Professional Networking Social Media Website SQL Analysis, PostgreSQL

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-- Website SQL Analysis Conducted by Miranda Zhao

CREATE TABLE Lpayment
    ("date" timestamp,"memberid" varchar(11),"amount" int)
;
    
INSERT INTO Lpayment
    ("date","memberid","amount")
VALUES
    ('2019-01-03','A',33.9),
    ('2019-01-04','C',19.3),
    ('2019-01-07','A',29.1),
    ('2019-01-05','B',32.2),
    ('2019-01-07','C',35.5)
;

CREATE TABLE Luser
    ("date" timestamp,"memberid" varchar(11))
;
    
INSERT INTO Luser
    ("date","memberid")
VALUES
    ('2019-01-01','A'),
    ('2019-01-01','A'),
    ('2019-01-04','A'),
    ('2019-01-04','C'),
    ('2019-01-03','A'),
    ('2019-01-03','B'),
    ('2019-01-05','B'),
    ('2019-01-06','B'),
    ('2019-01-06','B'),
    ('2019-01-04','C'),
    ('2019-01-05','C'),
    ('2019-01-05','D'),
    ('2019-01-07','A'),
    ('2019-01-07','C')
;

CREATE TABLE Ldate
    ("date" timestamp)
;
    
INSERT INTO Ldate
    ("date")
VALUES
    ('2019-01-01'),
    ('2019-01-02'),
    ('2019-01-03'),
    ('2019-01-04'),
    ('2019-01-05'),
    ('2019-01-06'),
    ('2019-01-07'),
    ('2019-01-08')
;

--Q1: Total Revenue & Payers: Create a table to display total revenue & number of buyers on each day
SELECT Lpayment.date AS Date, sum(amount) AS Revenue, count(memberid) AS Num_of_Buyers
From Lpayment
RIGHT JOIN Ldate on Lpayment.date = Ldate.date
where Lpayment.date is not null
GROUP BY Lpayment.date
ORDER BY 1,2,3;

--Q2: Top payers: Create a table to display the top one payer on each day

WITH cte AS
(
   SELECT *,
         ROW_NUMBER() OVER (PARTITION BY date ORDER BY amount DESC) AS rn
   FROM Lpayment
)
SELECT date, memberid
FROM cte
WHERE rn = 1
ORDER BY 1,2;

--Q3: A distribution of # days active within a week: Create a table to show how many members are active for 1 day, 2days, 3days,…7days during 01/01-07/01.

WITH cte1 AS 
(WITH cte AS
(
    SELECT DISTINCT date, memberid
    FROM Luser
    WHERE date BETWEEN '2019-01-01' AND '2019-01-07'
    GROUP BY memberid, date
) 
    SELECT memberid, COUNT(memberid) as days_active
    FROM cte
    GROUP by memberid)
    
    SELECT days_active, COUNT(days_active) as count
    FROM  cte1
    GROUP BY days_active
    ORDER BY 1,2;

--Q4: Active but not payers: Create a table to display people who were active but did not pay on each day
with cte AS 
(select Luser.date, Luser.memberid, amount from Luser
left join Lpayment 
on Luser.date =  Lpayment.date and Luser.memberid =  Lpayment.memberid)
select date, memberid from cte
where amount is null
ORDER BY 1,2;


absolute service time: 0,45 sec