MySQL: Count two date columns and group by day - mysql

I need to draw a line chart that will visualize both the orders and pickups for each day between certain dates. The order and pickup dates are stored in unixtime. My table looks something like this:
id order_date pickup_date
-------------------------------
1 1472749664 1472133376
2 1472551372 1472567548
3 1472652545 1472901368
4 1473154659 1473512323
5 1473512923 1475229824
6 1475586643 1475652635
What I am after is something like this
date orders pickups
-------------------------------
01-09-2016 1 0
02-09-2016 4 1
03-09-2016 3 2
04-09-2016 7 1
05-09-2016 0 0
06-09-2016 1 1
07-09-2016 6 3
08-09-2016 0 0
08-09-2016 3 5
10-09-2016 2 4
I know I can count based on one column, for example:
SELECT
COUNT(id) AS orders,
FROM_UNIXTIME(order_dates, '%d-%m-%Y') AS date
FROM orders
GROUP BY request_date
But I not sure how to count two columns and group them for each day.

You could use a query like this:
SELECT sum(orders) as orders, sum(pickups) as pickups, date
FROM (
SELECT
COUNT(id) AS orders, 0 as pickups,
FROM_UNIXTIME(`order_date`, '%d-%m-%Y') AS date
FROM orders
GROUP BY order_date
UNION
SELECT
0 AS orders, COUNT(id) as pickups,
FROM_UNIXTIME(`pickup_date`, '%d-%m-%Y') AS date
FROM orders
GROUP BY pickup_date ) ut
GROUP BY date
Here is a fiddle.

Related

How to find which number of orders for a customer

Given a table orders
id
customer_id
created_at
1
1
2022-09-01
2
2
2022-09-02
3
1
2022-09-03
4
1
2022-09-04
5
2
2022-09-04
How do I produce a column that describes which number in the series for the customers the order is?
Example
id
customer_id
created_at
order number
1
1
2022-09-01
1
2
2
2022-09-02
1
3
1
2022-09-03
2
4
1
2022-09-04
3
5
2
2022-09-5
2
You can use a window function for that. With a cumulative count over a partition by customer id, you get exactly the order number you need:
select orders.*,
count(*) over (partition by customer_id order by id) order_number
from orders
order by id;
In MySQL 5.7 you could do this:
select customer_id,
(select count(*)
from orders
where customer_id = main.customer_id and id <= main.id)
from orders main;

SQL query to find the repetitions of data

I've a sample data
id date user_id customer_id status
1 2022-06-23 1 12 no response
2 2022-06-23 1 12 no response
3 2022-06-24 1 12 no response
4 2022-06-23 2 15 no response
4 2022-06-23 2 15 successful
5 2022-06-23 3 16 call later
I need to fetch those kind of records where a user_id called the same customer_id on the same day got only no response status more than once but not any other statuses.
The result for the above example would be
id
1
2
You can use aggregation and the conditions in the HAVING clause:
SELECT user_id, customer_id, date
FROM tablename
GROUP BY user_id, customer_id, date
HAVING COUNT(*) > 1 AND SUM(status <> 'no response') = 0
If you want the respective rows of the table use the above query with the operator IN:
SELECT *
FROM tablename
WHERE (user_id, customer_id, date) IN (
SELECT user_id, customer_id, date
FROM tablename
GROUP BY user_id, customer_id, date
HAVING COUNT(*) > 1 AND SUM(status <> 'no response') = 0
);
See the demo.

Get monthly counts on multiple dates

I have a table that looks like this
id
date registered
date cancelled
1
2021-01-01
2021-03-02
2
2021-01-05
2021-01-21
3
2021-02-04
2021-02-25
4
2021-02-16
2021-03-26
How do I generate a query in mysql that will give me counts of cancelled and registered for each month.
I can do it for just one of the dates but don't know how to combine for both dates.
For eg for a single date I would do this.
SELECT date_format(`users`.`dateregistered`,_utf8'%Y-%m') AS `DateREegistered`, count(0) AS `Registration Count`
FROM `users`
GROUP BY date_format(`users`.`dateregistered`,_utf8'%Y-%m')
But I want something like this
Date
Registered Count
Cancelled Count
2021-01
2
1
2021-02
2
1
2021-03
0
2
Please let me know if you have any ideas.
You can join the distinct months appearing in date registered and date registered to the table and use conditional aggregation:
SELECT t.Date,
SUM(t.Date = date_format(dateregistered, '%Y-%m')) `Registered Count`,
SUM(t.Date = date_format(datecancelled, '%Y-%m')) `Cancelled Count`
FROM (
SELECT date_format(dateregistered, '%Y-%m') Date FROM users
UNION
SELECT date_format(datecancelled, '%Y-%m') FROM users
) t INNER JOIN users u
ON t.Date IN (date_format(dateregistered, '%Y-%m'), date_format(datecancelled, '%Y-%m'))
GROUP BY t.Date
See the demo.
Results:
Date
Registered Count
Cancelled Count
2021-01
2
1
2021-02
2
1
2021-03
0
2

How to find whether the same customers who ordered this month also ordered the next month?

I have an orders table
Order_id User_id Order_date
1 32 2020-07-19
2 24 2020-07-21
3 27 2020-07-27
4 24 2020-08-14
5 32 2020-08-18
6 32 2020-08-19
7 58 2020-08-20
Now I want to find how many of the users ordered in 1st month also ordered in the next month. In this case, user_id's 32,24,27 ordered in 7th month but only 24 and 32 ordered in the next month.
I want the result to be like :
Date Retained_Users Total_users
2020-07 Null 3
2020-08 2 3
I'm lost here. Can someone please help me with this?
In MySQL 8.0, you can do this with window functions:
select
order_month,
count(distinct case when cnt_orders_last_month > 0 then user_id end) retained_users,
count(distinct user_id) total_users
from (
select
user_id,
date_format(order_date, '%Y-%m-01') as order_month,
count(*) over(
partition by user_id
order by date(date_format(order_date, '%Y-%m-01'))
range between interval 1 month preceding and interval 1 day preceding
) cnt_orders_last_month
from mytable
) t
group by order_month
The logic lies in the range specification of the window function; it orders record by month, and counts how many orders the customer placed last month. Then all that is left to do is aggregate and count distinct users.
Demo on DB Fiddle

Group BY product using MAX(price) or MAX(date) according to time interval

I've been searching for answers for 2 day and still nothing. Please, help me.
I have a database with products, product's prices and the date when this prices were registered:
product_id | price | date
-------------------------
1 | 8.95 | 2012-12-01
2 | 3.40 | 2012-12-01
1 | 9.05 | 2012-12-19
3 | 2.34 | 2012-12-24
3 | 2.15 | 2012-12-01
1 | 8.80 | 2012-12-19
1 | 8.99 | 2012-12-02
2 | 3.45 | 2012-12-02
Observe that is possible to have different price values for a product on the same day (rows 3 and 6). This is because there are many suppliers for a single product. There is a supplier column on database too, but I found it irrelevant for the solution. You can add it to the solution if I'm wrong.
Basically what I want is to write a query that returns two combined sets of data, as follow:
First set is made by minimum price of products inserted in the last month. As today is jan, 15, query should read rows 3, 4 and 6, apply the minimum price, and return only rows 4 and 6, both with minimum price for that product on the last month.
Second set is made by last products inserted, with no price registered on last month. i.e, for products not shown in the first set, query should search for the last inserted ones.
I hope that is clear. Ask me more if it isn't.
The query result for this database should be:
product_id | price | date
-------------------------
1 | 8.80 | 2012-12-19 <-Min price for product 1 on last month
3 | 2.34 | 2012-12-24 <-Min price for product 3 on last month
2 | 3.45 | 2012-12-02 <-No reg for product 2 on last month, show last reg.
I've tried everything: UNION, (DATE_SUB(CURDATE(), INTERVAL 1 MONTH), MIN(price), MAX(date) etc, etc. Nothing works. I don't know where to search now, please help me.
(SELECT product_id, MIN(price), date
FROM products
WHERE date + INTERVAL 1 MONTH > NOW()
GROUP BY product_id)
UNION
(SELECT product_id, price, MAX(date)
FROM products
WHERE product_id NOT IN (SELECT product_id
FROM products
WHERE date + INTERVAL 1 MONTH > NOW()
GROUP BY product_id)
GROUP BY product_id)
This should work but I'm not sure it's the most optimized way to do it.
something like this will do the trick:
SELECT * FROM (
SELECT DISTINCT b.product_id, IF (c.min IS NULL,(SELECT ROUND(e.price,2) FROM products AS e WHERE e.product_id = b.product_id ORDER BY e.date DESC LIMIT 1 ),c.min) AS min, IF (c.date IS NULL,(SELECT f.date FROM products AS f WHERE f.product_id = b.product_id ORDER BY f.date DESC LIMIT 1 ),c.date) AS date, IF(c.min IS NULL,'<-No reg for product 2 on last month, show last reg.','<-Min price for product 1 on last month') as text FROM products AS b
LEFT JOIN
(SELECT a.product_id, round(MIN(a.price),2) AS min, a.date FROM products AS a WHERE a.date BETWEEN DATE_SUB(CURDATE(), INTERVAL 1 MONTH) AND CURDATE() GROUP BY a.product_id) AS c
ON (b.product_id = c.product_id)
) AS d
ORDER BY d.text, d.product_id
Gives output:
product_id|min|date|text
1|8.80|2012-12-19|<-Min price for product 1 on last month
3|2.34|2012-12-24|<-Min price for product 1 on last month
2|3.45|2012-12-02|<-No reg for product 2 on last month, show last reg.
Break it down into several sub-queries:
Products with prices in the last month, min price
join in date for that price
UNION
Products with no-prices in the last month, max date
join in price on that date
SQL Fiddle
Here
Query
SELECT MINPRICE.product_id, P.date, MINPRICE.price
FROM
(
-- Min price in last 31 days
SELECT product_id, MIN(price) AS price
FROM Prices
WHERE DATEDIFF(CURDATE(), date) < 31
GROUP BY product_id
) MINPRICE
-- Join in to get the date that the price occured on
INNER JOIN Prices P ON
P.product_id = MINPRICE.product_id
AND
P.price = MINPRICE.price
UNION
SELECT MAXDATE.product_id, MAXDATE.date, P.price
FROM
(
-- Product with no price in last 31 days - get most recent date
SELECT product_id, MAX(date) AS date
FROM Prices
WHERE product_id NOT IN
(
SELECT product_id
FROM Prices
WHERE DATEDIFF(CURDATE(), date) < 31
)
) MAXDATE
-- join in price on that date
INNER JOIN Prices P ON
P.product_id = MAXDATE.product_id
AND
P.date = MAXDATE.date
Not that I tested but you can try...
SELECT * FROM
(SELECT * FROM (
SELECT * FROM table ORDER BY date DESC)
as tmp GROUP BY product_id) t1
LEFT JOIN
(SELECT * FROM (
SELECT * FROM table WHERE date => CURDATE() ORDER BY price)
as tmp2 GROUP BY product_id) t2
ON t1.product_id = t2.product_id