I have a table and in that bank transactions are recorded. I want to pull out the users who made transaction just once in a month. That means their transaction count == 1 for that month.
I am recording card_number, timestamp etc..
SELECT *
FROM transactions
GROUP BY card_number, month
HAVING COUNT(card_number) = 1
Check This
select Card_Number,
DATEPART(m, [TimeStamp] )MonthNo,
COUNT(1) TxnCount
from Transactions
group by Card_Number,
DATEPART(m, [TimeStamp] )
having COUNT(1)=1
I am assuming that your table structure is somthing like below.
Card_Number int,
TimeStamp datetime
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)
I have an SQL structure like this:
Create Table Transactions (
Id integer primary key not null auto_increment,
ResourceId varchar(255),
Price Integer,
TransactionTime date
);
I would like to get the time (TransactionTime) along with the average of 3 days price. For example, the 3 day average of the 22nd will be the average of the 20th, 21st, and 22nd.
Thanks so much.
Presumably, you want this information on each row and for a given resource. If so:
select t.*,
avg(price) over (partition by resourceid
order by transactiontime
range between interval 2 day preceding and current row
) as avg_3
from transactions t;
For SQL server:
SELECT AVG(Price), MAX(TransactionTime) FROM Transactions GROUP BY FLOOR(DATEDIFF(DAY, GETDATE(), TransactionTime) / 3);
You can use nested select:
select t.TransactionTime,
(select sum(t1.Price) / 3
from Transactions t1 where t1.Data in (t.Data, t.Date-2);) as avg3;
from Transactions t;
I have a query that give me a table like this:
Person | Date_IN | Date_OUT | Structure
During a year a person ENTER and EXIT many times, ENTER and EXIT could be also the same day.
I'd like to count, for a specific day of year, how many person were IN each structure.
The final goal is to have, for a given period (1st march --> 31st march), the sum of total person for each day for each structure.
I believe the following would work. It assumes that you have a table of dates (consists of one column which contains all the dates between 1950 and 2050) and you simply join it with the person check in/out table:
SELECT dates.date, Structure, COUNT(DISTINCT Person) Persons_on_That_Date
FROM dates
LEFT JOIN turndata ON dates.date BETWEEN Date_IN AND Date_OUT
WHERE dates.date BETWEEN '2018-03-01' AND '2018-03-31'
GROUP BY dates.date, Structure
ORDER BY Structure, dates.date
Demo Here
Note: the above assumes that the out date is inclusive (the person is counted as inside on that date). If out date is exclusive then the ON clause becomes:
... ON Date_IN <= dates.date AND dates.date < Date_OUT
Please use below query, data is grouped by structure for particular timeframe.
SELECT structure, COUNT(DISTINCT person) as no_of_person
FROM table_name
WHERE DATE(Date_IN) BETWEEN '2018-08-01' AND '2018-08-31'
GROUP BY structure
You say there can be no multiple date_in for the same day and person, because a person is in at least one day. So for a given date we only must look at the latest event per person until then to see whether the person is/was in that day.
These are the steps:
create a data set for the requiered days on-the-fly
join with the table and get the last date_in until that day per person
join with the table again to get the last records
aggregate per day and count persons present
This is:
select
data.day
sum(t.date_in is not null and (t.date_out is null or t.date_out = data.day)) as count_in
from
(
select days.day, t.person, max(t.date_in) as max_date_in
from (select date '2018-03-01' as day union all ...) days
left join t on t.date_in <= days.day
group by days.day, t.person
) data
left join t on t.person = data.person and t.date_in = data.max_date_in
group by data.day
order by data.day;
I am trying to query a table. There are 3 important fields: attendant_id, client_id, and date.
Each time an attendant works with a client, they add an entry which includes their id, the client's id, and the date. Occasionally, an attendant will work with more than one client on the same day. I would like to capture when this happens. Here is what I have so far:
SELECT *
FROM timesheet_lines tsl1
WHERE EXISTS
(
SELECT *
FROM timesheet_lines tsl2
WHERE tsl1.date = tsl2.date
AND tsl1.attendant_id = tsl2.attendant_id
AND tsl1.client_id <> tsl2.client_id
AND tsl1.date between '2014-04-01' AND '2014-06-30'
LIMIT 2,5
)
I only want to display results where an attendant worked with at least 2 different clients. I don't expect it to be possible to have more than 5 on a single day. This is why I am using LIMIT 2,5.
I am also only interested in April through June of this year.
I think I may have the right syntax, but the query seems to be taking forever to run. Is there a faster query? There should be only about 42000+ entries all together for this particular date range. I am not expecting to get more than about 500-600 results that meet the criteria.
I ended up using the following:
create TEMPORARY table tempTSL1
(date1 date, start1 time, end1 time, attend1 varchar(50), client1 varchar(50), type1 tinyint);
insert into tempTSL1(date1, start1, end1, attend1, client1, type1)
select date, start_time, end_time, attendant_id, client_id, type
from timesheet_lines
WHERE
timesheet_lines.date BETWEEN '2014-04-01' AND '2014-06-30'
and timesheet_lines.type IN (1,2,5,6);
create TEMPORARY table tempTSL2
(date2 date, start2 time, end2 time, attend2 varchar(50), client2 varchar(50), type2 tinyint);
insert into tempTSL2(date2, start2, end2, attend2, client2, type2)
select date, start_time, end_time, attendant_id, client_id, type
from timesheet_lines
WHERE
timesheet_lines.date BETWEEN '2014-04-01' AND '2014-06-30'
and timesheet_lines.type IN (1,2,5,6);
SELECT *
FROM tempTSL1
WHERE (attend1,date1) IN (
SELECT attend2
,date2
FROM tempTSL2 tsl2
GROUP BY attend2
,date2
HAVING COUNT(date2) > 1
)
GROUP BY attend1
,client1
,date1
HAVING COUNT(client1) = 1
ORDER BY date1,attend1,start1
You are likely making it much more complex than it needs to be. Try something like this:
SELECT attendant_id
,client_id
,date
FROM timesheet_lines
WHERE (attendant_id,date) IN (
SELECT attendant_id
,date
FROM timesheet_lines tsl1
GROUP BY attendant_id
,date
HAVING COUNT(date) > 1
)
GROUP BY attendant_id
,client_id
,date
HAVING COUNT(client_id) = 1
The subquery returns results only of attendants performing multiple activities on the same date. The top query will pull from the same table, matching the attendant and dates of activity, and filter the result set to items where there is only 1 client in the grouping. Example:
attendant_id client_id date
1 A 2014-01-01
1 B 2014-01-01
2 C 2014-01-01
2 D 2014-01-02
Will return:
attendant_id client_id date
1 A 2014-01-01
1 B 2014-01-01
Untested, but I think it should be in line with what you are looking for, assuming the following two statements are true:
You are not trying to capture two different attendants working the same client on the same day
An attendant can only perform one activity per client per day
If the second point is not true, then you will need to incorporate additional fields into the subquery (such as an activity_id or something).
Hope this helps.
I am using PostgreSQL. I need to get the dates for the first 5 transactions of every user on my DB.
Transaction - trans.id, trans.date, trans.cust_id, trans.value
Customer - cust.id, cust.created_at
I need to get the date of the first 5 transactions for all the customers.
Try this query:
SELECT cust_id, date
FROM (
SELECT cust_id,
date,
row_number() OVER (partition by cust_id
ORDER BY date, id ) rn
FROM Transaction
) as alias
WHERE rn <= 5
ORDER BY 1,2
demo: http://sqlfiddle.com/#!15/cfd2e/4