I need to compare rows in the same table of a query.
Here is an example of the table:
id checkin checkout
1 01/15/13 01/31/13
1 01/31/13 05/20/13
2 01/15/13 05/20/13
3 01/15/13 01/19/13
3 01/19/13 05/20/13
4 01/15/13 02/22/13
5 01/15/13 03/01/13
I compare the checkout date to today's date, if it is before today's date then I want to return the result. However, similar to id's 1 and 3, they have multiple records. If one of the records associated with the same id have a record that has a checkout date after today's date then I don't want to return any of their records. I only want to return a record of each id where every record is before today's date in the checkout field.
For this purpose, analytic functions are the best approach:
select id, checkin, checkout
from (select t.*, max(checkout) over (partition by id) as maxco
from t
) t
where maxco <= trunc(sysdate)
This assumes that the data is stored as date values and not as strings (otherwise, the max will return the wrong value).
select id, checking from
Table
where checkout < CURRENT_DATE --Postgresql date of today, Oracle should have an equivalent now
and id not in (select id from Table where checkout >= CURRENT_DATE);
This should give you all the results of records that are before in time and that all with the same id are also before in time.
SELECT Ta.*
FROM TABLE Ta,
(SELECT MAX(checkout) checkout, ID FROM TABLE GROUP BY ID) Tb
WHERE Ta.ID = Tb.ID
AND sysdate >= Ta.checkout -- checks for date in current record
AND sysdate >= Tb.checkout -- checks for dates in all records
Related
I have a google bigquery table with orders with a DATE column and other columns related to the orders. The starting date of the dataset is from 2021-01-01 (yyyy-mm-dd).
My aim is to filter on the DATE column from last year and this year to the previous iso week. For this, I used the ISOWEEK to create a new column:
WITH
last_week_last_year AS (
SELECT
DATE,
EXTRACT(ISOWEEK FROM DATE) AS isoweek,
FROM
`orders`
WHERE
EXTRACT(ISOWEEK FROM DATE) = EXTRACT(ISOWEEK FROM CURRENT_DATE())-1
GROUP BY 1, 2
ORDER BY DATE
)
SELECT * FROM last_week_last_year
This query results as the following table:
The issue is that when I filter on the original orders table by the DATE from the last_week_last_year table I get all the orders back instead of just the filtered version.
My method to filter is WHERE DATE IN (SELECT DATE FROM last_week_last_year) as seen below.
SELECT
*
FROM
`orders`
WHERE
DATE IN (SELECT DATE FROM last_week_last_year)
ORDER BY DATE DESC;
A snapshot of resulting table. It contains all of the records from 2021-01-01 until the latest day.
How can I make sure that on the latter query the table is filtered based on the first query's dates in DATE column?
I can't get my head around whether this is even possible, but I feel like I might have done it before and lost that bit of code. I am trying to craft a select statement that contains an inner join on a subquery to show the number of days between two dates from the same table.
A simple example of the data structure would look like:
Name ID Date Day Hours
Bill 1 3/3/20 Thursday 8
Fred 2 4/3/20 Monday 6
Bill 1 8/3/20 Tuesday 2
Based on this data, I want to select each row plus an extra column which is the number of days between the date from each row for each ID. Something like:
Select * from tblData
Inner join (datediff(Select Top(1) Date from tblData where Date < Date), Date) And ID = ID)
or for simplicity:
Select * from tblData
Inner join (datediff(Select Top(1) Date from tblData where Date < 8/3/20), 8/3/20) And ID = 1)
The resulting dataset would look like:
Name ID Date Day Hours DaysBtwn
Bill 1 3/3/20 Thursday 8 4 (Assuming there was an earlier row in the table)
Fred 2 4/3/20 Monday 6 5 (Assuming there was an earlier row in the table)
Bill 1 8/3/20 Tuesday 2 5 (Based on the previous row date being 3/3/20 for Bill)
Does this make sense and am I trying to do this the wrong way? I want to do this for about 600000 rows in table and therefore efficiency is the key, so if there is a better way to do this, i'm open to suggestions.
You can use lag():
select t.*, datediff(day, lag(date) over(partition by id order by date), date) diff
from mytable t
I think you just want lag():
select t.*,
datediff(day,
lag(date) over (partition by name order by date),
date
) as diff
from tblData t;
Note: If you want to filter the data so rows in the result set are used for the lag() but not in the result set, then use a subquery:
select t.*
from (select t.*,
datediff(day,
lag(date) over (partition by name order by date),
date
) as diff
from tblData t
) t
where date < '2020-08-03';
Also note the use of the date constant as a string in YYYY-MM-DD format.
I need help in business days calculation.
I've two tables
1) One table ACTUAL_TABLE containing order date and contact date with timestamp datatypes.
2) The second table BUSINESS_DATES has each of the calendar dates listed and has a flag to indicate weekend days.
using these two tables, I need to ensure business days and not calendar days (which is the current logic) is calculated between these two fields.
My thought process was to first get a range of dates by comparing ORDER_DATE with TABLE_DATE field and then do a similar comparison of CONTACT_DATE to TABLE_DATE field. This would get me a range from the BUSINESS_DATES table which I can then use to calculate count of days, sum(Holiday_WKND_Flag) fields making the result look like:
Order# | Count(*) As DAYS | SUM(WEEKEND DATES)
100 | 25 | 8
However this only works when I use a specific order number and cant' bring all order numbers in a sub query.
My Query:
SELECT SUM(Holiday_WKND_Flag), COUNT(*) FROM
(
SELECT
* FROM
BUSINESS_DATES
WHERE BUSINESS.Business BETWEEN (SELECT ORDER_DATE FROM ACTUAL_TABLE
WHERE ORDER# = '100'
)
AND
(SELECT CONTACT_DATE FROM ACTUAL_TABLE
WHERE ORDER# = '100'
)
TEMP
Uploading the table structure for your reference.
SELECT ORDER#, SUM(Holiday_WKND_Flag), COUNT(*)
FROM business_dates bd
INNER JOIN actual_table at ON bd.table_date BETWEEN at.order_date AND at.contact_date
GROUP BY ORDER#
Instead of joining on a BETWEEN (which always results in a bad Product Join) followed by a COUNT you better assign a bussines day number to each date (in best case this is calculated only once and added as a column to your calendar table). Then it's two Equi-Joins and no aggregation needed:
WITH cte AS
(
SELECT
Cast(table_date AS DATE) AS table_date,
-- assign a consecutive number to each busines day, i.e. not increased during weekends, etc.
Sum(CASE WHEN Holiday_WKND_Flag = 1 THEN 0 ELSE 1 end)
Over (ORDER BY table_date
ROWS Unbounded Preceding) AS business_day_nbr
FROM business_dates
)
SELECT ORDER#,
Cast(t.contact_date AS DATE) - Cast(t.order_date AS DATE) AS #_of_days
b2.business_day_nbr - b1.business_day_nbr AS #_of_business_days
FROM actual_table AS t
JOIN cte AS b1
ON Cast(t.order_date AS DATE) = b1.table_date
JOIN cte AS b2
ON Cast(t.contact_date AS DATE) = b2.table_date
Btw, why are table_date and order_date timestamp instead of a date?
Porting from Oracle?
You can use this query. Hope it helps
select order#,
order_date,
contact_date,
(select count(1)
from business_dates_table
where table_date between a.order_date and a.contact_date
and holiday_wknd_flag = 0
) business_days
from actual_table a
Using SQL I need to return a smooth set of results (i.e. one per day) from a dataset that contains 0-N records per day.
The result per day should be the most recent previous value even if that is not from the same day. For example:
Starting data:
Date: Time: Value
19/3/2014 10:01 5
19/3/2014 11:08 3
19/3/2014 17:19 6
20/3/2014 09:11 4
22/3/2014 14:01 5
Required output:
Date: Value
19/3/2014 6
20/3/2014 4
21/3/2014 4
22/3/2014 5
First you need to complete the date range and fill in the missing dates (21/3/2014 in you example). This can be done by either joining a calendar table if you have one, or by using a recursive common table expression to generate the complete sequence on the fly.
When you have the complete sequence of dates finding the max value for the date, or from the latest previous non-null row becomes easy. In this query I use a correlated subquery to do it.
with cte as (
select min(date) date, max(date) max_date from your_table
union all
select dateadd(day, 1, date) date, max_date
from cte
where date < max_date
)
select
c.date,
(
select top 1 max(value) from your_table
where date <= c.date group by date order by date desc
) value
from cte c
order by c.date;
May be this works but try and let me know
select date, value from test where (time,date) in (select max(time),date from test group by date);
I have a table like this:
Id Begin_Date End_date
1 01-JAN-12 05-JAN-12
1 01-FEB-12 01-MAR-12
1 15-FEB-12 05-MAR-12
For a given Id, it gives a set of date ranges. Let's say that if a date is between the begin and end date for that Id, then that Id is "on". Otherwise, "off"
The problem here is these last two rows -- the date ranges overlap and contradict each other. The second row claims that the 1 was "on" between 01-FEB-12 and 01-MAR-123, but the third row claims that 1 was off before before 14-FEB-12. Similarly, the second row claims that 1 was off on 02-MAR-12, but row 3 claims it was on.
The reconciliation logic I'd like to apply is that, in cases of contradictions, pick the earliest possible begin date and the earliest possible end date after it. The result would therefore be:
Id Begin_Date End_date
1 01-JAN-12 05-JAN-12
1 01-FEB-12 01-MAR-12
I was able to pull this off with the lag analytical function, but I ran into difficulty with other use cases. Take this input data set.
Id Begin_Date End_date
1 01-JAN-12 10-JAN-12
1 5-JAN-12 8-JAN-12
1 12-JAN-12 15-JAN-12
1 1-JAN-12 14-JAN-12
What I expect here as output is:
Id Begin_Date End_date
1 01-JAN-12 8-JAN-12
1 01-JAN-12 14-JAN-12
...because the first row is the earliest begin date, and its end date is the earliest end date after that. The next row is the earliest begin date after the previous end date, and the end date of that row is the earliest end date after that. There are no begin dates after 14-JAN-12, so I'm done.
I'm having very little luck solving this problem. One approach I tried was getting the rank partitioned by id and compare it to the max rank. I then used the lag function to compare to previous ranks. However, this strategy totally fails for use cases above.
Any suggestions?
Well, the critical requirement rests on this:
The reconciliation logic I'd like to apply is that, in cases of
contradictions, pick the earliest possible begin date and the earliest
possible end date after it.
sqlfiddle here
CREATE TABLE table1
(
id INT,
DateStart DATE,
DateEnd DATE
);
INSERT INTO table1
VALUES
(1, TO_DATE('20110101','YYYYMMdd'), TO_DATE('20110110','YYYYMMdd'));
INSERT INTO table1
VALUES
(2, TO_DATE('20110105','YYYYMMdd'), TO_DATE('20110108','YYYYMMdd'));
INSERT INTO table1
VALUES
(3, TO_DATE('20110112','YYYYMMdd'), TO_DATE('20110115','YYYYMMdd'));
INSERT INTO table1
VALUES
(4, TO_DATE('20110101','YYYYMMdd'), TO_DATE('20110114','YYYYMMdd'));
INSERT INTO table1
VALUES
(5, TO_DATE('20110206','YYYYMMdd'), TO_DATE('20110208','YYYYMMdd'));
INSERT INTO table1
VALUES
(6, TO_DATE('20110201','YYYYMMdd'), TO_DATE('20110207','YYYYMMdd'));
The select statement:
SELECT ID, DATESTART, DATEEND
FROM
(
SELECT ID, TYPE, DATES AS DATESTART,
LEAD(DATES) OVER (ORDER BY DATES) AS DATEEND
FROM
(
SELECT ID, TYPE,DATES,
LAG(ID) OVER (ORDER BY DATES) AS LASTID,
LAG(TYPE) OVER (ORDER BY DATES) AS LASTTYPE,
LAG(DATES) OVER (ORDER BY DATES) AS LASTDATES
FROM
(
SELECT ID,'START' AS TYPE,DATESTART AS DATES
FROM table1
UNION ALL
SELECT ID,'END',DATEEND
FROM table1
)
) H
WHERE TYPE != LASTTYPE OR LASTTYPE IS NULL
)
WHERE TYPE = 'START'
ORDER BY DATESTART
Here's a step by step for each subquery:
explode each row's date start and date end into one column
copy the last row using LAG and put it in current row
filter out the rows which is are in the middle (e.g. 1,2,3,4 remove 2,3)
get the end date in the next row because these are either first or last rows
extract only useful rows, those rows which has TYPE = START
For the second data set:
Id Begin_Date End_date
1 01-JAN-12 10-JAN-12
1 5-JAN-12 8-JAN-12
1 12-JAN-12 15-JAN-12
1 1-JAN-12 14-JAN-12
After your reconciliation logic, the result would be:
Id Begin_Date End_date
1 01-JAN-12 8-JAN-12 (includes the rows 1,2 and 4 -> minimum begin_date is 1-JAN, minimum end_date is 8-JAN)
1 12-JAN-12 15-JAN-12 (includes row 3)