I want the excluded result of rows which are having same minutes
Example: customers table
id date
1 2015-07-23 00:06:56
2 2015-07-23 00:11:38
3 2015-07-23 01:10:16
4 2015-07-23 01:10:13
5 2015-07-24 01:13:26
6 2015-07-24 01:13:13
I want the query to exclude id's: 3, 4, 5 & 6 because (3 & 4) and (5 & 6) are having same minutes
So expected result is:
id date
1 2015-07-23 00:06:56
2 2015-07-23 00:11:38
Thanks in advance!
Try this out. Subquery finds out the hour and minutes that are duplicated. Outer query selects all records except the one returned by the subquery.
select *
from customers
where date_format(dt, '%H%i') not in
(
select date_format(dt, '%H%i')
from customers a
group by date_format(dt, '%H%i')
having count(*) > 1
)
Example: http://sqlfiddle.com/#!9/88a24/4
Filter out duplicate hour + minute by day
select *
from test
where date_format(dt, '%Y%j%H%i') not in
(
select date_format(dt, '%Y%j%H%i')
from test a
group by date_format(dt, '%Y%j%H%i')
having count(*) > 1
)
Example: http://sqlfiddle.com/#!9/fa2da/2
Another example: http://sqlfiddle.com/#!9/7eeaf/1 (using data from updated question)
You can form a group using the year, day, hour and minute portions of the date column, and then only retain groups which have one record. This filters out all records which have duplicate minutes. Then I JOIN this temporary table back to your customers table for the result. This solution does not use a subquery.
SELECT id, date
FROM customers c
INNER JOIN
(
SELECT id
FROM customers
GROUP BY
EXTRACT(YEAR FROM date),
EXTRACT(MONTH FROM date),
EXTRACT(DAY FROM date),
EXTRACT(HOUR FROM date),
EXTRACT(MINUTE FROM date),
HAVING COUNT(*) = 1
) t
ON c.id = t.id
Try this query
select * from tablename
group by date
having count(*)=1
Related
I have 2 tables in Mysql. I want to regroup and count the Number of Orderid per month for each customer. If there is no order, I would like to add 0.
Customer Table
CustomerID
1
2
3
Order Table
OrderId CustomerID Date
1 1 2022-01-02
2 1 2022-01-04
3 2 2022-02-03
4 2 2022-03-03
Expect results
CustomerID Date CountOrderID
1 2022-01 2
2 2022-01 1
3 2022-01 0
1 2022-02 0
2 2022-02 1
3 2022-02 0
1 2022-03 0
2 2022-03 1
3 2022-03 0
How I can do this in Mysql?
SELECT customer.CustomerID,
year_month.y_m AS `Date`,
COUNT(order.OrderId) AS CountOrderID
FROM customer
CROSS JOIN (
SELECT DISTINCT DATE_FORMAT(`date`, '%Y-%m') AS y_m
FROM order
) AS year_month
LEFT JOIN order ON order.CustomerID = customer.CustomerID
AND DATE_FORMAT(order.`date`, '%Y-%m') = year_month.y_m
GROUP BY 1, 2;
If order table does not contains for some year and month then according row won't present in the output. If you need in it then you'd generate calendar table instead of year_month subquery.
you can reduce the number of cte's I added more here to explain the steps:
first you need the format year and month, for that I used DATE_FORMAT() function
since you need to have all the combination of dates and the year month you need a cross join. This will produce all the distinct dates with all the distinct customer id's. In other words all the pairs between dates and customer id
once you have a table with all the combinations you need to pass the actual data with the left join this will produce null where you actually don't have rows and hence will produce 0 when the count is performed
the last step is simply count function
with main as (
select distinct DATE_FORMAT(date,'%Y-%m') as year_month from order
),
calendar as (
select * from customer
cross join main
),
joining_all as (
select
calendar.*,
order. OrderId
left join order
on calendar.CustomerID = order.CustomerID
and calendar.year_month = DATE_FORMAT(order.date,'%Y-%m')
)
select
CustomerID,
year_month as Date,
count(OrderId) as CountOrderID
from joining_all
group by 1,2
maybe the shorter version can work with the code below. if runs into syntax you can use the one above
with main as (
select distinct DATE_FORMAT(date,'%Y-%m') as year_month from order
cross join customer
)
select
main.CustomerID,
main.year_month as Date,
count(order.OrderId) as CountOrderID
from main
left join order
on main.CustomerID = order.CustomerID
and main.year_month = DATE_FORMAT(order.date,'%Y-%m')
group by 1,2
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 have id, date,time(doenst have date) columns, and I am just wondering how to select rows where id repeated at least two other times(repeated three times) in last hour..
enter image description here
use GROUP BY query to get duplicate ID
select *
from 'table'
where date >= DATE_SUB(NOW(),INTERVAL 1 HOUR)
group by id
having count(id) >= 2
select * from 'table' where date >= DATE_SUB(NOW(),INTERVAL 1 HOUR) group by id having count(id) >= 2
Will select records within last hour, group them by id and filter which group have more than 1 records
Here's my "customers" table:
To get number of enquiries per for a particular month and year, I'm using following query:
SELECT YEAR(customer_date) AS Year, MONTH(customer_date) AS Month, COUNT(customer_id) AS Count FROM customers WHERE customer_product = 6 GROUP BY YEAR(customer_date), MONTH(customer_date)
I get following result:
You can see that as there is no enquery in the April month, so no row fetched for month number 4. But I want 0 value in Count column if there is no record found in that particular month and year.
This is what I want:
One option uses a calendar table to represent all months and years, even those which do not appear in your data set:
SELECT
t1.year,
t2.month,
COUNT(c.customer_id) AS Count
FROM
(
SELECT 2017 AS year UNION ALL
SELECT 2018
) t1
CROSS JOIN
(
SELECT 1 AS month UNION ALL
SELECT 2 UNION ALL
SELECT 3 UNION ALL
SELECT 4 UNION ALL
SELECT 5 UNION ALL
SELECT 6 UNION ALL
SELECT 7 UNION ALL
SELECT 8 UNION ALL
SELECT 9 UNION ALL
SELECT 10 UNION ALL
SELECT 11 UNION ALL
SELECT 12
) t2
LEFT JOIN customers c
ON t1.year = YEAR(c.customer_date) AND
t2.month = MONTH(c.customer_date)
WHERE
c.customer_product = 6
GROUP BY
t1.year,
t2.month
ORDER BY
t1.year,
t2.month;
Note: The above query can probably be made faster by actually creating dedicated calendar tables in your MySQL schema.
The following index on the customers table might help:
CREATE INDEX idx ON customers(customer_product, customer_id);
This might make the join between the calendar tables and customers faster, assuming that the customer_product = 6 condition is restrictive.
I have a mysql database with vehicles records. I need a fast query that will return the newest records of those records that were updated within the last 4 minutes. For example vehicle "A" may be updated several times a minute so it will appear many times within the last 4min. Same with vehicle B C etc. I need only the most recent entries for each vehicle within a 4 min window. I have tried like this
SELECT *
FROM yourtable AS a
WHERE a.ts =
(SELECT MAX(ts)
FROM yourtable AS b
WHERE b.ts > NOW() - INTERVAL 5 MINUTE
AND b.name = a.name)
but it takes too long to produce results >10seconds.
You don't need the self-join.
select max(ts), name from Table1
where ts > NOW() - INTERVAL 5 MINUTE
group by name
To get all the rows for the latest updates and not only the name and timestamp:
SELECT t.*
FROM
TableX AS t
JOIN
( SELECT name
, MAX(ts) AS maxts
FROM TableX
WHERE ts > NOW() - INTERVAL 4 MINUTE
GROUP BY name
) AS grp
ON grp.name = t.name
AND grp.maxts = t.ts
You'll need at least an index on the timestamp column for this query.