Completed Sql6

Problem1_GamePlay2.sql

@@ -0,0 +1,4 @@
+# Write your MySQL query statement below
+select a.player_id, a.device_id 
+from activity a
+where a.event_date in (select min(b.event_date) from activity b where a.player_id=b.player_id)

Problem2_GamePlay3.sql

@@ -0,0 +1,5 @@
+# Write your MySQL query statement below
+select 
+player_id,event_date,
+sum(games_played) over(partition by player_id order by event_Date) as games_played_so_far
+from activity;

Problem3_shortestDist.sql

@@ -0,0 +1,6 @@
+# Write your MySQL query statement below
+select 
+round(min(sqrt(pow(p2.x-p1.x,2)+pow(p2.y-p1.y,2))),2) as shortest
+from point2d p1 
+inner join point2d p2
+on p1.x!=p2.x or p1.y!=p2.y

Problem4_Combine2Tables.sql

@@ -0,0 +1,4 @@
+# Write your MySQL query statement below
+select p.firstname,p.lastname,a.city,a.state
+from Person P
+left join address a on p.personid=a.personid

Problem5_CustomersWithIncreasingPurchase.sql

@@ -0,0 +1,14 @@
+# Write your MySQL query statement below
+with cte as (
+select distinct customer_id,year(order_date) as year,
+sum(price) over(partition by customer_id,year(order_date)) as price
+from Orders order by customer_id,year)
+
+select 
+distinct c1.customer_id
+from cte c1
+left join cte c2 on c1.customer_id=c2.customer_id
+and c1.year +1= c2.year
+and c1.price < c2.price
+group by c1.customer_id
+having count(*)-count(c2.customer_id)=1;