My problem: I have table with price and date. I need to have an average price from last 7 existing days. E.g.: I have prices from today, yesterday, 30 days ago, 43 days ago etc. I need an average not from the last 7 days, but from the last 7 existing days.
My code:
SELECT AVG(price)
FROM table
GROUP BY date
ORDER BY date DESC LIMIT 7
But I have 7 average price from every day.
Maybe someone has another idea
Use a subquery to get the last 7 existing days, get the earliest of those dates, then join that with the table.
SELECT AVG(price)
FROM table AS t1
JOIN (SELECT MIN(dateday) AS mindate
FROM
(SELECT DATE(date) AS dateday
FROM table
GROUP BY dateday
ORDER BY dateday DESC LIMIT 7
) AS x
) AS t2
ON t1.date >= t2.mindate
use avg function and subquery
select avg(price)
from
(SELECT date,price
FROM table
ORDER BY date desc limit 7
) as t
Related
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)
What I am trying to do is to get the record count for each day of the last 7 days,
Let's say I have 3 records today, 4 records yesterday, 2 records two days ago, etc.
I'd like to have something like that:
[12/06/2021] 1
[11/06/2021] 4
[10/06/2021] 3
[09/06/2021] 6
[08/06/2021] 7
[07/06/2021] 2
[06/06/2021] 7
(Or get only the count, it's OK too.)
I have a field - message_datetime that saves the datetime.
Is there a way to do this in one query?
What I've done:
select CAST(message_datetime AS DATE),count(message_datetime) from messages group by CAST(message_datetime AS DATE) WHERE message_datetime
It worked but I wanted the last 7 days. Thanks
I was waiting for you to post your own effort, but since somebody has already "jumped the gun" and started to post answers, so:
Assuming the name of the table is my_table, then try:
select date(message_datetime) as message_date, count(*) as cnt from my_table
where datediff(curdate(), date(message_datetime)) < 7
group by date(message_datetime)
order by date(message_datetime) desc
Update
Following the Strawberry's suggestion, here is an updated query that should be more performant if message_datetime is an indexed column:
select date(message_datetime) as message_date, count(*) as cnt from my_table
where message_date_time >= date_sub(curdate(), interval 6 day)
group by date(message_datetime)
order by date(message_datetime) desc
i have this table
desc t1;
with this datas
mysql> select* from t1;
now i do this
mysql> select day_date,count(day_date) from t1 group by (date(day_date));
how can I count last 7 days data from SQL and group them by day/date (excluding today).
I should be able to use the result as $resultday1, $resultday2, $resultday3 etc.
If there was 10 total SQL entries in day 1 (yesterday) $resultday1 should show "10".
and the days should be last 7 only, and today/current day should not consider.
The following PHP SQL script shows the total count only
SELECT COUNT(1) FROM orders WHERE username='jondoe'
database is a list of referrals made by a registered user in previous days.
a single table contains all user's referral details, table name "orders" as per above example.
This is the exact query as you want
SELECT
COUNT(*), DATE(order_date) order_date
FROM
orders
WHERE
order_date < CURDATE()
AND order_date > DATE_SUB(NOW(), INTERVAL 7 DAY)
GROUP BY order_date
ORDER BY order_date DESC;
I want to visualize my entries by counting how many have been created at the same day.
SELECT dayname(created_at), count(*) FROM logs
group by day(created_at)
ORDER BY created_at desc
LIMIT 7
So I get something like:
Thursday 4
Wednesday 12
Monday 4
Sunday 1
Saturday 20
Friday 23
Thursday 10
But I also want to have the Tuesday in there with 0 so I have it for one week.
Is there a way to do this with full mysql or do I need to update the result before I can give it to the chart?
EDIT:
This is the final query:
SELECT
DAYNAME(date_add(NOW(), interval days.id day)) AS day,
count(logs.id) AS amount
FROM days LEFT OUTER JOIN
(SELECT *
FROM logs
WHERE TIMESTAMPDIFF(DAY,DATE(created_at),now()) < 7) logs
on datediff(created_at, NOW()) = days.id
GROUP BY days.id
ORDER BY days.id desc;
The table days includes numbers from 0 to -6
You only need a table of offsets which could be a real table or something built on the fly like select 0 ofs union all select -1 ....
create table days (ofs int);
insert into days (ofs) values
(0), (-1), (-2), (-3),
(-4), (-5), (-6), (-7);
select
date_add('20160121', interval days.ofs day) as created_at,
count(data.id) as cnt
from days left outer join logs data
on datediff(data.created_at, '20160121') = days.ofs
group by days.ofs
order by days.ofs;
http://sqlfiddle.com/#!9/3e6bc7/1
For performance it would probably be better to limit the search in the data (logs) table:
select
date_add('20160121', interval days.ofs day) as created_at,
count(data.id) as cnt
from days left outer join
(select * from logs where created_at between <start> and <end>) data
on datediff(data.created_at, '20160121') = days.offset
group by days.offset
order by days.offset;
One downside is that you do have to parameterize this with a fixed anchor date in a couple of expressions. It might be better to have a table of real dates sitting in a table somewhere so you don't have to do the calculations.
Use RIGHT JOIN to a dates table, so you can request data for each and all days, no matter if some days have data or not, simply, mull days will show as CERO or NULL.
You can create a dates table, some sort of calendar table.
id_day | day_date |
--------------------
1 | 2016-01-01 |
2 | 2016-01-02 |
.
.
365 | 2016-12-31 |
With this table, you can relate date, then extract day, month, week, whatever you want with MYSQL DATE AND TIME FUNCTIONS
SELECT t2.dayname(day_date), count(t1.created_at) FROM logs t1 right join dates_table t2 on t1.created_at=t2.day_date group by t2.day_date ORDER BY t1.created_at desc LIMIT 7
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.