adding sql problem

shreyashji · shreyashji · commit 26fd7581ac1e · 2022-03-10T21:42:04.000+05:30
diff --git a/40.AK-SQL Cross Join  Use Cases  Master Data  Performance Data.sql b/40.AK-SQL Cross Join  Use Cases  Master Data  Performance Data.sql
@@ -0,0 +1,59 @@
+#create table productss (id int,name varchar(10));
+#insert into productss VALUES (1, 'A'),(2, 'B'),(3, 'C'),(4, 'D'),(5, 'E');
+#create table colors (color_id int,color varchar(50));
+#insert into colors values (1,'Blue'),(2,'Green'),(3,'Orange');
+#create table sizes(size_id int,size varchar(10));
+#insert into sizes values (1,'M'),(2,'L'),(3,'XL');
+#create table transactions(order_id int,product_name varchar(20),color varchar(10),size varchar(10),amount int);
+#insert into transactions values (1,'A','Blue','L',300),(2,'B','Blue','XL',150),(3,'B','Green','L',250),(4,'C','Blue','L',250),
+#(5,'E','Green','L',270),(6,'D','Orange','L',200),(7,'D','Green','M',250);
+select * from productss;
+select * from colors;
+select * from sizes;
+select * from transactions;
+#cross join
+select * from productss,colors;
+select p.*,c.* from productss p,colors c;
+
+#use case of cross join
+#prepare master data
+select * from transactions;
+#give me the sales for each product size,color
+select product_name,color,size,sum(amount) as total_amount
+ from transactions
+group by product_name,color,size;
+
+#if not sold show zero
+with master_data_cte as (
+select p.name as product_name,c.color,s.size from productss p,colors c,sizes s)
+, sales_cte as (select product_name,color,size,sum(amount) as total_amount
+from transactions
+group by product_name,color,size)
+select md.product_name,md.color,md.size , isnull(s.total_amount,0) as total_amount from master_data_cte md 
+left join 
+sales_cte s on md.product_name = s.product_name
+and md.color = s.color and md.size = s.size
+order by total_amount;
+
+#use case 2 : prepare large no . of  rows for performance testing
+#unique order id
+select row_number() over(order by t.order_id) as order_id,
+t.Product_Name,t.color,t.size, 
+Case when row_number() over(order by t.order_id)%3=0 then 'L' else 'XL' end size,t.amount
+from transactions t,superstore_orders o;
+
+#more rows by joining again with transaction
+select row_number() over(order by t.order_id) as order_id,
+t.Product_Name,t.color,t.size, 
+Case when row_number() over(order by t.order_id)%3=0 then 'L' else 'XL' end size,t.amount
+from transactions t,superstore_orders o,transactions t1;
+#table structure will be created
+select * into transactions_test from transactions where 1=2;
+select * from transactions_test;
+#now insert
+insert into transactions_test
+select row_number() over(order by t.order_id) as order_id,
+t.Product_Name,t.color,t.size, 
+Case when row_number() over(order by t.order_id)%3=0 then 'L' else 'XL' end size,t.amount
+from transactions t,superstore_orders o,transactions t1;
+
diff --git a/41.cs-Highest Energy Consumption.sql b/41.cs-Highest Energy Consumption.sql
@@ -0,0 +1,18 @@
+#Highest Energy Consumption
+#Find the date with the highest total energy consumption from the Meta/Facebook
+#data centers. Output the date along with the total energy consumption across all data centers.
+
+WITH amazon_datacentre as 
+(select * from fb_eu_energy
+UNION
+select * from fb_asia_energy
+UNION
+select * from fb_na_energy)
+select date,tot_consumption from ( 
+select date,tot_consumption,
+dense_rank() over(order by tot_consumption desc) rnk from 
+(select date,sum(consumption) as tot_consumption from amazon_datacentre
+group by date) b1 )b2
+where b2.rnk=1
+order by date
+;
diff --git a/42. AK -Most Asked SQL JOIN based Interview Question no. of Records after 4 types of JOINs.sql b/42. AK -Most Asked SQL JOIN based Interview Question no. of Records after 4 types of JOINs.sql
@@ -0,0 +1,31 @@
+#Most Asked SQL JOIN based Interview Question | # of Records after 4 types of JOINs
+#CREATE TABLE t1(id1 int );
+#CREATE TABLE t2(id2 int );
+
+#insert into t1 values (1);
+#insert into t1 values (1);
+
+#insert into t2 values (1);
+#insert into t2 values (1);
+#insert into t2 values (1);
+#delete from t2;
+
+select * from t1;
+select * from t2;
+#how join works on duplicate keys
+#inner join,it will go and join with all 3 records from right table 2*3-6 
+select * from t1
+inner join t2 on t1.id1=t2.id2;
+ #result on inner join+not matching records from left
+select * from t1
+left join t2 on t1.id1=t2.id2;
+
+#right= non matching records from left,but here all are matching so same no of records like above
+#inner join+not matching records from right
+select * from t1
+right join t2 on t1.id1=t2.id2;
+
+#matching records from oth table & non matching records from both table,but there is no unmatching record,so result will be same
+#ssms only
+select * from t1
+full outer join t2 on t1.id1 = t2.id2;
diff --git a/43. AK- all joins most asked question.sql b/43. AK- all joins most asked question.sql
@@ -0,0 +1,39 @@
+#Most Asked SQL JOIN based Interview Question | # of Records after 4 types of JOINs
+CREATE TABLE t3(id int );
+CREATE TABLE t4(id int );
+
+insert into t3 values (1);
+insert into t3 values (1);
+insert into t3 values (2);
+
+insert into t4 values (1);
+insert into t4 values (1);
+insert into t4 values (1);
+insert into t4 values (3);
+
+#delete from t2;
+select * from t3;
+select * from t4;
+
+select * from t3
+inner join t4 on t3.id=t4.id;
+ #result on inner join+not matching records from left
+select * from t3
+left join t4 on t3.id=t4.id;
+
+#right= non matching records from left
+#inner join+not matching records from right
+select * from t3
+right join t4 on t3.id=t4.id;
+
+#matching records from oth table & non matching records from both table
+#ssms only
+select * from t3
+full outer join t4 on t3.id = t4.id;
+
+#mysql full outer join
+SELECT * FROM t3
+LEFT JOIN t4 ON t3.id = t4.id
+UNION 
+SELECT * FROM t3
+right JOIN t4 ON t3.id = t4.id
diff --git a/44. AK- all joins with diff values.sql b/44. AK- all joins with diff values.sql
@@ -0,0 +1,47 @@
+#Most Asked SQL JOIN based Interview Question | # of Records after 4 types of JOINs
+#CREATE TABLE t5(id int );
+#CREATE TABLE t6(id int );
+
+#insert into t5 values (1);
+#insert into t5 values (1);
+#insert into t5 values (2);
+#insert into t5 values (2);
+#insert into t5 values (4);
+
+#insert into t6 values (1);
+#insert into t6 values (1);
+#insert into t6 values (1);
+#insert into t6 values (3);
+#insert into t6 values (2);
+#insert into t6 values (2);
+
+#delete from t2;
+select * from t5;
+select * from t6;
+
+select * from t5
+inner join t6 on t5.id=t6.id;
+#result on inner join+not matching records from left,6+ 2*1=(2) 2 records of 2
+ 
+#all are matching in left table so sresult will be same as inner join
+select * from t5
+left join t6 on t5.id=t6.id;
+
+#right= non matching records from left
+#inner join+not matching records from right
+#6+1*2 =2 records of 2+ for non matching 3 1 record will be there =9 records
+select * from t5
+right join t6 on t5.id=t6.id;
+
+#matching records from oth table & non matching records from both table
+#ssms only
+# 3*2 +2 +1 + 2*2 + 4 is not matching so 1 record for this + 3 not matching = 12 records
+select * from t3
+full outer join t4 on t3.id = t4.id;
+
+#mysql full outer join
+SELECT * FROM t3
+LEFT JOIN t4 ON t3.id = t4.id
+UNION 
+SELECT * FROM t3
+right JOIN t4 ON t3.id = t4.id
diff --git a/45. AK- ALL JOINS with nulls.sql b/45. AK- ALL JOINS with nulls.sql
@@ -0,0 +1,52 @@
+#Most Asked SQL JOIN based Interview Question | # of Records after 4 types of JOINs
+#CREATE TABLE t7(id int );
+#CREATE TABLE t8(id int );
+
+#insert into t7 values (1);
+#insert into t7 values (1);
+#insert into t7 values (2);
+#insert into t7 values (2);
+#insert into t7 values (4);
+#insert into t7 values (null);
+
+#insert into t8 values (1);
+#insert into t8 values (1);
+#insert into t8 values (1);
+#insert into t8 values (3);
+#insert into t8 values (2);
+#insert into t8 values (2);
+#insert into t8 values (null);
+
+#delete from t2;
+select * from t7;
+select * from t8;
+
+#null cant be compare with other null,so null also not joined with another null
+
+select * from t7
+inner join t8 on t7.id=t8.id;
+#result on inner join+not matching records from left,6+ 2*1=(2) 2 records of 2
+ 
+#all are matching in left table so sresult will be same as inner join
+# 2*3 + 4+ 1 not matching+1 not matching =12
+select * from t7
+right join t8 on t7.id=t8.id;
+
+#right= non matching records from left
+#inner join+not matching records from right
+#2*3 +4+ 1 not m+1 not m
+select * from t7
+left join t8 on t7.id=t8.id;
+
+#matching records from oth table & non matching records from both table
+#ssms only
+# 3*2 +2 +1 + 2*2 + 4 is not matching so 1 record for this + 3 not matching = 12 records
+select * from t7
+full outer join t8 on t7.id=t8.id;
+
+#mysql full outer join
+SELECT * FROM t3
+LEFT JOIN t4 ON t3.id = t4.id
+UNION 
+SELECT * FROM t3
+right JOIN t4 ON t3.id = t4.id
