I want to know which hour period is the most frequent time a person visits, 2nd most frequent, 3rd most and so on. So for example 1 hour period starts at the top of the hour until the next top Eg 07:00:00-07:59:59 would be the 7am hour period.
CREATE TABLE visits (
id primary key
date_visited datetime not null,
cus_name varchar(32) not null
);
I am a little confused because I don't get how I can group by an hour
SELECT count(*)
FROM visits
GROUP BY date_visited
ORDER BY count(*) DESC
You can use HOUR function:
SELECT hour(date_visited), count(*) as number of visits
FROM visits
GROUP BY hour(date_visited)
ORDER BY count(*) DESC
If you want to give numbers to the hours according to number of visits then you can use the analytical function (MySql 8.0 or higher) as follows:
SELECT hour(date_visited), count(*) as number of visits,
row_number() over (order by count(*) desc) as num
FROM visits
GROUP BY hour(date_visited)
ORDER BY num
Related
I have the following table of orders for users like the following:
CREATE TABLE orders (
order_id UUID,
user_id UUID,
date date,
order_type integer
);
I want to write SQL to do the following:
For every order want to compute the number of orders the user has within a previous week (7 days).
Write the following, but it counts the number of orders for each user but not for two levels of groupby.
SELECT order_id, user_id,
COUNT(order_id) OVER(PARTITION BY user_id ORDER BY date ROWS BETWEEN 7 PRECEDING AND CURRENT ROW) as num_orders_7days
FROM orders
You should use RANGE clause instead of ROWS with the proper date intervals:
SELECT order_id, user_id, date,
COUNT(order_id) OVER (
PARTITION BY user_id
ORDER BY date
RANGE BETWEEN INTERVAL 7 day PRECEDING AND INTERVAL 1 day PRECEDING
) as num_orders_7days
FROM orders;
See the demo.
I have two tables, "Gate_Logs" and "Employee".
The "Gate_Logs" table has three columns.
Employee ID - A 4 Digit Unique Number assigned to every employee
Status – In or Out
Timestamp – A recorded timestamp
The "Employee" Table has
Employee ID
Level
Designation
Joining Date
Reporting Location
Reporting Location ID - Single Digit ID
I want to find out which employee had the highest weekly work time over the past year, and I am trying to get this data for each individual location. I want to look at the cumulative highest. Let's say Employee X at Location L worked 60 hours in a particular week, which was the highest at that location, so X will be the person I wanted to query.
Please provide any pointers on how I can proceed with this, have been stuck at it for a while.
SQL version 8.0.27
It can use window function LAG to pair In/Out records
periods - pair in/out records
sumup_weekly - compute weekly work hours for each employee
rank_weekly - rank employees per location per week
and finally select those rank one
WITH periods AS (
SELECT
`employee_id`,
`status` to_status,
`timestamp` to_timestamp,
LAG(`status`) OVER w AS fr_status,
LAG(`timestamp`) OVER w AS fr_timestamp
FROM gate_log
WINDOW w AS (PARTITION BY `employee_id` ORDER BY `timestamp` ASC)
),
sumup_weekly AS (
SELECT
`employee_id`,
WEEKOFYEAR(fr_timestamp) week,
SUM(TIMESTAMPDIFF(SECOND, fr_timestamp, to_timestamp)) seconds
FROM periods
WHERE fr_status = 'In' AND to_status = 'Out'
GROUP BY `employee_id`, `week`
),
rank_weekly AS (
SELECT
e.`employee_id`,
e.`location_id`,
w.`week`,
SEC_TO_TIME(w.`seconds`) work_hours,
RANK() OVER(PARTITION BY e.`location_id`, w.`week` ORDER BY w.`seconds` DESC) rank_hours
FROM sumup_weekly w
JOIN employee e ON w.`employee_id` = e.`employee_id`
)
SELECT *
FROM rank_weekly
WHERE rank_hours = 1
DEMO
Here is my query:
select count(1) from
(select count(1) num, user_id from pos_transactions pt
where date(created_at) <= '2020-6-21'
group by user_id
having num = 1) x
It gives me the number of users who have had 1 transaction until 2020-6-21. Now I want to group it also per date(created_at). I mean, I want to get a list of dates (such as 2020-6-21, 2020-6-22 etc ..) plus the number of users who have had 1 transaction in that date (day).
Any idea how can I do that?
EDIT: The result of query above is correct, the issue is, it's manually now. I mean, I have to increase 2020-6-21 by hand. I want to make it automatically. In other words, I want a list of all dates (from 2020-6-21 til now) contains the number of users who have had 1 transaction until that date.
If you want the number of users who had one transaction on each day, then you need to aggregate by the date as well:
select dte, count(*)
from (select date(created_at) as dte, user_id
from pos_transactions pt
where date(created_at) <= '2020-6-21'
group by dte, user_id
having count(*) = 1
) du
group by dte;
There is a task: develop a fragment of the Web site that provides work with one table.
Attributes of the table:
Day of the week,
Time of the beginning of the lesson,
Subject name,
Number of the audience,
Full name of the teacher.
We need to make a query: determine the day of the week with the largest number of entries, if there are more than one maximum (ie, they are the same), then output them all. I did the query as follows:
SELECT COUNT (*) cnt, day
FROM schedule
GROUP BY day
ORDER BY cnt DESC
LIMIT 1;
But if there are several identical maxima, then only one is displayed. How to write a query which returns them all?
You can use your query as a subquery in the HAVING clause, e.g.:
SELECT day, count(*) as cnt
FROM schedule
GROUP BY day
HAVING count(*) = (
SELECT count(*) as cnt
FROM schedule
GROUP BY day
ORDER BY cnt DESC
LIMIT 1
)
ORDER BY day
I have a table that has a unique key each time a user creates a case:
id|doctor_id|created_dt
--|---------|-----------
1|23 |datetimestamp
2|23 |datetimestamp
3|17 |datetimestamp
How can I select and return the average amount of entries a user has per month?
I have tried this:
SELECT avg (id)
FROM `cases`
WHERE created_dt BETWEEN DATE_SUB(CURDATE(),INTERVAL 90 DAY) AND CURDATE()
and doctor_id = 17
But this returns a ridiculously large value that cannot be true.
To clarify: I am trying to get something like doctor id 17 has an average of 2 entries per month into this table.
I think you were thrown off by the idea of "averaging". You don't want the average id, or average user_id. You want the average number of entries into the table, so you would use COUNT():
SELECT count(id)/3 AS AverageMonthlyCases
FROM `cases`
WHERE created_dt BETWEEN DATE_SUB(CURDATE(),INTERVAL 90 DAY) AND CURDATE()
group by doctor_id
Since you have a 90 day interval, you want to count the number of rows per 30 days, or the count/3.
SELECT AVG(cnt), user_id
FROM (
SELECT COUNT(id) cnt, user_id
FROM cases
WHERE created_dt BETWEEN <yourDateInterval>
GROUP BY user_id, year(created_dt), month(created_dt)
)
Since you need average number of entries, AVG function is not really applicable, because it is SUM()/COUNT() and obviously you do not need that (why would you need SUM of ids).
You need something like this
SELECT
doctor_id,
DATE(created_dt,'%m-%Y') AS month,
COUNT(id) AS visits
FROM `cases`
GROUP BY
`doctor_id`,
DATE(created_dt,'%m-%Y')
ORDER BY
`doctor_id` ASC,
DATE(created_dt,'%m-%Y') ASC
To get visits per month per doctor. If you want to average it, you can then use something like
SELECT
doctor_id,
SUM(visits)/COUNT(month) AS `average`
FROM (
SELECT
doctor_id,
DATE(created_dt,'%m-%Y') AS month,
COUNT(id) AS visits
FROM `cases`
GROUP BY
`doctor_id`,
DATE(created_dt,'%m-%Y')
ORDER BY
`doctor_id` ASC,
DATE(created_dt,'%m-%Y') ASC
) t1
GROUP BY
doctor_id
Obviously you can add your WHERE clauses, as this query is compatible for multiple years (i.e. it will not count January of 2013th and January of 2014th as one month).
Also, it takes into account if a doctor has "blank" months, where he did not have any patients, so it will not count those months (0 can destroy and average).
Use this, you'll group each doctor's total id, by month.
Select monthname(created_dt), doctor_id, count(id) as total from cases group by 1,2 order by 1
Also you can use GROUP_CONCAT() as nested query in order to deploy a pivot like table, where each column is each doctor_id.