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.