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.
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.
My table is like this:
root_tstamp
userId
2022-01-26T00:13:24.725+00:00
d2212
2022-01-26T00:13:24.669+00:00
ad323
2022-01-26T00:13:24.629+00:00
adfae
2022-01-26T00:13:24.573+00:00
adfa3
2022-01-26T00:13:24.552+00:00
adfef
...
...
2021-01-26T00:12:24.725+00:00
d2212
2021-01-26T00:15:24.669+00:00
daddfe
2021-01-26T00:14:24.629+00:00
adfda
2021-01-26T00:12:24.573+00:00
466eff
2021-01-26T00:12:24.552+00:00
adfafe
I want to get the number of users in the current year and in previous year like below using SQL.
Date Users previous_year
2022-01-01 10 5
2022-01-02 20 15
The code is written as follows.
select CAST(root_tstamp as DATE) as Date,
count(DISTINCT userid) as users,
count(Distinct case when CAST(root_tstamp as DATE) = dateadd(MONTH,-12,CAST(root_tstamp as DATE)) then userid end) as previous_year
FROM table1
But it returns 0 for previous_year values.
How can I fix that?
Possible solution for SQL Server:
WITH cte AS ( SELECT 2022 [year]
UNION ALL
SELECT 2021 )
SELECT cte.[year],
COUNT(DISTINCT test.userId) current_users_amount,
COUNT(DISTINCT CASE WHEN YEAR(test.root_tstamp) < cte.[year]
THEN test.userId
END) previous_users_amount
FROM test
JOIN cte ON YEAR(test.root_tstamp) <= cte.[year]
GROUP BY cte.[year]
https://dbfiddle.uk/?rdbms=sqlserver_2017&fiddle=88b78aad9acd965bdbac4c85a0b81927
This query (for MySql) returns unique number of userids where the root_timestamp is in the current year, by day, and the number of unique userids for the same day last year. If there is no record for a day in the current year nothing will be displayed for that day. If there are rows for the current year, but no rows for the same day last year, then NULL will be shown for that lastyear column.
SELECT cast(ty.root_tstamp as date) as Dte,
COUNT(DISTINCT ty.userId) as users_this_day,
count(distinct lysd.userid) as users_sameday_lastyear
FROM test ty
left join
test lysd
on cast(lysd.root_tstamp as date)=date_add(cast(ty.root_tstamp as date), interval -1 year)
WHERE YEAR(ty.root_tstamp) = year(current_date())
GROUP BY Dte
If you wish to show output rows for calendar days even if there are no rows in current year and/or last year, then you also need a calendar table to be introduced (let's hope that it is not what you need)
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
In an e-commerce website, I have a database that contains the following fields:
id
date_purchased
...
The field date_purchased has the current format : 2018-02-14 16:27:37(year-month-day hours-minutes-seconds)
I would like to get, for the year 2018 for example, ordered by ASC, the number of orders made for each hour.
I can't figure out how to order by a certain year, and count the number of orders made each hour of that year.
Something like :
SELECT count(*)
FROM table
WHERE (
SELECT *
FROM table
GROUP BY DATEPART(hour, [date_purchased]) ASC
)
GROUP BY year(date_purchased) ASC
I think the following would fulfill your requirements in mysql:
SELECT HOUR(date_purchased) hour_purchased, count(*) hour_Count
FROM yourtable
WHERE YEAR(date_purchased) = 2018
GROUP BY HOUR(date_purchased)
ORDER BY hour_count DESC
All I want to count entries based on date.(i.e entries with same date.)
My table is
You can see 5th and 6th entry have same date.
Now, the real problem as i think is the same date entry have different time so i am not getting what I want.
I am using this sql
SELECT COUNT( created_at ) AS entries, created_at
FROM wp_frm_items
WHERE user_id =1
GROUP BY created_at
LIMIT 0 , 30
What I am getting is this.
I want entries as 2 for date 2012-02-22
The reason you get what you get is because you also compare the time, down to a second apart. So any entries created the same second will be grouped together.
To achieve what you actually want, you need to apply a date function to the created_at column:
SELECT COUNT(1) AS entries, DATE(created_at) as date
FROM wp_frm_items
WHERE user_id =1
GROUP BY DATE(created_at)
LIMIT 0 , 30
This would remove the time part from the column field, and so group together any entries created on the same day. You could take this further by removing the day part to group entries created on the same month of the same year etc.
To restrict the query to entries created in the current month, you add a WHERE-clause to the query to only select entries that satisfy that condition. Here's an example:
SELECT COUNT(1) AS entries, DATE(created_at) as date
FROM wp_frm_items
WHERE user_id = 1
AND created_at >= DATE_FORMAT(CURDATE(),'%Y-%m-01')
GROUP BY DATE(created_at)
Note: The COUNT(1)-part of the query simply means Count each row, and you could just as well have written COUNT(*), COUNT(id) or any other field. Historically, the most efficient approach was to count the primary key, since that is always available in whatever index the query engine could utilize. COUNT(*) used to have to leave the index and retrieve the corresponding row in the table, which was sometimes inefficient. In more modern query planners this is probably no longer the case. COUNT(1) is another variant of this that didn't force the query planner to retrieve the rows from the table.
Edit: The query to group by month can be created in a number of different ways. Here is an example:
SELECT COUNT(1) AS entries, DATE_FORMAT(created_at,'%Y-%c') as month
FROM wp_frm_items
WHERE user_id =1
GROUP BY DATE_FORMAT(created_at,'%Y-%c')
You must eliminate the time with GROUP BY
SELECT COUNT(*) AS entries, created_at
FROM wp_frm_items
WHERE user_id =1
GROUP BY DATE(created_at)
LIMIT 0 , 30
Oops, misread it.
Use GROUP BY DATE(created_at)
Try:
SELECT COUNT( created_at ) AS entries, created_at
FROM wp_frm_items
WHERE user_id =1
GROUP BY DATE(created_at)
LIMIT 0 , 30