I have a transaction table where I have to find the first and second date of transaction of every customer. Finding first date is very simple where I can use MIN() func to find the first date but the second and in particular finding the difference between the two is getting very challenging and somehow I am not able to find out any feasible way:
select a.customer_id, a.transaction_date, a.Row_Count2
from ( select
transaction_date as transaction_date,
reference_no as customer_id,
row_number() over (partition by reference_no
ORDER BY reference_no, transaction_date) AS Row_Count2
from transaction_detail
) a
where a.Row_Count2 < 3
ORDER BY a.customer_id, a.transaction_date, a.Row_Count2
Gives me this :
What I want is , following columns:
||CustomerID|| ||FirstDateofPurchase|| ||SecondDateofPuchase|| ||Diff. b/w Second & First Date ||
You can use window functions LEAD/LAG to return results you are looking for
First try to find all the leading dates by reference number using LEAD, generate row number for each row using your original logic. You can then do difference on dates for row number value 1 row from the result set.
Ex (I'm not excluding same day transactions and treating them as separate and generating row number based on result set from your query above, you can easily change the sql below to consider these as one and remove them so that you get next date as second date):
declare #tbl table(reference_no int, transaction_date datetime)
insert into #tbl
select 1000, '2018-07-11'
UNION ALL
select 1001, '2018-07-12'
UNION ALL
select 1001, '2018-07-12'
UNIOn ALL
select 1001, '2018-07-13'
UNIOn ALL
select 1002, '2018-07-11'
UNIOn ALL
select 1002, '2018-07-15'
select customer_id, transaction_date as firstdate,
transaction_date_next seconddate,
datediff(day, transaction_date, transaction_date_next) diff_in_days
from
(
select reference_no as customer_id, transaction_date,
lead(transaction_date) over (partition by reference_no
order by transaction_date) transaction_date_next,
row_number() over (partition by reference_no ORDER BY transaction_date) AS Row_Count
from #tbl
) src
where Row_Count = 1
You can do this with CROSS APPLY.
SELECT td.customer_id, MIN(ca.transaction_date), MAX(ca.transaction_date),
DATEDIFF(day, MIN(ca.transaction_date), MAX(ca.transaction_date))
FROM transaction_detail td
CROSS APPLY (SELECT TOP 2 *
FROM transaction_detail
WHERE customer_id = td.customer_id
ORDER BY transaction_date) ca
GROUP BY td.customer_id
Related
This is a sample data file
Data Contains unique IDs with different latitudes and longitudes on multiple timestamps.I would like to select the rows of latest 30 days of coordinates for each unique ID.Please help me on how to run the query .This date is in Hive table
Regards,
Akshay
According to your example above (where no current year dates for id=2,3), you can numbering date for each id (order by date descending) using window function ROW_NUMBER(). Then just get latest 30 values:
--get all values for each id where num<=30 (get last 30 days for each day)
select * from
(
--numbering each date for each id order by descending
select *, row_number()over(partition by ID order by DATE desc)num from Table
)X
where num<=30
If you need to get only unique dates (without consider time) for each id, then can try this query:
select * from
(
--numbering date for each id
select *, row_number()over(partition by ID order by new_date desc)num
from
(
-- move duplicate using distinct
select distinct ID,cast(DATE as date)new_date from Table
)X
)Y
where num<=30
In Oracle this will be:
SELECT * FROM TEST_DATE1
WHERE DATEUPDT > SYSDATE - 30;
select * from MyTable
where
[Date]>=dateadd(d, -30, getdate());
To group by ID and perform aggregation, something like this
select ID,
count(*) row_count,
max(Latitude) max_lat,
max(Longitude) max_long
from MyTable
where
[Date]>=dateadd(d, -30, getdate())
group by ID;
i use sql server and i have this table :
ID Date Amount
I need to write a query that returns only users that have made at least 3 consecutive purchases, each one larger than the other.
I know that i need to use partition_id and row_number but i dont know how to do it
Thank you in advance
If you want three purchases in a row with increases in amount, then use lead() to get the next amounts:
select t.*
from (select t.*,
lead(amount, 1) over (partition by id order by date) as next_date,
lead(amount, 2) over (partition by id order by date) as next2_amount
from t
) t
where next_amount > amount and next2_amount > next_amount;
I originally missed the "greater than" part of the question. If you wanted purchases on three days in a row, then:
If you want three days in a row and there is at most one purchase per day, then you can use:
select t.*
from (select t.*,
lead(date, 2) over (partition by id order by date) as next2_date
from t
) t
where next2_date = dateadd(day, 2, date);
If you can have duplicates on a date, I would suggest this variant:
select t.*
from (select t.*,
lead(date, 2) over (partition by id order by date) as next2_date
from (select distinct id, date from t) t
) t
where next2_date = dateadd(day, 2, date);
Need a query to select "Running Total" as mentioned in the image
2017 year end total plus Every months new figure should add up to previous total.
https://i.stack.imgur.com/DL7p0.png "Example"
This following script will work for MSSQL and you can use the same logic for other databases as well-
WITH your_table(year,month,partersgrowth)
AS
(
SELECT '2019','jan', 100 UNION ALL
SELECT '2019','feb', 300 UNION ALL
SELECT '2019','mar', 400 UNION ALL
SELECT '2019','apr', 500 UNION ALL
SELECT '2018','Dec', 200
)
SELECT A.year,A.month,A.partersgrowth,
(
SELECT SUM(B.partersgrowth)
FROM your_table B
WHERE CAST(B.Year +'-'+B.month+'-01' AS DATE)
<= CAST(A.Year +'-'+A.month+'-01' AS DATE)
) Running_Total
FROM your_table A
ORDER BY CAST(A.Year +'-'+A.month+'-01' AS DATE)
using #mkRabbani solution.. you can simplify it like this:
;WITH your_table(year,month,partersgrowth)
AS
(
SELECT '2019','jan', 100 UNION ALL
SELECT '2019','feb', 300 UNION ALL
SELECT '2019','mar', 400 UNION ALL
SELECT '2019','apr', 500 UNION ALL
SELECT '2018','Dec', 200
)
select *, sum(partersgrowth) over ( order by [year],[month]) as running_total
from your_table
EDIT: As pointed out by comment below.. you want to order by a proper date in the sum part ( I would use order by year and then the month number rather than the month name)
If you use MSSQL you can run the following code
with cte as (
select t.*, lag(partersGrowth) over(partition by year order by month asc rows ROWS UNBOUNDED PRECEDING) prevTotal
from Table )
select years, month, partersGrowth, prevTotal || '+' partersGrowth as "Need Running Total"
from cte
I have the sample data set below which list the water meters not working for specific reason for a certain range period (jan 2016 to december 2018).
I would like to have a query that retrieves the last maximum and minimum consecutive period where the meter was not working within that range of period.
any help will be greatly appreciated.
You have two options:
select code, to_char(min_period, 'yyyymm') min_period, to_char(max_period, 'yyyymm') max_period
from (
select code, min(period) min_period, max(period) max_period,
max(min(period)) over (partition by code) max_min_period
from (
select code, period, sum(flag) over (partition by code order by period) grp
from (
select code, period,
case when add_months(period, -1)
= lag(period) over (partition by code order by period)
then 0 else 1 end flag
from (select mrdg_acc_code code, to_date(mrdg_per_period, 'yyyymm') period from t)))
group by code, grp)
where min_period = max_min_period
Explanation:
flag rows where period is not equal previous period plus one month,
create column grp which sums flags consecutively,
group data using code and grp additionaly finding maximal start of period,
show only rows where min_period = max_min_period
Second option is recursive CTE available in Oracle 11g and above:
with
data(period, code) as (
select to_date(mrdg_per_period, 'yyyymm'), mrdg_acc_code from t
where mrdg_per_period between 201601 and 201812),
cte (period, code) as (
select to_char(period, 'yyyymm'), code from data
where (period, code) in (select max(period), code from data group by code)
union all
select to_char(data.period, 'yyyymm'), cte.code
from cte
join data on data.code = cte.code
and data.period = add_months(to_date(cte.period, 'yyyymm'), -1))
select code, min(period) min_period, max(period) max_period
from cte group by code
Explanation:
subquery data filters only rows from 2016 - 2018 additionaly converting period to date format. We need this for function add_months to work.
cte is recursive. Anchor finds starting rows, these with maximum period for each code. After union all is recursive member, which looks for the row one month older than current. If it finds it then net row, if not then stop.
final select groups data. Notice that period which were not consecutive were rejected by cte.
Though recursive queries are slower than traditional ones, there can be scenarios where second solution is better.
Here is the dbfiddle demo for both queries. Good luck.
use aggregate function with group by
select max(mdrg_per_period) mdrg_per_period, mrdg_acc_code,max(mrdg_date_read),rea_Desc,min(mdrg_per_period) not_working_as_from
from tablename
group by mrdg_acc_code,rea_Desc
This is a bit tricky. This is a gap-and-islands problem. To get all continuous periods, it will help if you have an enumeration of months. So, convert the period to a number of months and then subtract a sequence generated using row_number(). The difference is constant for a group of adjacent months.
This looks like:
select acc_code, min(period), max(period)
from (select t.*,
row_number() over (partition by acc_code order by period_num) as seqnum
from (select t.*, floor(period / 100) * 12 + mod(period, 100) as period_num
from t
) t
where rea_desc = 'METER NOT WORKING'
) t
group by (period_num - seqnum);
Then, if you want the last one for each account, you can use a subquery:
select t.*
from (select acc_code, min(period), max(period),
row_number() over (partition by acc_code order by max(period desc) as seqnum
from (select t.*,
row_number() over (partition by acc_code order by period_num) as seqnum
from (select t.*, floor(period / 100) * 12 + mod(period, 100) as period_num
from t
) t
where rea_desc = 'METER NOT WORKING'
) t
group by (period_num - seqnum)
) t
where seqnum = 1;
I have a database of transactions, accounts, profit/loss, and date. I need to find the dates which the largest profit occurs by account. I have already found a way to find these actually max/min values but I can't seem to be able to pull the actual date from it. My code so far is like this:
Select accountnum, min(ammount)
from table
where date > '02-Jan-13'
group by accountnum
order by accountnum
Ideally I would like to see account num, the min or max, and then the date which this occurred on.
Try something like this to get the min and max amount for each customer and the date it happened.
WITH max_amount as (
SELECT accountnum, max(amount) amount, date
FROM TABLE
GROUP BY accountnum, date
),
min_amount as (
SELECT accountnum, min(amount) amount, date
FROM TABLE
GROUP BY accountnum, date
)
SELECT t.accountnum, ma.amount, ma.date, mi.amount, ma.date
FROM table t
JOIN max_amount ma
ON ma.accountnum = t.accountnum
JOIN min_amount mi
ON mi.accountnum = t.accountnum
If you want the data for just this year you could add a where clause to the end of the statement
WHERE t.date > '02-Jan-13'
The easiest way to do this is using window/analytic functions. These are ANSI standard and most databases support them (MySQL and Access being two notable exceptions).
Here is one way:
select t.accountnum, min_amount, max_amount,
min(case when amount = min_amount then date end) as min_amount_date,
min(case when amount = min_amount then date end) as max_amount_date,
from (Select t.*,
min(amount) over (partition by accountnum) as min_amount,
max(amount) over (partition by accountnum) as max_amount
from table t
where date > '02-Jan-13'
) t
group by accountnum, min_amount, max_amount;
order by accountnum
The subquery calculates the minimum and maximum amount for each account, using min() as a window function. The outer query selects these values. It then uses conditional aggregation to get the first date when each of those values occurred.
;with cte as
(
select accountnum, ammount, date,
row_number() over (partition by accountnum order by ammount desc) rn,
max(ammount) over (partition by accountnum) maxamount,
min(ammount) over (partition by accountnum) minamount
from table
where date > '20130102'
)
select accountnum,
ammount as amount,
date as date_of_max_amount,
minamount,
maxamount
from cte where rn = 1