SQL Oracle How to convert String Week into date - sql

I have dates stored like String in database.
The format is 'yyyy-ww' (example: '2015-43').
I need to get the first day of the week.
I tried to convert this string into date but there is no 'ww' option for the function "to_date".
Do you have an idea to perform this convertion?
EDIT
Test results based on the answers -
Thanks for your anwsers, but I have many problems to apply your solutions to my context:
select
TRUNC ( 2015 + ((43 - 1) * 7), 'IW' )
from dual
==> Error : ORA-01722: invalid number
select
TRUNC(to_date('2015','YYYY')+ to_number('01') *7, 'IW')
from dual
==> 2015-02-02 00:00:00
I waited for a date in january
select
trunc(to_date(regexp_substr('2015-01', '\d+',1,2), 'YYYY') + regexp_substr('2015-01', '\d+') * 7, 'IW') dt2
from dual
==> 0039-09-14 00:00:00
select
regexp_substr('2015-01', '\d+',1,2) as res1,
regexp_substr('2015-01', '\d+') * 7 as res2
from dual
==> res1 = 01
==> res2 = 14105

try to use by truncate
with t as (
select '16-2010' dt from dual
)
--
--
select dt,
trunc(to_date(regexp_substr(dt, '\d+',1,2), 'YYYY') + regexp_substr(dt, '\d+') * 7, 'IW') dt2
from t

I have dates stored like String in database.
You should never do that. It is a bad design. you should store date as DATE and not as a string. For all kinds of requirements for date manipulations Oracle provides the required DATE functions and format models. As and when needed, you could extract/display the way you want.
I need to get the first day of the week.
TRUNC (dt, 'IW') returns the Monday on or before the given date.
Anyway, in your case, you have the literal as YYYY-WW format. You could first extract the year and week number and combine them together to get the date using TRUNC.
TRUNC ( year + ((week_number - 1) * 7)
, 'IW
)
So, the above should give you the Monday of the week number passed for that year.
SQL> WITH DATA AS
2 ( SELECT '2015-43' str FROM dual
3 )
4 SELECT TRUNC(to_date(SUBSTR(str, 1, 4),'YYYY')+ to_number(SUBSTR(str, instr(str, '-',1)+1))*7, 'IW')
5 FROM DATA
6 /
TRUNC(TO_
---------
23-NOV-15
SQL>

Similar to Lalit's, however, I think I've corrected the math (his seemed to be off a bit when I tested .. )
with w_data as (
select sysdate + level +200 d from dual connect by level <= 10
),
w_weeks as (
select d, to_char(d,'yyyy-iw') c
from w_data
)
SELECT d, c, trunc(d,'iw'),
TRUNC(
to_date(SUBSTR(c, 1, 4)||'0101','yyyymmdd')-8+to_char(to_date(SUBSTR(c, 1, 4)||'0101','yyyymmdd'),'d')
+to_number(SUBSTR(c, instr(c, '-',1)+1)-1)*7 ,'IW')
FROM w_weeks;
The extra columns help show the dates before, and after.

I would do the following:
WITH d1 AS (
SELECT '2015-43' AS mydate FROM dual
)
SELECT TRUNC(TRUNC(TO_DATE(REGEXP_SUBSTR(mydate, '^\d{4}'), 'YYYY'), 'YEAR') + (COALESCE(TO_NUMBER(REGEXP_SUBSTR(mydate, '\d+$')), 0)-1) * 7, 'IW')
FROM d1
The first thing the above query does is get the first four digits of the string 2015-43 and truncates that to the closest year (if you convert convert 2015 using TO_DATE() it returns a date within the current month; that is SELECT TO_DATE('2015', 'YYYY') FROM dual returns 01-FEB-2015; we need to truncate this value to the YEAR in order to get 01-JAN-2015). I then add the number of weeks minus one times seven and truncate the whole thing by IW. This returns a date of 01-OCT-2015 (see SQL Fiddle here).

According ISO the 4th of January is always in week 1, so your query should look like
Select
TRUNC(TO_DATE(REGEXP_SUBSTR(your_column, '^\d{4}')||'-01-04', 'YYYY-MM-DD')
+ 7*(REGEXP_SUBSTR(your_column, '\d$')-1), 'IW')
from your_table;
However, there is a problem. ISO year used for Week number can be different than actual year. For example, 1st Jan 2008 was in ISO week number 53 of 2007.
I think a proper working solution you get only when you generate ISO weeks from date value.
WITH w AS
(SELECT TO_CHAR(DATE '2010-01-04' + LEVEL * INTERVAL '7' DAY, 'IYYY-IW') AS week_number,
TRUNC(DATE '2010-01-04' + LEVEL * INTERVAL '7' DAY, 'IW') AS first_day
FROM dual
CONNECT BY DATE '2010-01-04' + LEVEL * INTERVAL '7' DAY < SYSDATE)
SELECT your_Column, first_day
FROM w your_table
JOIN w ON week_number = your_Column;
Your date range must bigger than 2010-01-04 and not bigger than current day.

This is what I used:
select
to_date(substr('2017/01',1,4)||'/'||to_char(to_number(substr('2017/01',6,2)*7)-5),'yyyy/ddd') from dual;

Related

Get whole month dates based on day name

I want to get all dates of a month based on day name like if I want to get all FRIDAY dates from current date to a specific end date. I got a solution in SQL but can't find any solution in ORACLE.
You can use hierarchy query with next_day function as follows:
-- input variable
with dataa (start_date, end_date) as
(select date '2020-11-01', date '2020-12-15' from dual)
-- your query starts from here
select next_day((level-1)*7 + (start_date - 1), 'FRIDAY')
FROM DATAA
connect by
next_day((level-1)*7 + (start_date - 1),'FRIDAY') <= end_date
Db<>fiddle here
You can use hierarchical query as
WITH t AS
(
SELECT TRUNC(sysdate,'MM')+level-1 AS Fridays
FROM dual
WHERE TO_CHAR(TRUNC(sysdate,'MM')+level-1,'Dy','NLS_DATE_LANGUAGE=American')='Fri'
CONNECT BY level <= LAST_DAY(TRUNC(sysdate)) - TRUNC(sysdate,'MM') + 1
)
SELECT *
FROM t
WHERE Fridays BETWEEN :data1 AND :date2
Demo
Using connect by then filtering by Friday is a simple solution. In the example below I am using July 1st as the start date and December 1st as the end date.
SELECT DATE '2020-7-1' + LEVEL AS friday_dates
FROM DUAL
WHERE TO_CHAR (DATE '2020-7-1' + LEVEL, 'DY') = 'FRI'
CONNECT BY LEVEL <= DATE '2020-12-1' - DATE '2020-7-1';

Oracle SQl Dev, how to calc num of weekdays between 2 dates

Does anyone know how can I calculate the number of weekdays between two date fields? I'm using oracle sql developer. I need to find the average of weekdays between multiple start and end dates. So I need to get the count of days for each record so I can average them out. Is this something that can be done as one line in the SELECT part of my query?
This answer is similar to Nicholas's, which isn't a surprise because you need a subquery with a CONNECT BY to spin out a list of dates. The dates can then be counted while checking for the day of the week. The difference here is that it shows how to get the weekday count value on each line of the results:
SELECT
FromDate,
ThruDate,
(SELECT COUNT(*)
FROM DUAL
WHERE TO_CHAR(FromDate + LEVEL - 1, 'DY') NOT IN ('SAT', 'SUN')
CONNECT BY LEVEL <= ThruDate - FromDate + 1
) AS Weekday_Count
FROM myTable
The count is inclusive, meaning it includes FromDate and ThruDate. This query assumes that your dates don't have a time component; if they do you'll need to TRUNC the date columns in the subquery.
You could do it the following way :
Lets say we want to know how many weekdays between start_date='01.08.2013' and end_date='04.08.2013' In this example start_date and end_date are string literals. If your start_date and end_date are of date datatype, the TO_DATE() function won't be needed:
select count(*) as num_of_weekdays
from ( select level as dnum
from dual
connect by (to_date(end_date, 'dd.mm.yyyy') -
to_date(start_date, 'dd.mm.yyyy') + 1) - level >= 0) s
where to_char(sysdate + dnum, 'DY',
'NLS_DATE_LANGUAGE=AMERICAN') not in ('SUN', 'SAT')
Result:
num_of_weekdays
--------------
2
Checkout my complete working function code and explanation at
https://sqljana.wordpress.com/2017/03/16/oracle-calculating-business-days-between-two-dates-in-oracle/
Once you have created the function just use the function as part of the SELECT statement and pass in the two date columns for Start and End dates like this:
SELECT Begin_Date, End_Date, fn_GetBusinessDaysInterval(Begin_Date, End_Date) AS BusinessDays FROM YOURTABLE;

Select Dates where time is less or equal to '12:00' Oracle

I require a query that selects rows where the time is less or equal to 12:00
I had something like this in mind:
SELECT daterow FROM datecolumn WHERE daterow <= TO_DATE('12:00, HH24:MI')
However i get an error:
ORA-01843: not a valid month
How would i go about to get all rows that have a time less than 12:00 mid-day?
Try this,
SELECT daterow FROM datecolumn WHERE daterow <= TO_DATE('12:00', 'HH24:MI');
Try This:
SELECT daterow FROM datecolumn
WHERE TO_DATE(daterow,'HH24:MI') <= TO_DATE('12:00', 'HH24:MI');
In order to select all rows where time portion of the daterow column value is less than or equal to mid-day 12:00 you can use to_char() function to extract hour and minutes and to_number() to convert it to a number for further comparison:
-- sample of data. Just for the sake of demonstration
SQL> with t1(col) as(
2 select sysdate - to_dsinterval('P0DT3H') from dual union all
3 select sysdate - to_dsinterval('P0DT2H') from dual union all
4 select sysdate - to_dsinterval('P0DT1H') from dual union all
5 select sysdate + to_dsinterval('P0DT3H') from dual union all
6 select sysdate + to_dsinterval('-P2DT0H') from dual
7 )
8 select to_char(col, 'dd.mm.yyyy hh24:mi:ss') as res
9 from t1 t
10 where to_number(to_char(col, 'hh24mi')) <= 1200
11 ;
Result:
RES
-------------------
26.08.2013 08:10:59
26.08.2013 09:10:59
26.08.2013 10:10:59
24.08.2013 11:10:59
Sorry, but <= TO_DATE('12:00', 'HH24:MI') does not work. It does not extract the hour and minute from each date and compares it to 12:00. Instead it constructs the date representing high noon on the fisrt of the current month and compares each date to this date.
If you want to extract something from a date, use the extract function.
Attention: When using extract on a date, and want to extract hours, minutes or seconds, you have first to convert the date to a timestamp.
Example:
SELECT
extract(hour FROM cast(A AS TIMESTAMP)) AS h,
extract(MINUTE FROM cast(A AS TIMESTAMP)) AS m
FROM
DEMO
;
You can find a complete example on sqlfiddle. The example also shows that the to_date method doesn't work.

Finding day of the week in oracle

to_char(sysdate,'Day') returns the current day of the week. What I want is to get the date of the most recent "sunday" that passed. Of course we can go for a complex query to get it done. But is there any simple way I'm not aware of?
You can do it with
SELECT NEXT_DAY(SYSDATE-8, 'SUN') FROM DUAL;
here
SYSDATE-8
returns the day before 8 days &
NEXT_DAY(mydate, 'SUN')
returns the next sunday to it
You can do it with a query as simple as this:
SELECT dt
FROM ( SELECT SYSDATE - LEVEL - 1 dt
FROM DUAL
CONNECT BY LEVEL < 8)
WHERE TO_CHAR ( dt, 'd' ) = 1
I think NEXT_DAY(SYSDATE-6, 'SUN') would work. Assuming current day as sunday, I go back 6 days (i.e. last monday) and so when I look for the next sunday, I'd get the sysdate itself. Whereas, next_day(sysdate-8,'SUN') could be used to get the last day of previous week. Thanks for your efforts.
SELECT dt
FROM ( SELECT to_date('5/23/2014') - LEVEL + 1 dt
FROM DUAL
CONNECT BY LEVEL < 8)
WHERE TO_CHAR ( dt, 'd' ) = 1
I changed LEVEL - 1 dt to LEVEL + 1 dt to get it to work to find the previous Sunday.

How can I extract and display the date range based on Quarter in Oracle?

I'm building a summary table as part of a stored procedure and I have two columns. The first column needs to show the start and the second column needs to show the end of a date range that is based from an input parameter that is a number designating the quarter. I was able to extract the following from AskTom but I have some questions.
Open C1 FOR
SELECT ( SELECT TRUNC (SYSDATE, 'Q')-1+1 AS 'StartOf' FROM DUAL ),
SELECT ( SELECT TRUNC(ADD_MONTHS (SYSDATE, +3), 'Q')-2 AS 'EndOf' FROM DUAL )
FROM DUAL;
Question 1. Will the Math here account for LeapYears... I don't think it will but I'm not sure how to handle that.
Question 2. How do I add the input parameter, 'inQuarter' as the specific quarter? I've tried putting it in place of sysdate but I need to reformat it into date first I think?
Thanks in advance for any responses.
Tom Kyte has givven you the answer:
http://asktom.oracle.com/pls/asktom/f?p=100:11:0::::P11_QUESTION_ID:250288900346075009
Open C1 FOR
select add_months( dt, (inQuarter-1)*3 ),
last_day(add_months( dt, (inQuarter-1)*3+2 ) )
from (
select to_date( inYear || '0101', 'yyyymmdd' ) dt
from dual)
You can convert a numeric year and numeric quarter parameter to a DATE
SELECT add_months( trunc( to_date( to_char( <<numeric year>> ),
'YYYY' ),
'YYYY' ),
3 * <<numeric quarter>> ) first_of_quarter,
add_months( trunc( to_date( to_char( <<numeric year>> ),
'YYYY' ),
'YYYY' ),
4 * <<numeric quarter>> ) - 1 last_of_quarter,
FROM dual
The time component of both dates will be midnight on the last day of the quarter (i.e. 24 hours before the beginning of the next quarter). You may want the last of the quarter to be 23:59:59 on the last day of the quarter if you want the range to be inclusive of all possible dates in the quarter.
I suggest:
Open C1 FOR
SELECT TRUNC (d_inQuarter, 'Q') AS "StartOf",
TRUNC(ADD_MONTHS (d_inQuarter, +3), 'Q') AS "EndOf"
FROM (SELECT add_months(to_date(to_char(i_yr)||'-01-01','YYYY-MM-DD'), (i_q-1)*3)
AS d_inQuarter FROM DUAL);
- with integer parameters i_yr and i_q representing the year and quarter, respectively.
Note that EndOf will represent midnight on the first day of the next quarter, so any selection should be based on conditions < "EndOf", not <= "EndOf". (This should ensure that all times on the last day of the quarter are included.)