seed value, all other insertion following first will take seed/reseed value and add increment factor to form the identity value

Lets demonstrate it with the help of a temporary table, here we are creating a table with identity column and set the seed value to 1 and increment factor to 1

create table #tempReseed

(id int identity(1,1),

col1 varchar(10));

Command(s) completed successfully.

Now before we execute any insert statement on the table we are Reseed the Identity value to 15

dbcc CHECKIDENT('#tempReseed',reseed,15)

Checking identity information: current identity value 'NULL'.

DBCC execution completed.

Now lets insert a records and check what value will get inserted to id column.

insert into #tempReseed(col1) values('ABC')

(1 row(s) affected)

Querying the table after insertion.

select * from #tempReseed

id          col1

----------- ----------

15          ABC

Here we found the id is set with reseed value i.e. 15

Lets again reseed the identity value, this time to 25

dbcc CHECKIDENT('#tempReseed',reseed,25)

Checking identity information: current identity value '15'.

DBCC execution completed.

Lets fire an insertion after reseeding identity value to 25

insert into #tempReseed(col1) values('DEF')

(1 row(s) affected)

Lets query the table to find what gets inserted this time.

select * from #tempReseed 

id          col1

----------- ----------

15          ABC

26          DEF

Cool!!! This time Reseed value was added with the incremented factor i.e. 1 an new value formed was 25+1 =26 which gets inserted.

Conclusion: First insertion will always takes the seed value, all other insertion following first will take seed/reseed value and add increment factor to form the identity value.

Code and enjoy … very sync :)

Post Reference: Vikram Aristocratic Elfin Share

SCHEMABINDING: You need to use schema qualified names for everything

A simple definition: Schema binding is a mode of ensuring that the objects referenced within a function or view, do not changes their definition in any way tat would crack the binded objects.

With SCHEMABINDING the obstacle is that you need to use schema qualified names for everything. You will get lots of error messages like this

Lets stimulate the same error with the below sample. Here we are creating table without any qualified schema

create table test1

(col1 int,

col2 varchar(10),

col3 varchar(10))

go

Since our table is ready lets try to create a view and refer the above table in it.

 create view vw_test1

with schemabinding

as

select col1,col2 from test1

Msg 4512, Level 16, State 3, Procedure vw_test1, Line 10

Cannot schema bind view 'vw_test1' because name 'test1' is invalid for schema binding. Names must be in two-part format and an object cannot reference itself.

Here we can see our code fall in error, since we have not specified any schema for our view. Lets add a schema and bind our above view to the newly created schema.

create view dbo.vw_test1

with schemabinding

as

select col1,col2 from test1

Msg 4512, Level 16, State 3, Procedure vw_test1, Line 10

Cannot schema bind view 'dbo.vw_test1' because name 'test1' is invalid for schema binding. Names must be in two-part format and an object cannot reference itself.

So you need everything i.e every object to be schema bounded.

Lets drop the table and recreate it with full qualified name

drop table test1

Command(s) completed successfully.

create schema test

Command(s) completed successfully.

create table test.test1

(col1 int,

col2 varchar(10),

col3 varchar(10))

go

Command(s) completed successfully.

Lets now create view with fully qualified name

create view dbo.vw_test1

with schemabinding

as

select col1,col2 from test.test1

Command(s) completed successfully.

col1        col2       col3

----------- ---------- ----------

(0 row(s) affected)

Conclusion: You need to create every object under full qualified name i.e. defined with schema while using SCHEMABINDING option.

Leave everything and do your coding; that’s the way to happiness. Just for SQL lovers :)

Post Reference: Vikram Aristocratic Elfin Share

Find A, B, C with a condition where A+B+C=30 Chose A, B, C only from given list of number (1,3,5, 7,9,11)

Thanks Sushree for whatapping me this question, but unfortunately the answer is NO we cannot chose  any number from the given list to form A+B+C=30 :D

As a Techie as soon as I got this problem from my cute friend, the only reflection popped up in my mind...Oh!! Put pen to paper write a small block of code and hit upon the answer.

I am so stupid, I didn’t even thought that the given numbers are all ODD in nature, and when you add three odd number you will never going to get even result and here we are expecting to get 30 i.e. even output. :D

But anyways to my best level of stupidity I wrote a small piece of code, which ultimately proves that it is not possible J

DECLARE @no1 INT

       ,@no2 INT

       ,@no3 INT

       ,@no4 INT

CREATE TABLE #temp (

       id INT identity(1, 1)

       ,GivenNo INT

       );

INSERT INTO #temp

SELECT 1 UNION

SELECT 3 UNION

SELECT 5 UNION

SELECT 7 UNION

SELECT 9 UNION

SELECT 11 UNION

SELECT 13 UNION

SELECT 15

WHILE (1 = 1)

BEGIN

       SELECT @no1 = GivenNo

       FROM #temp

       WHERE id = FLOOR(10 * RAND());

       SELECT @no2 = GivenNo

       FROM #temp

       WHERE id = FLOOR(10 * RAND());

       SELECT @no3 = GivenNo

       FROM #temp

       WHERE id = FLOOR(10 * RAND());

       IF isnull(@no1, 0) + isnull(@no2, 0) + isnull(@no3, 0) = 30

       BEGIN

              PRINT cast(@no1 AS VARCHAR(10));

              PRINT cast(@no2 AS VARCHAR(10));

              PRINT cast(@no3 AS VARCHAR(10));

              BREAK

       END

END

DROP TABLE #temp

Expecting a tea with you  :)

Post Reference: Vikram Aristocratic Elfin Share