Posts Tagged ‘MySQL’

How to use Variable in MySQL Like Clause ?

October 24, 2018 Leave a comment

Here is an example to use SQL variable in the mysql select query like clause

SET @name = 'advancewebsoftware';

SELECT * from `user` WHERE name LIKE CONCAT('%', @name, '%');

Hope this help!


Categories: Sql Server Tags: , ,

MYSQL : How to skip first 10 records from a SELECT query and take next 10?

October 4, 2017 Leave a comment

We know there are skip and take operators available in linq to skip and take

The Take operator is used to return a given number of rows from a object list and the Skip operator skips over a specified number of rows in object list.

var employee = (from emp in employeeContext.EMPLOYEEs

where emp.SALARY < 1300

select emp).Skip(2).Take(5);

Is there a way in MySQL to achieve the same like have the first 10 result from a SELECT query skipped and select next 10?

Answer : Yes we can use LIMIT with two parameter

Introduction to MySQL LIMIT clause

The LIMIT clause is used in the SELECT statement to constrain the number of rows in a result set. The LIMIT clause accepts one or two arguments. The values of both arguments must be zero or positive integers.

The following illustrates the LIMIT clause syntax with two arguments:

Let’s examine the LIMIT clause parameters:

  • The offset specifies the offset of the first row to return. The offset of the first row is 0, not 1.
  • The count specifies the maximum number of rows to return.

When you use the LIMIT clause with one argument, this argument will be used to determine the maximum number of rows to return from the beginning of the result set.

SELECT *  FROM table LIMIT count;

Use LIMIT with two parameters. For example, to return results 11-60 (where result 1 is the first row), use:

SELECT * FROM foo LIMIT 10, 50
Hope this help !

Get Row Number in MySQL

SQL Server

To return a row number with record we have ROW_NUMBER() in SQL which  returns the sequential number of a row within a partition of a result set, starting at 1 for the first row in each partition.


OVER ( [ PARTITION BY value_expression , … [ n ] ] order_by_clause )


Here I am explaining how to return row number in my sql

For Example we have following table

  fruit  VARCHAR(32),
  amount DECIMAL

(NULL, 'apple', 12.75), 
(NULL, 'orange', 1.89), 
(NULL, 'pear', 19.23),
(NULL, 'banana', 4.25), 
(NULL, 'cherry', 123.75), 
(NULL, 'plum', 23.15);
SELECT fruit, amount FROM tblSales ORDER BY amount DESC;

If we run this query it will result like
| fruit  | amount |
| cherry |    124 |
| plum   |     23 |
| pear   |     19 |
| apple  |     13 |
| banana |      4 |
| orange |      2 |

Now here is the query with ranking

SET @rank=0;
SELECT @rank:=@rank+1 AS rank, fruit, amount FROM tblSales
 ORDER BY amount DESC;

| rank | fruit  | amount |
|    1 | cherry |    124 |
|    2 | plum   |     23 |
|    3 | pear   |     19 |
|    4 | apple  |     13 |
|    5 | banana |      4 |
|    6 | orange |      2 |
Hope it helps!

Age calculation in mysql Select query

Following is the way to calculate age based on date in mysql select query

SELECT name, 

| name | birth | CURDATE() | age | 
| Fluffy | 1993-02-04 | 2003-08-19 | 10 | 
| Claws | 1994-03-17 | 2003-08-19 | 9 |
| Buffy | 1989-05-13 | 2003-08-19 | 14 | 
| Fang | 1990-08-27 | 2003-08-19 | 12 | 
| Bowser | 1989-08-31 | 2003-08-19 | 13 | 
| Chirpy | 1998-09-11 | 2003-08-19 | 4 | 
| Whistler | 1997-12-09 | 2003-08-19 | 5 | 
| Slim | 1996-04-29 | 2003-08-19 | 7 | 
| Puffball | 1999-03-30 | 2003-08-19 | 4 | 

Hope this is helpful !

Function to find dates in current week in mysql

MySQL YEARWEEK() returns year and week number for a given date.

Syntax: YEARWEEK(date_value,Mode)

Parameters or Arguments

A date or datetime value from which to extract the year and week.
Optional. It is used to specify what day the week starts on. It can be one of the following:

mode Explanation Week Value
0 First day of the week is Sunday 0-53
1 First day of the week is Monday and the first week has more than 3 days 0-53
2 First day of the week is Sunday 1-53
3 First day of the week is Monday and the first week has more than 3 days 1-53
4 First day of the week is Sunday and the first week has more than 3 days 0-53
5 First day of the week is Monday 0-53
6 First day of the week is Sunday and the first week has more than 3 days 1-53
7 First day of the week is Monday 1-53

Example : select YEARWEEK(now());

The following statement will return the year and week number of the date 2016-05-19.

SELECT YEARWEEK(‘2016-05-19’);




Hope this help!

Comma separated values in mysql


GROUP_CONCAT function concatenates values within each group defined by GROUP BY clause.

Syntax GROUP_CONCAT([DISTINCT] exp [ORDER BY sorting] [SEPARATOR ‘sep’])
Quick Example SELECT GROUP_CONCAT(city) FROM cities GROUP BY state;
Separator Comma (,) by default, ” eliminates separator
NULL Values Skipped
Max Length 1024 by default, specified by group_concat_max_len system variable


     city VARCHAR(70),
     state CHAR(2)
   INSERT INTO cities VALUES ('San Francisco', 'CA');
   INSERT INTO cities VALUES ('San Diego', 'CA');
   INSERT INTO cities VALUES ('Los Angeles', 'CA');
   INSERT INTO cities VALUES ('Austin', 'TX');
   INSERT INTO cities VALUES ('Houston', 'TX');

   SELECT state, GROUP_CONCAT(city)
   FROM cities
   GROUP BY state;
The result:
CA San Francisco,San Diego,Los Angeles
TX Austin,Houston
SQL Server:
SELECT state,
STUFF((SELECT ',' + city FROM cities WHERE state = c.state FOR XML PATH('')),1 ,1 ,'')
FROM cities c
GROUP BY state

Hope this help !
Rahul Bhatia

MySQL Error: The user specified as a definer does not exist

October 17, 2014 Leave a comment

Error :

the user specified as a definer (‘root’@’%’) does not exist


Solution :

grant all on *.* to ‘root’@‘%’ identified by ‘password’ with grant

Categories: Sql Server Tags: , , ,

Cannot drop database because it is currently in use.

This is a very common error when DROP Database is command is executed and the database is not dropped.

The following commands will raise above error:
USE AdventureWorks;
DROP DATABASE AdventureWorks;

The following commands will not raise an error and successfully drop the database:

USE Master;
DROP DATABASE AdventureWorks;

If you are still getting the error after you try using
use master
drop database (databaseName)

Close SQL Server Management Studio completely. Open it again and connect as normal. Now you will be able to drop the database with
use master
drop database (databaseName)

Hope it helps!