SQL Server: Keyboard Shortcuts – Part 1

Writing clean code is an important part in development activity. People sometimes ignore this may be because of some or other reason, which creates difficult to understand the code in later stage when it need changes. Here are some quick keyboard shortcuts for SQL Server Management Studio, which helps to write clean code and speed up development works.

So the below shortcuts related to code editing and execution in SSMS.

Purpose	Keyboard Shortcut
Open New Query Window	CTRL + N
Comment selected text	CTRL + K then CTRL + C
Make selected text to UPPER case	CTRL + SHIFT + U
Make selected text to LOWER case	CTRL + SHIFT + L
Display estimated execution plan	CRTL + L
Execute query	F5 (Or) ALT + X (Or) CTRL + E
Show / hide result pane	CTRL + R
Toggle between Query & Result pane	F6

 

So have quick and clean coding…

Fixing Error: 8134 (Devided By Zero Error Encountered) in SQL Server

Problem:

While performing mathematical operations it throws Divided by zero error.

Msg 8134, Level 16, State 1, Line 5 Divide by zero error encountered.

 

This situation often arises in production databases if the script has not been tested with sufficient data before putting the script to production database. This happens when a number is divided by 0 (zero).

Solution:

There can be many ways to handle this error. Here are some of my workarounds in SQL Server.

1. Using NULLIF & ISNULL/ COALESCE
2. Using CASE
3. Using ARITHABORT & ANSI_WARNINGS

Method: 1

SELECT ISNULL(Number1 / NULLIF(Number2, 0), 0) AS [Result]
FROM tbl_err_8134

In this method uses NULLIF. In this case when the divisor is 0 (Zero) it will return NULL to the divisor, so the result will also became NULL. Then by IFNULL it returns 0 as the result is NULL here.

Method: 2

SELECT CASE WHEN Number2 = 0 THEN 0 ELSE Number1 / Number2 END AS [Result]
FROM tbl_err_8134

In this method uses CASE. Here when the divisor is 0 (Zero) it will return 0 as result or else the result will be division of two numbers.

Method: 3

SET ARITHABORT OFF
SET ANSI_WARNINGS OFF
GO

SELECT ISNULL(Number1 / Number2, 0) AS [Result]
from tbl_err_8134

Here when ARITHABORT & ANSI_WARNINGS are set to OFF it will continue processing and will return NULL as a result. To know more about ARITHABORT you can follow this link.

Download the complete script file here.

Finding Nth highest number in SQL Server

Many times you face this question while working with real time data and most interviews. In this post we will go through different ways to find the records with Nth highest number. Use this script to populate some sample data to test the results.

CREATE TABLE PLAYER
(
PLAYERID INT IDENTITY(100,2) NOT NULL,
NAME VARCHAR(100) NOT NULL,
DOB DATETIME,
COUNTRY VARCHAR(3),
PRIZEMONEY MONEY NOT NULL
)
GO

INSERT INTO PLAYER(NAME,DOB,COUNTRY,PRIZEMONEY) VALUES
('Andre Agassi','4/29/1970','USA',$31152975),  ('Rafael Nadal','6/3/1986','ESP',$27224163), ('Boris Becker','11/22/1967','GER',$25080956),
('Yevgeny Kafelnikov','2/18/1974','RUS',$23883797), ('Ivan Lendl','3/7/1960','USA',$21262417),  ('Stefan Edberg','1/19/1966','SWE',$20630941),
('Goran Ivanisevic','9/13/1971','CRO',$19876579),  ('Michael Chang','2/22/1972','USA',$19145632), ('Lleyton Hewitt','2/24/1981','AUS',$18312036),
('Andy Roddick','8/30/1982','USA',$17109084), ('Novak Djokovic','5/22/1987','SRB',$15984098), ('Gustavo Kuerten','9/10/1976','BRA',$14807000),
('Jonas Bjorkman','3/23/1972','SWE',$14600323),  ('Marat Safin','1/27/1980','RUS',$14373291), ('Jim Courier','8/17/1970','USA',$14034132),
('Carlos Moya','8/27/1976','ESP',$13382822), ('Nikolay Davydenko','6/2/1981','RUS',$1339499),  ('Michael Stich','10/18/1968','GER',$129528),
('Juan Carlos Ferrero','2/12/1980','ESP',$12588898),  ('John McEnroe','2/16/1959','USA',$12552132), ('Thomas Muster','10/2/1967','AUT',$12225910),
('Tim Henman','9/6/1974','GBR',$11635542), ('Sergio Bruguera','1/16/1971','ESP',$11632199),  ('Patrick Rafter','12/28/1972','AUS',$11127058),
('Thomas Enqvist','3/13/1974','SWE',$10461641),  ('Petr Korda','1/23/1968','CZE',$10448900), ('Alex Corretja','4/11/1974','ESP',$10411354),
('Todd Woodbridge','4/2/1971','AUS',$10095245), ('Richard Krajicek','12/6/1971','NED',$10077425), ('Roger Federer','8/8/1981','SUI',$53362068),
('Pete Sampras','8/12/1971','USA',$43280489)
GO

Here is 4 different ways to find the same result.

--Define Level of N
DECLARE @N INT = 3

--Method-1: Without using any functions
SELECT NAME,DOB,COUNTRY,PRIZEMONEY FROM PLAYER WHERE PRIZEMONEY =
(SELECT DISTINCT PRIZEMONEY FROM PLAYER p1 WHERE (@N-1)=
(SELECT COUNT(DISTINCT p2.PRIZEMONEY) FROM PLAYER p2 WHERE p2.PRIZEMONEY > p1.PRIZEMONEY ))

--Method-2: Using TOP
SELECT NAME,DOB,COUNTRY,PRIZEMONEY FROM PLAYER WHERE PRIZEMONEY =
(SELECT TOP 1 PRIZEMONEY FROM PLAYER
WHERE PRIZEMONEY in (SELECT DISTINCT TOP (@N) PRIZEMONEY FROM PLAYER ORDER BY PRIZEMONEY DESC)
ORDER BY PRIZEMONEY)

--Method-3: Using MIN() function
SELECT NAME,DOB,COUNTRY,PRIZEMONEY FROM PLAYER WHERE PRIZEMONEY =
(SELECT MIN([t].[amt]) FROM
(SELECT DISTINCT TOP (@N) PRIZEMONEY AS [amt] FROM PLAYER ORDER BY PRIZEMONEY DESC) AS [t])

--Method-4: Using Dense_Rank()
;WITH cte
AS (
SELECT NAME,DOB,COUNTRY,PRIZEMONEY,  DENSE_RANK() OVER (ORDER BY PRIZEMONEY DESC) AS [rnk]
FROM PLAYER t
)
SELECT NAME,DOB,COUNTRY,PRIZEMONEY
FROM cte
WHERE [rnk] = @N

Also let me know your suggestions on this.

CHECK constraint with User Defined Function in SQL Server

This post describes how to use result of an user defined function with CHECK constraint in SQL Server.  For demonstration, considered a situation where it is not  allowed to insert or update records where calculated age of a person is less than 18 years as per his/her Date of Birth.

So for this first we need to create the function before creating the CHECK constraint. Here is a function which will return age as per the date of birth provided.

/*
This function will take Date Of Birth as input parameter,
and returns Age in Years to the caller
*/
CREATE FUNCTION [dbo].[fnGetAge](@DateOfBirth DATETIME)
RETURNS SMALLINT
AS
BEGIN
DECLARE @Age SMALLINT
SET @Age =(DATEDIFF(YY, @DateOfBirth, GETDATE())-
(CASE
WHEN GETDATE() >= DATEADD(YY, DATEDIFF(YY, @DateOfBirth, GETDATE()), @DateOfBirth) THEN 0
ELSE 1
END))
RETURN @Age
END;
GO

Now create a table where CHECK constraint will refer to this function to check if the age of the person meets the required criteria or not (minimum 18 Years in this case).

--Create Customer table
CREATE TABLE Customers
(
CustID INT IDENTITY(1,1) NOT NULL,
CustName VARCHAR(100) NOT NULL,
DateOfBirth DATETIME NOT NULL,
Email VARCHAR(100),
CONSTRAINT pkCustomers PRIMARY KEY(CustID),
--Calculate & check if age of customer is atleast 18 Years
CONSTRAINT chkCheckAge CHECK(dbo.fnGetAge(DateOfBirth) >= 18)
)
GO

--Populate table with some sample data.
INSERT INTO Customers(CustName, DateOfBirth, Email)
VALUES ('ABC','19810726','abc@cust.info'),
('XYZ','19840510','xyz@cust.info'),
('MNO','19720417','mno@cust.info')
GO
--Result Message
--(3 row(s) affected)

Now try to insert a record where calculated age is less than 18 years and see what happens.

--Try to insert a record where Age less than 18 Years(as per provided Date of Birth)
INSERT INTO Customers(CustName, DateOfBirth, Email)
VALUES ('TEST','20010315','test@cust.info')
GO

/*
--Error Message
Msg 547, Level 16, State 0, Line 2
The INSERT statement conflicted with the CHECK constraint "chkCheckAge".
The conflict occurred in database "SQLJourney", table "dbo.Customers", column 'DateOfBirth'.
The statement has been terminated.
*/

As the age does not meet the required criteria in defined CHECK constraint, it doesn’t allow to insert this record to the table.

Note:

CHECK constraint evaluates the provided expression while UPDATE operation as well.

You cannot DROP the function till the table (which refers to that function) exists in the database.

Generate Report with Alternate Order of an Attribute (say Gender) in SQL Server

Some times we may need to generate a report which display employees/customers in alternate order of any particular attribute.This post describes for employees with alternate order of Gender i.e. first a Male employee then a Female employee and so on or vice versa.

In SQL Server this is just a simpler way. There could be may ways to do this. Here is my set based approach with ROW_NUMBER() function.

DECLARE @EMPLOYEE TABLE
(
NAME VARCHAR(100) NOT NULL,
GENDER VARCHAR(1) NOT NULL
)

--Populate sample data
INSERT INTO @EMPLOYEE(NAME,GENDER) VALUES
('Anup','M'), ('Sheetal','F')
,('Sonam','F'), ('Payal','F')
,('Parul','F'), ('Peter','M')
,('Rahul','M'), ('Roxy','F')
,('Preeti','F'), ('Manav','M')

SELECT * FROM @EMPLOYEE

--Result
;WITH CTE(SEQ,ENAME,GENDER)
AS
(
SELECT ROW_NUMBER() OVER(PARTITION BY GENDER ORDER BY GENDER), NAME, GENDER FROM @EMPLOYEE
)
SELECT ENAME,GENDER FROM CTE
ORDER BY SEQ, GENDER DESC
GO

Also put your valuable suggestions on this.

SQL Server Reporting Tool – SqlAnswersQuery

Here is a FREE tool SqlAnswersQuery you can use for SQL Server reporting purposes.  In fact this tool is also compatible with other databases like Microsoft Access, Oracle, MySQL and DB2

You can download it from http://www.sqlanswers.com/Software/SAQ/Default.aspx.

I found it interesting. Let me know your experience with this tool.

My Windows is 32-bit or 64-bit Operating System?

While installing a application to my machine, I come up to check my Windows bit count. So here is the quick post on how to find the Windows Bit Count.

The machine I was working was with Windows XP operating system and the steps are:

Go to Start >> Programs >> Accessories >> System Tools >> System Information.

(OR)

Start >> Run >> type “winmsd.exe” >> OK

Now under “System Summery” find an item named “System type”. If its value is x86-based then the CPU is 32-bit or if it’s value is x64-based then CPU is 64-bit (the image below is of a 32-bit machine).

For more information on this, visit http://support.microsoft.com/kb/827218.

Insert Excel Data into a SQL Server table using OPENROWSET

There are many ways to load excel data into SQL Server database, using DTS, BULK import, SSIS and many others. In this post I will go with OPENROWSET. This is very useful when we need ad-hoc connection to an OLE DB source.

Here below is the image of the excel file taken for demonstrate.

Here is the script to import the file into SQL Server database using OPENROWSET.

SELECT  exl.*
INTO #myExcelData
FROM OPENROWSET ('Microsoft.Ace.OLEDB.12.0'
,'Excel 12.0; Database=C:\Import\ExcelDataImport.xlsx; HDR=YES'
,'SELECT * FROM [Sheet1$]') AS exl
GO

Now see the data imported into our table.

SELECT * FROM #myExcelData
GO

Here I have taken Excel 2007 as source and a SQL Server temp table as destination. Let me know if this is useful to you.

Fixing Error Message 1807 could not obtain exclusive lock on database ‘model’ in SQL Server

Some days back I got this question from one of my friend, when he was trying to create a New Database, he is getting the below error message.

Msg 1807, Level 16, State 3, Line 1

Could not obtain exclusive lock on database ‘model’. Retry the operation later.

Msg 1802, Level 16, State 4, Line 1

CREATE DATABASE failed. Some file names listed could not be created. Check related errors.

The Situation:

He was trying to create table on Model database first then he tried to create a New Database. As Model database is used as a template for a database being created, SQL Server tries to obtain an Exclusive Lock on Model database. So if one session is using model database SQL Server can’t obtain exclusive lock on model.

If you are using SQL Server Management Studio try these steps:

Step1: Open a New Query (say SQL Query1.sql) and select “model” from database dropdown box.

Step2: Now in another Query (say SQL Query2.sql) run the below query:

select * from sys.dm_tran_locks

Observation: You will find result like below.

(Click on the image to enlarge)

This indicates Shared Lock has been Granted on resource_database_id = 3 (database_id: 3 is for model)

Step3: Now try to create database either from Object Explorer of SQL Server Management Studio or by CREATE DATABASE statement.

Step4: This time run query in step-3 and observe the result.

(Click on the image to enlarge)

This indicates an Exclusive lock on model database but has not been granted yet (it’s in WAIT state).

Solution:

Make sure to disconnect all the sessions which uses model database then CREATE DATABASE statement later. If this problem still remains, restart management studio and try again.

This is tested in SQL Server 2008, also leave your suggestions.

UNIQUE Key Constraint with Multiple NULL values in SQL Server

This was a really interesting situation when we development team members were discussing over this topic. It was a situation where it was not possible to allow multiple NULL values into a column with UNIQUE Key defined in a table. Here I will give the problem first then we will go with the alternate solution for this.

Problem:

In SQL Server it is not allowed to insert multiple NULL values into a UNIQUE Key column. Below is the situation:

--Create employee table
CREATE TABLE EMPLOYEE
(EMPLOYEEID INT NOT NULL
CONSTRAINT PK_EMPLOYEE PRIMARY KEY CLUSTERED
,FNAME VARCHAR(100)
,LNAME VARCHAR(100)
,UNIQUEID INT UNIQUE)
GO
--Insert data values (used row constructor)
INSERT INTO EMPLOYEE(EMPLOYEEID, FNAME, LNAME, UNIQUEID)VALUES
(1,'RAHUL','MISHRA','112342115'),
(2,'SHEEL','KURANI','1725421455'),
(3,'SHEETAL','BAJAJ','1423721455'),
(4,'KAUSHIK','NARANG','1123721955'),
(5,'ADRIAN','THOULISS','1452342145')
GO
--Insert data value with NULL values to UNIQUE key column
INSERT INTO EMPLOYEE(EMPLOYEEID, FNAME, LNAME)
VALUES(6,'SAUGAT','KIOHLI')
--See below the result
SELECT EMPLOYEEID, FNAME, LNAME, UNIQUEID FROM EMPLOYEE
GO

When I try to enter another record with NULL value to UNIQUEID column, it does not allow me to do that and throws an error.

--Insert data value with NULL values to UNIQUE key column
INSERT INTO EMPLOYEE(EMPLOYEEID, FNAME, LNAME)
VALUES(7,'VAID','GURU')

--ERROR MESSAGE
--Msg 2627, Level 14, State 1, Line 2</pre>
--Violation of UNIQUE KEY constraint 'UQ__EMPLOYEE__04FF5B33023D5A04'.
--Cannot insert duplicate key in object 'dbo.EMPLOYEE'.
--The statement has been terminated.

Solution:

Here is the alternate solution for the above situation. Instead of define an UNIQUE Key to UNIQUEID column I will create another COMPUTED Column with UNIQUE Key. See how it is done:

--ALTERNATE SOLUTION
--Drop the UNIQUE key
ALTER TABLE EMPLOYEE
DROP CONSTRAINT UQ__EMPLOYEE__04FF5B33023D5A04

--Add a computed column with UNIQUE key
ALTER TABLE EMPLOYEE
ADD ALTUNIQUEID AS CASE WHEN UNIQUEID IS NULL THEN EMPLOYEEID ELSE UNIQUEID END
GO
ALTER TABLE EMPLOYEE
ADD CONSTRAINT UKey_QNIQUEID UNIQUE (ALTUNIQUEID)
GO

--Now insert multiple records with NULL to UNIQUEID column
INSERT INTO EMPLOYEE(EMPLOYEEID, FNAME, LNAME)
VALUES(7,'VAID','GURU'),(8,'TARUN','KISHORE')
GO
--See below the result
SELECT * FROM EMPLOYEE
GO

Your valuable comments are most appreciated.