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Archive for the ‘Interview Questions’ Category

SQL Server – Resolve Collation Conflict with Temp Table

Posted by Prashant on December 2, 2012


When one of our projects deployed to production application was throwing a strange error and while debugging it is found that one of the stored procedure getting failed because of collation conflict issue.

The reason behind this error is the stored procedure used a temp table to join a table on string column. Temp tables are created on TempDB, so if you don’t specify collation for a string columns they inherit the collation of TempDB. If the collation of user database doesn’t match with the collation of TempDB and you are comparing and/or joining on string columns, it may cause this collation conflict issue.

Resolution:

There are many ways to resolve this issue. However the best way as per me is to change to specify the collation of the string column of temp table to database default, so that it will inherit the collation of .user database.

CREATE TABLE #tmpTable(IDCol INT, StringCol VARCHAR(100) COLLATE DATABASE_DEFAULT)

Smart coding 🙂

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SQL Server: SET vs SELECT in Variable Assignment

Posted by Prashant on July 14, 2010


Most often we are using SET and SELECT interchangeably while assigning values to variable, however there are differences between them. In this post we will discuss some diff between SET and SELECT.

  • SET can assign value to only one variable at a time, where as SELECT can assign values to multiple variables at once.

Example:

--Declare variable
DECLARE @myVar1 INT, @myVar2 VARCHAR(10);

--Assign value to variable using SET
SET @myVar1 = 1234;
SET @myVar2 = 'SQL Journey';

--Extract varible values
SELECT @myVar1, @myVar2

--Assign value to variable using SELECT
SELECT @myVar1 = 1234, @myVar2 = 'SQL Journey';

--Extract varible values
SELECT @myVar1, @myVar2
  • SET operation will raise an error if when result of a query returning multiple values is used to assign value to a variable. However  SELECT will not raise any error and assign the last record of query result to the variable. So, Microsoft recommends SET for variable assignment

Example:

CREATE TABLE #TestVar(myID int, myComments VARCHAR(30))
INSERT INTO #TestVar VALUES(1, 'A'),(2, 'B'),(3, 'C'),(4, 'D'),(4, 'E')

--Using SET
--Declare variable
DECLARE @myVar3 VARCHAR(30);
--Assign query result (returning multiple values) to variable
SET @myVar3 = (SELECT myComments FROM #TestVar WHERE myID = 6)
--Extract varible values
SELECT @myVar3


--Msg 512, Level 16, State 1, Line 8
--Subquery returned more than 1 value.
--This is not permitted when the subquery follows =, !=, <, <= , >, >= or
--when the subquery is used as an expression


--Using SELECT
--Declare variable
DECLARE @myVar4 VARCHAR(30);
--Assign query result (returning multiple values) to variable
SELECT @myVar4 = myComments FROM #TestVar WHERE myID = 4;
--Extract varible values
SELECT @myVar4;
  • In the above case if the query result will not return any values then the variable value will be NULL either using SELECT or SET.
  • While talking in terms of performance variable assignment using SET will take less time as compared to SELECT.

Posted in Interview Questions, SQL Server | Tagged: , , , , , , | Leave a Comment »

SQL Server: Remove Non-Alphanumeric Characters from a Column

Posted by Prashant on July 13, 2010


In database we often need to clean the non-alphanumeric characters in some column of a table. Here is the code I use to remove non-alphanumeric characters in Sql Server.

Create the table with sample data:

CREATE TABLE STRSTORE(COMMENT VARCHAR(2000))

INSERT INTO STRSTORE(COMMENT) VALUES
('1.We^lcom&<e> to Sql{Journey}.&'),
('2=https://sqljourney.wordpress.com/'),
('3#[S$tring] &Cle<@anUp'),
('4-We* are'' do+ne~')

Remove non-alphanumeric characters from the column.

WHILE @@ROWCOUNT > 1
UPDATE STRSTORE
SET COMMENT = REPLACE(COMMENT, SUBSTRING(COMMENT, PATINDEX('%[^a-zA-Z0-9 ]%', COMMENT), 1), '')
WHERE PATINDEX('%[^a-zA-Z0-9 ]%', COMMENT) <> 0

SELECT	*
FROM STRSTORE

Let me also know if you have some better ideas.

Posted in Interview Questions, SQL Server | Tagged: , , , , , , , , , | 1 Comment »

SQL Server: Keyboard Shortcuts – Part 1

Posted by Prashant on July 8, 2010


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…

Posted in Interview Questions, SQL Server | Tagged: , , , | 1 Comment »

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

Posted by Prashant on July 2, 2010


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.

Posted in Interview Questions, SQL Server | Tagged: , , , , , , | 6 Comments »

Finding Nth highest number in SQL Server

Posted by Prashant on June 29, 2010


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.

Posted in Interview Questions, SQL Server | Tagged: , , , , , , , , | Leave a Comment »

CHECK constraint with User Defined Function in SQL Server

Posted by Prashant on June 25, 2010


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.

Posted in Interview Questions, SQL Server | Tagged: , , , , , , , | 4 Comments »

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

Posted by Prashant on June 23, 2010


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.

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Insert Excel Data into a SQL Server table using OPENROWSET

Posted by Prashant on June 16, 2010


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.

Posted in Interview Questions, SQL Server, SSIS | Tagged: , , , , , , | 4 Comments »

UNIQUE Key Constraint with Multiple NULL values in SQL Server

Posted by Prashant on June 12, 2010


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.

Posted in Interview Questions, SQL Server | Tagged: , , , , , | 2 Comments »

 
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