I am trying to limit the amount of data I pull in prior to processing/analysing it in python.
Mainly due to memory constraints.
Each transaction results in ~3-4 different transaction_events.
-----------
trx_id timestamp
trx_1 | 2021.01.01 15:45:40
trx_1_2 | 2021.01.01 15:45:40
trx_1_3 | 2021.01.01 15:45:40
trx_2 | 2021.02.01 14:15:40
trx_2_2 | 2021.02.01 14:15:40
trx_2_3 | 2021.02.01 14:15:40
All I need is 1 record per timestamp.
-----------
trx_id timestamp
trx_1 | 2021.01.01 15:45:40
trx_2 | 2021.02.01 14:15:40
I've already tried the following suggestions:
On the ORACLE community forum
and
select distinct(date) return the same date several time
I've tried various variations too
SELECT DISTINCT TRUNC(timestamp, 'DD')
SELECT DISTINCT TRUNC(timestamp)
SELECT DISTINCT to_char(timestamp, 'yyyy-mm-dd')
However with no results.
You can use the ROW_NUMBER analytic function and partition by the first 5 characters of the TRX_ID and the timestamp:
SELECT trx_id, ts
FROM (
SELECT t.*,
ROW_NUMBER() OVER (
PARTITION BY SUBSTR( trx_id, 1, 5 ), ts ORDER BY trx_id
) AS rn
FROM table_name t
)
WHERE rn = 1;
Which, for your sample data:
CREATE TABLE table_name ( trx_id, ts ) AS
SELECT 'trx_1', TIMESTAMP '2021-01-01 15:45:40' FROM DUAL UNION ALL
SELECT 'trx_1_2', TIMESTAMP '2021-01-01 15:45:40' FROM DUAL UNION ALL
SELECT 'trx_1_3', TIMESTAMP '2021-01-01 15:45:40' FROM DUAL UNION ALL
SELECT 'trx_2', TIMESTAMP '2021-02-01 14:15:40' FROM DUAL UNION ALL
SELECT 'trx_2_2', TIMESTAMP '2021-02-01 14:15:40' FROM DUAL UNION ALL
SELECT 'trx_2_3', TIMESTAMP '2021-02-01 14:15:40' FROM DUAL;
Outputs:
TRX_ID
TS
trx_1
2021-01-01 15:45:40.000000000
trx_2
2021-02-01 14:15:40.000000000
If you can have other TRX_ID with different length patterns then you can look for the second underscore character and get the substring before that:
SELECT trx_id, ts
FROM (
SELECT t.*,
ROW_NUMBER() OVER (
PARTITION BY CASE INSTR( trx_id, '_', 1, 2 )
WHEN 0
THEN trx_id
ELSE SUBSTR( trx_id, 1, INSTR( trx_id, '_', 1, 2 ) - 1 )
END,
ts
ORDER BY trx_id
) AS rn
FROM table_name t
)
WHERE rn = 1;
db<>fiddle here
You can use aggregation:
select min(trx_id), timestamp
from t
group by timestamp;
Here is a db<>fiddle.
Related
The problem I am facing is how to find distinct time periods from multiple time periods with overlap in Teradata ANSI SQL.
For example, the attached tables contain multiple overlapping time periods, how can I combine those time periods into 3 unique time periods in Teradata SQL???
I think I can do it in python with the loop function, but not sure how to do it in SQL
ID
Start Date
End Date
001
2005-01-01
2006-01-01
001
2005-01-01
2007-01-01
001
2008-01-01
2008-06-01
001
2008-04-01
2008-12-01
001
2010-01-01
2010-05-01
001
2010-04-01
2010-12-01
001
2010-11-01
2012-01-01
My expected result is:
ID
start_Date
end_date
001
2005-01-01
2007-01-01
001
2008-01-01
2008-12-01
001
2010-01-01
2012-01-01
From Oracle 12, you can use MATCH_RECOGNIZE to perform a row-by-row comparison:
SELECT *
FROM table_name
MATCH_RECOGNIZE(
PARTITION BY id
ORDER BY start_date
MEASURES
FIRST(start_date) AS start_date,
MAX(end_date) AS end_date
ONE ROW PER MATCH
PATTERN (overlapping_ranges* last_range)
DEFINE overlapping_ranges AS NEXT(start_date) <= MAX(end_date)
)
Which, for the sample data:
CREATE TABLE table_name (ID, Start_Date, End_Date) AS
SELECT '001', DATE '2005-01-01', DATE '2006-01-01' FROM DUAL UNION ALL
SELECT '001', DATE '2005-01-01', DATE '2007-01-01' FROM DUAL UNION ALL
SELECT '001', DATE '2008-01-01', DATE '2008-06-01' FROM DUAL UNION ALL
SELECT '001', DATE '2008-04-01', DATE '2008-12-01' FROM DUAL UNION ALL
SELECT '001', DATE '2010-01-01', DATE '2010-05-01' FROM DUAL UNION ALL
SELECT '001', DATE '2010-04-01', DATE '2010-12-01' FROM DUAL UNION ALL
SELECT '001', DATE '2010-11-01', DATE '2012-01-01' FROM DUAL;
Outputs:
ID
START_DATE
END_DATE
001
2005-01-01 00:00:00
2007-01-01 00:00:00
001
2008-01-01 00:00:00
2008-12-01 00:00:00
001
2010-01-01 00:00:00
2012-01-01 00:00:00
db<>fiddle here
Update: Alternative query
SELECT id,
start_date,
end_date
FROM (
SELECT id,
dt,
SUM(cnt) OVER (PARTITION BY id ORDER BY dt) AS grp,
cnt
FROM (
SELECT ID,
dt,
SUM(type) OVER (PARTITION BY id ORDER BY dt, ROWNUM) * type AS cnt
FROM table_name
UNPIVOT (dt FOR type IN (start_date AS 1, end_date AS -1))
)
WHERE cnt IN (1,0)
)
PIVOT (MAX(dt) FOR cnt IN (1 AS start_date, 0 AS end_date))
Or, an equivalent that does not use UNPIVOT, PIVOT or ROWNUM and works in both Oracle and PostgreSQL:
SELECT id,
MAX(CASE cnt WHEN 1 THEN dt END) AS start_date,
MAX(CASE cnt WHEN 0 THEN dt END) AS end_date
FROM (
SELECT id,
dt,
SUM(cnt) OVER (PARTITION BY id ORDER BY dt) AS grp,
cnt
FROM (
SELECT ID,
dt,
SUM(type) OVER (PARTITION BY id ORDER BY dt, rn) * type AS cnt
FROM (
SELECT r.*,
ROW_NUMBER() OVER (PARTITION BY id ORDER BY dt ASC, type DESC) AS rn
FROM (
SELECT id, 1 AS type, start_date AS dt FROM table_name
UNION ALL
SELECT id, -1 AS type, end_date AS dt FROM table_name
) r
) p
) s
WHERE cnt IN (1,0)
) t
GROUP BY id, grp
Update 2: Another Alternative
SELECT id,
MIN(start_date) AS start_date,
MAX(end_Date) AS end_date
FROM (
SELECT t.*,
SUM(CASE WHEN start_date <= prev_max THEN 0 ELSE 1 END)
OVER (PARTITION BY id ORDER BY start_date) AS grp
FROM (
SELECT t.*,
MAX(end_date) OVER (
PARTITION BY id ORDER BY start_date
ROWS BETWEEN UNBOUNDED PRECEDING AND 1 PRECEDING
) AS prev_max
FROM table_name t
) t
) t
GROUP BY id, grp
db<>fiddle Oracle PostgreSQL
This is a gaps and islands problem. Try this:
with u as
(select ID, start_date, end_date,
case
when start_date <= lag(end_date) over(partition by ID order by start_date, end_date) then 0
else 1 end as grp
from table_name),
v as
(select ID, start_date, end_date,
sum(grp) over(partition by ID order by start_date, end_date) as island
from u)
select ID, min(start_date) as start_Date, max(end_date) as end_date
from v
group by ID, island;
Fiddle
Basically you can identify "islands" by comparing start_date of current row to end_date of previous row (ordered by start_date, end_date), if it precedes it then it's the same island. Then you can do a rolling sum() to get the island numbers. Finally select min(start_date) and max(end_date) from each island to get the desired output.
This may work ,with little bit of change in function , I tried it in Dbeaver :
select ID,Start_Date,End_Date
from
(
select t.*,
dense_rank () over(partition by extract (year from Start_Date) order BY End_Date desc) drnk
from testing_123 t
) temp
where temp.drnk = 1
ORDER BY Start_Date;
Try this
WITH a as (
SELECT
ID,
LEFT(Start_Date, 4) as Year,
MIN(Start_Date) as New_Start_Date
FROM
TAB1
GROUP BY
ID,
LEFT(Start_Date, 4)
), b as (
SELECT
a.ID,
Year,
New_Start_Date,
End_Date
FROM
a
LEFT JOIN
TAB1
ON LEFT(a.New_Start_Date, 4) = LEFT(TAB1.Start_Date, 4)
)
select
ID,
New_Start_Date as Start_Date,
MAX(End_Date)
from
b
GROUP BY
ID,
New_Start_Date;
Example: https://dbfiddle.uk/?rdbms=mysql_8.0&fiddle=97f91b68c635aebfb752538cdd752ace
I have the following data which I want to select as follows:
How can I modify the query to select the output as shown below?
select primary_id, timestamp, secondary_id,... from tablename where
timestamp <= to_timestamp('2020-07-29 00:00:00', 'YYYY-MM-DD HH24:MI:SS') and
timestamp < to_timestamp('2020-07-29 04:00:00', 'YYYY-MM-DD HH24:MI:SS')
order by timestamp, secondary_id;
primary_id timestamp secondary_id attribute1 attribute2 ... -- I want to get
-------------------------------------------------------------------
1 2020/01/20 10 ... ... ... -- <- this
2 2020/02/28 10 ... ... ...
3 2020/03/01 10 ... ... ... -- <- and this
4 2020/04/08 20 ... ... ... -- <- this
5 2020/05/31 20 ... ... ...
6 2020/06/30 20 ... ... ...
7 2020/06/31 20 ... ... ...
8 2020/07/31 20 ... ... ... -- <- and this
You can use window functions to rank records having the same secondary_id by ascending and descending timestamp, and then use that information to filter in the first and last record per group:
select primary_id, timestamp, secondary_id, ...
from (
select
t.*,
row_number() over(partition by secondary_id order by timestamp asc ) rn_asc,
row_number() over(partition by secondary_id order by timestamp desc) rn_desc
from tablename t
where
timestamp <= timestamp '2020-07-29 00:00:00'
and timestamp < timestamp '2020-07-29 04:00:00'
) t
where 1 in (rn_asc, rn_desc)
order by timestamp, secondary_id;
Note that you don't need to_timestamp() to convert these literal strings: you can use literal dates instead.
This also works when the value of secondary_id can be repeated in another group of rows, it simply checks if the current id is different from the previous or next row:
select *
from (
select
t.*,
lag(secondary_id) over(order by timestamp asc ) lag_id,
lead(secondary_id) over(order by timestamp asc) lead_id
from tablename t
where timestamp <= timestamp '2020-07-29 00:00:00'
and timestamp < timestamp '2020-07-29 04:00:00'
) t
where lag_id is null
or lead_id is null
or lag_id <> secondary_id
or lead_id <> secondary_id
order by timestamp, secondary_id;
Should be quite efficient as there's the same ORDER BY for both LEAD & LAG.
Please use below query,
select primary_id, timestamp, secondary_id,... from
(select primary_id, timestamp, secondary_id,...,
row_number() over (partition by secondary_id order by timestamp) as rnk1,
row_number() over (partition by secondary_id order by timestamp desc) as rnk2
from tablename where
timestamp <= to_timestamp('2020-07-29 00:00:00', 'YYYY-MM-DD HH24:MI:SS') and
timestamp < to_timestamp('2020-07-29 04:00:00', 'YYYY-MM-DD HH24:MI:SS') ) qry
where rnk1=1 and rnk2 = 1
order by timestamp, secondary_id;
You could use first_value and last_value. These are analytics functions and can be used like in the demo below.
with demo_data ( primary_id, secondary_id, timestamp)as
( select 1, 10, date '2020-01-01' from dual
union all
select 2 ,10, date '2020-01-28' from dual
union all
select 3, 10, date '2020-02-03' from dual
union all
select 4, 20, date '2020-03-02' from dual
union all
select 5, 20, date '2020-03-15' from dual
)
, grouped_data as
( select primary_id,
secondary_id,
timestamp,
decode(first_value(primary_id) over(partition by secondary_id order by timestamp ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING ), primary_id, 'Y', 'N') first_row_in_group,
decode(last_value(primary_id) over(partition by secondary_id order by timestamp ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING), primary_id, 'Y', 'N') last_row_in_group
from demo_data
)
select primary_id, secondary_id, timestamp
from grouped_data s
where first_row_in_group = 'Y' or last_row_in_group = 'Y'
/
I have a problem with choosing from the list of absences, those that follow one another and grouping them into periods.
date_from (data_od) date_to(data_do)
--------------------------
18/08/01 - 18/08/15
18/08/16 - 18/08/20
18/08/21 - 18/08/31
18/09/01 - 18/09/08
18/05/01 - 18/05/31
18/06/01 - 18/06/30
18/03/01 - 18/03/18
18/02/14 - 18/02/28
above is a list of absences, and the result of which should be a table:
date_from (data_od) date_to(data_do)
--------------------------
18/08/01 18/09/08
18/05/01 18/06/30
18/02/14 18/03/18
For now, I did something like this, but I only research in twos :(
SELECT u1.data_od,u2.data_do
FROM l_absencje u1 CROSS APPLY
(SELECT * FROM l_absencje labs
WHERE labs.prac_id=u1.prac_id AND
TRUNC(labs.data_od) = TRUNC(u1.data_do)+1
ORDER BY id DESC FETCH FIRST 1 ROWS ONLY
) u2 where u1.prac_id=1067 ;
And give me that:
18/08/01 18/08/20 bad
18/08/16 18/08/31 bad
18/08/21 18/09/08 bad
18/05/01 18/06/30 good
18/02/14 18/03/18 good
You can use a combination of the LAG(), LEAD() and LAST_VALUE() analytic functions:
SQL Fiddle
Oracle 11g R2 Schema Setup:
CREATE TABLE absences ( date_from, date_to ) AS
SELECT DATE '2018-08-01', DATE '2018-08-15' FROM DUAL UNION ALL
SELECT DATE '2018-08-16', DATE '2018-08-20' FROM DUAL UNION ALL
SELECT DATE '2018-08-21', DATE '2018-08-31' FROM DUAL UNION ALL
SELECT DATE '2018-09-01', DATE '2018-09-08' FROM DUAL UNION ALL
SELECT DATE '2018-05-01', DATE '2018-05-31' FROM DUAL UNION ALL
SELECT DATE '2018-06-01', DATE '2018-06-30' FROM DUAL UNION ALL
SELECT DATE '2018-03-01', DATE '2018-03-18' FROM DUAL UNION ALL
SELECT DATE '2018-02-14', DATE '2018-02-28' FROM DUAL;
Query 1:
SELECT *
FROM (
SELECT CASE
WHEN date_to IS NOT NULL
THEN LAST_VALUE( date_from ) IGNORE NULLS
OVER( ORDER BY ROWNUM )
END AS date_from,
date_to
FROM (
SELECT CASE date_from
WHEN LAG( date_to ) OVER ( ORDER BY date_to )
+ INTERVAL '1' DAY
THEN NULL
ELSE date_from
END AS date_from,
CASE date_to
WHEN LEAD( date_from ) OVER ( ORDER BY date_from )
- INTERVAL '1' DAY
THEN NULL
ELSE date_to
END AS date_to
FROM absences
)
)
WHERE date_from IS NOT NULL
AND date_to IS NOT NULL
Results:
| DATE_FROM | DATE_TO |
|----------------------|----------------------|
| 2018-02-14T00:00:00Z | 2018-03-18T00:00:00Z |
| 2018-05-01T00:00:00Z | 2018-06-30T00:00:00Z |
| 2018-08-01T00:00:00Z | 2018-09-08T00:00:00Z |
I have a table that looks like this:
+--------------------+---------+
| Month (date) | amount |
+--------------------+---------+
| 2016-10-01 | 20 |
| 2016-08-01 | 10 |
| 2016-07-01 | 17 |
+--------------------+---------+
I'm looking for a query (sql statement) which satisfies the following conditions:
Give me the value of the previous month.
If there is no value for the previous month lock back in time until one can be found.
If there is just a value for the current month give me this value.
In the example table the row I'm looking for would be this:
+--------------------+---------+
| 2016-08-01 | 10 |
+--------------------+---------+
Has anyone a idea for a non complex select query?
Thanks in advance,
Peter
You may need the following:
SELECT *
FROM ( SELECT *
FROM test
WHERE TRUNC(SYSDATE, 'month') >= month
ORDER BY CASE
WHEN TRUNC(SYSDATE, 'month') = month
THEN 0 /* if current month, ordered last */
ELSE 1 /* previous months are ordered first */
END DESC,
month DESC /* among previous months, the greatest first */
)
WHERE ROWNUM = 1
Another way using MAX
WITH tbl AS (
SELECT TO_DATE('2016-10-01', 'YYYY-MM-DD') AS "month", 20 AS amount FROM dual
UNION
SELECT TO_DATE('2016-08-01', 'YYYY-MM-DD') AS "month", 10 AS amount FROM dual
UNION
SELECT TO_DATE('2016-07-01', 'YYYY-MM-DD') AS "month", 5 AS amount FROM dual
)
SELECT *
FROM tbl
WHERE TRUNC("month", 'MONTH') = NVL((SELECT MAX(t."month")
FROM tbl t
WHERE t."month" < TRUNC(SYSDATE, 'MONTH')),
TRUNC(SYSDATE, 'MONTH'));
I would use row_number():
select t.*
from (select t.*,
row_number() over (order by (case when to_char(dte, 'YYYY-MM') = to_char(sysdate, 'YYYY-MM') then 1 else 2 end) desc,
dte desc
) as seqnum
from t
) t
where seqnum = 1;
Actually, you don't need row_number() for this:
select t.*
from (select t.*
from t
order by (case when to_char(dte, 'YYYY-MM') = to_char(sysdate, 'YYYY-MM') then 1 else 2 end) desc,
dte desc
) t
where rownum = 1;
It's not the nicest query but it should work.
select amount, date from (
select amount, date, row_number over(partition by HERE_PUT_ID order by
case trunc(date, 'month') when trunc(sysdate, 'month') then to_date('00010101', 'yyyymmdd') else trunc(date, 'month') end
desc) r)
where r = 1;
I guess you have some id in table so put id column instead of HERE_PUT_ID if you want query for whole table just delete: partition by HERE_PUT_ID
I added more data for testing, and an "id" column (a more realistic scenario) to show how this would work. If there is no "id" in your data, simply delete any reference to it from the solution.
Notes - month is a reserved Oracle word, don't use it as a column name. The solution assumes the date column contains dates that are already truncated to the beginning of the month. The trick in "order by" in the dense_rank last is to assign a value (ANY value!) when the month is the current month; by default, the value assigned to all other months is NULL, which by default come after any non-null value in an ascending order.
You may want to test the various solutions for efficiency if execution time is important.
with
inputs ( id, mth, amount ) as (
select 1, date '2016-10-01', 20 from dual union all
select 1, date '2016-08-01', 10 from dual union all
select 1, date '2016-07-01', 17 from dual union all
select 2, date '2016-10-01', 30 from dual union all
select 2, date '2016-09-01', 25 from dual union all
select 3, date '2016-10-01', 20 from dual union all
select 4, date '2016-08-01', 45 from dual union all
select 4, date '2016-06-01', 30 from dual
)
-- end of TEST DATA - the solution (SQL query) is below this line
select id,
max(mth) keep(dense_rank last order by
case when mth = trunc(sysdate, 'mm') then 0 end, mth) as mth,
max(amount) keep(dense_rank last order by
case when mth = trunc(sysdate, 'mm') then 0 end, mth) as amount
from inputs
group by id
order by id -- ORDER BY is optional
;
ID MTH AMOUNT
--- ---------- -------
1 2016-08-01 10
2 2016-09-01 25
3 2016-10-01 20
4 2016-08-01 45
You could sort the data in the direction you want to:
with MyData as
(
SELECT to_date('2016-10-01','YYYY-MM-DD') MY_DATE, 20 AMOUNT FROM DUAL UNION
SELECT to_date('2016-08-01','YYYY-MM-DD') MY_DATE, 10 AMOUNT FROM DUAL UNION
SELECT to_date('2016-07-01','YYYY-MM-DD') MY_DATE, 17 AMOUNT FROM DUAL
),
MyResult AS (
SELECT
D.*
FROM MyData D
ORDER BY
DECODE(
12*TO_CHAR(MY_DATE,'YYYY') + TO_CHAR(MY_DATE,'MM'),
12*TO_CHAR(SYSDATE,'YYYY') + TO_CHAR(SYSDATE,'MM'),
-1,
12*TO_CHAR(MY_DATE,'YYYY') + TO_CHAR(MY_DATE,'MM'))
DESC
)
SELECT * FROM MyResult WHERE RowNum = 1
I'm trying to figure out a way of identifying a "run" of results (successive rows, in order) that meet some condition. Currently, I'm ordering a result set, and scanning by eye for particular patterns. Here's an example:
SELECT the_date, name
FROM orders
WHERE
the_date BETWEEN
to_date('2013-09-18',..) AND
to_date('2013-09-22', ..)
ORDER BY the_date
--------------------------------------
the_date | name
--------------------------------------
2013-09-18 00:00:01 | John
--------------------------------------
2013-09-19 00:00:01 | James
--------------------------------------
2013-09-20 00:00:01 | John
--------------------------------------
2013-09-20 00:00:02 | John
--------------------------------------
2013-09-20 00:00:03 | John
--------------------------------------
2013-09-20 00:00:04 | John
--------------------------------------
2013-09-21 16:00:01 | Jennifer
--------------------------------------
What I want to extract from this result set is all the rows attributed to John on 2013-09-20. Generally what I'm looking for is a run of results from the same name, in a row, >= 3. I'm using Oracle 11, but I'm interested to know if this can be achieved with strict SQL, or if some kind of analytical function must be used.
You need multiple nested window functions:
SELECT *
FROM
(
SELECT the_date, name, grp,
COUNT(*) OVER (PARTITION BY grp) AS cnt
FROM
(
SELECT the_date, name,
SUM(flag) OVER (ORDER BY the_date) AS grp
FROM
(
SELECT the_date, name,
CASE WHEN LAG(name) OVER (ORDER BY the_date) = name THEN 0 ELSE 1 END AS flag
FROM orders
WHERE
the_date BETWEEN
TO_DATE('2013-09-18',..) AND
TO_DATE('2013-09-22', ..)
) dt
) dt
) dt
WHERE cnt >= 3
ORDER BY the_date
Try this
WITH ORDERS
AS (SELECT
TO_DATE ( '2013-09-18 00:00:01',
'YYYY-MM-DD HH24:MI:SS' )
AS THE_DATE,
'John' AS NAME
FROM
DUAL
UNION ALL
SELECT
TO_DATE ( '2013-09-19 00:00:01',
'YYYY-MM-DD HH24:MI:SS' )
AS THE_DATE,
'James' AS NAME
FROM
DUAL
UNION ALL
SELECT
TO_DATE ( '2013-09-20 00:00:01',
'YYYY-MM-DD HH24:MI:SS' )
AS THE_DATE,
'John' AS NAME
FROM
DUAL
UNION ALL
SELECT
TO_DATE ( '2013-09-20 00:00:02',
'YYYY-MM-DD HH24:MI:SS' )
AS THE_DATE,
'John' AS NAME
FROM
DUAL
UNION ALL
SELECT
TO_DATE ( '2013-09-20 00:00:03',
'YYYY-MM-DD HH24:MI:SS' )
AS THE_DATE,
'John' AS NAME
FROM
DUAL
UNION ALL
SELECT
TO_DATE ( '2013-09-20 00:00:04',
'YYYY-MM-DD HH24:MI:SS' )
AS THE_DATE,
'John' AS NAME
FROM
DUAL
UNION ALL
SELECT
TO_DATE ( '2013-09-21 16:00:01',
'YYYY-MM-DD HH24:MI:SS' )
AS THE_DATE,
'Jennifer' AS NAME
FROM
DUAL)
SELECT
B.*
FROM
(SELECT
TRUNC ( THE_DATE ) THE_DATE,
NAME,
COUNT ( * )
FROM
ORDERS
WHERE
THE_DATE BETWEEN TRUNC ( TO_DATE ( '2013-09-18',
'YYYY-MM-DD' ) )
AND TRUNC ( TO_DATE ( '2013-09-22',
'YYYY-MM-DD' ) )
GROUP BY
TRUNC ( THE_DATE ),
NAME
HAVING
COUNT ( * ) >= 3) A,
ORDERS B
WHERE
A.NAME = B.NAME
AND TRUNC ( A.THE_DATE ) = TRUNC ( B.THE_DATE );
OUTPUT
9/20/2013 12:00:01 AM John
9/20/2013 12:00:02 AM John
9/20/2013 12:00:03 AM John
9/20/2013 12:00:04 AM John