adding sql problem

Author: shreyashji

36.20 How to find strings with lower case characters Case Insensitive Collate.sql

@@ -4,4 +4,5 @@
 select upper(FirstName) as uper_first,FirstName  from dbo.Emp 
 where upper(FirstName) COLLATE Latin1_Genral_CS_AS != FirstName;
 
-select SERVERPROPERTY('Collation')# CI MEANS CASE SENSITIVE
+select SERVERPROPERTY('Collation')# CI MEANS CASE SENSITIVE
+

36.23 How to find n consecutive date records Sales for at least n consecutive days.sql

@@ -0,0 +1,21 @@
+#How to find n consecutive date records | Sales for at least n consecutive days.
+#DATEADD(DAYS,SPECIFY WHAT VALUE OF DATE(HOW many days want to subtract), from which you want to subtract)
+select  order_date ,
+DATEADD(D, - row_number() over(order by order_date), order_date) as col1
+from superstore_orders;
+
+#subtract row no from oderd date col,so we get same value or date for consecutive records,perfrom count on col1 it has same value,use cte 
+with cte as (
+select  order_date ,
+DATEADD(D, - row_number() over(order by order_date), order_date) as col1
+from superstore_orders)
+select count(*) as cnt,MIN(order_date) as startdate as od,max(order_date) as enddate FROM cte group by col1
+
+#atleast 3 consecutive date
+with cte as (
+select  order_date ,
+DATEADD(D, - row_number() over(order by order_date), order_date) as col1
+from superstore_orders)
+select startdate,enddate from 
+(select count(*) as cnt,MIN(order_date) as startdate as od,max(order_date) as enddate FROM cte group by col1) as  tbla
+where cnt>=3

36.24 No sales for n consecutive days Identify date gaps.sql

@@ -0,0 +1,28 @@
+#No sales for n consecutive days | Identify date gaps
+select order_date from superstore_orders
+order by order_date;
+
+#we have to get the next record ,so we use lead(next consecutive date), #01-04-20	01-06-18 gap
+select order_date,
+lead(order_date)  over(order by order_date) as Lead_date
+from superstore_orders
+order by order_date;
+
+#now we need to find difference value
+#ssms only
+select order_date,
+datediff(day, order_date, (lead(order_date)  over(order by order_date) ) ) as Gap
+from superstore_orders
+order by order_date;
+
+#need to find whenn there is no sales,put above in subquerry
+select * from (
+select order_date,
+datediff(day, order_date, (lead(order_date)  over(order by order_date) ) ) as Gap
+from superstore_orders) as NoSales
+where  NoSales.gap > 1 # greatedr than 1
+
+
+
+
+

36.25 How to find First and Last day of week.sql

@@ -0,0 +1,9 @@
+#How to find First and Last day of week
+
+select datediff(wk,'2021-08-05',GetDate());# get 1 week differnce from this
+
+select dateadd(wk,datediff(wk,'2021-08-01',GetDate()),'2021-08-01');
+dateadd(wk,datediff(wk,'2021-08-01',GetDate()) + 1 ,'2021-08-01'); # to rrive at the last week
+
+
+select datediff('2021-08-05',date)

36.26 How to find top n salaries in a department and sum the remaining as 'Others'.sql

@@ -0,0 +1,13 @@
+select emp_name,department_id,salary from
+(select emp_name,department_id,salary
+,rank() over(partition by department_id order by salary desc) as maxsal
+ from empm) as empsal
+ where empsal.maxsal <= 2 #top 2 highest salary
+union all  #union have impact on performance,union all 
+select 'others' as emp_name,department_id,sum(salary) from
+(select department_id,salary
+,rank() over(partition by department_id order by salary desc) as maxsal
+ from empm) as empsal
+ where empsal.maxsal > 2 #>2 does not lie in top 2 salary of the dapartment
+ group by department_id
+ #union all combines duplicate record in two queries if exist,union combines only distict record

36.27 Compare monthly sales with previous month, same month previous year, first month of year.sql

@@ -0,0 +1,39 @@
+#In this video, we write a SQL to compare monthly sales -
+#1) With previous month
+#2) same month previous year
+#3) first month of year
+
+#Compare monthly sales with previous month, same month previous year, first month of year
+select order_date,sales
+from superstore_orders;
+
+#sales of each month from different years ,LAG: bring data from previous row
+select  year(order_date),Month(order_date),sum(sales) , 
+lag(sum(sales)) OVER(order by year(order_date),Month(order_date) )
+from superstore_orders
+where year(order_date) in (2012,2013)
+group by year(order_date),Month(order_date);
+
+#to get month by month comparison,compare monthly data with revious month data 
+select  year(order_date),Month(order_date),sum(sales),
+lag(sum(sales)) OVER(order by year(order_date),Month(order_date) )
+from superstore_orders
+where year(order_date) in (2012,2013)
+group by year(order_date),Month(order_date);
+
+#2compare same month from previous year,default offset is 1,lag(sum(sales),offset)
+select  year(order_date),Month(order_date), sum(sales) -
+lag(sum(sales),12) OVER(order by year(order_date),Month(order_date) )
+from superstore_orders
+where year(order_date) in (2012,2013)
+group by year(order_date),Month(order_date);
+
+#compare data of 1st month of same year,eg. august,subtract 7
+#october (10)-1=9
+#fetch month(order_date) and -1 from currnt month
+#3.compare monthly sales from 1st moth of that year
+select  year(order_date),Month(order_date), sum(sales),
+lag(sum(sales),Month(order_date) - 1) OVER(order by year(order_date),Month(order_date) )
+from superstore_orders
+where year(order_date) in (2012,2013)
+group by year(order_date),Month(order_date);

36.28 How to find number of emails from the same domain.sql

@@ -0,0 +1,12 @@
+#How to find number of emails from the same domain
+
+select * from entries;
+#extract domian from charindex ,get position of occurence of '@'
+select CharIndex('@',email),email from entries;
+
+#extract the string after @,total length of string - first 6 char
+select right(email,len(email)- CharIndex('@',email)) from entries;
+
+#How to find number of emails from the same domain
+select count(*) , right(email,len(email)- CharIndex('@',email)) from entries
+group by right(email,len(email)- CharIndex('@',email));