How do I determine how many days in a month, from the first to the current day (could be anywhere into the month). So if I have a field that gives 6/01/12 5:32:13 PM and 6/07/12 5:33:04 PM how do I get the difference?
I think something like this should do it:
SELECT TRUNC(SYSDATE) - TRUNC(SYSDATE,'MONTH') + 1 FROM DUAL
SELECT EXTRACT(DAY FROM sysdate) FROM dual;
Related
I'm working on a project with a db that contains a date column for patient visits in the format of %m-%d-yyyy and need to sort so that it only pulls the rows where that date is within the last two weeks. I've tried a few different functions of convert, to_date, and can't seem to get anything to work.
I'm still very new to SQL and I don't know if this is a special case because I'm working with an oracle db
Not the full code, because it has dozens of queries and multiple joins (would that affect the date syntax?) but this is the format I'm trying for...
create table "Visits"
insert into "Visits" (
'John Doe',
'5/24/2021',
'Story about the visit',
'More room for story if needed')
select
"User_Name",
"Visit_Date",
"Visit_Narrative",
"Visit_Narrative_Overflow"
from "Visits"
where "Visits"."Visit_Date" >= TRUNC(SYSDATE) - 14
I'm working on a project with a db that contains a date column for patient visits in the format of %m-%d-yyyy
No, you don't have it in the format mm.dd.yyyy. A DATE data type value is stored within the database as a binary value in 7-bytes (representing century, year-of-century, month, day, hour, minute and second) and this has no format.
need to sort so that it only pulls the rows where that date is within the last two weeks.
You want a WHERE filter:
If you want to have the values that happened in the last 14 days then TRUNCate the current date back to midnight and subtract 14 days:
SELECT visit_date
FROM patient
WHERE visit_date >= TRUNC(SYSDATE) - INTERVAL '14' DAY
or
SELECT visit_date
FROM patient
WHERE visit_date >= TRUNC(SYSDATE) - 14
(or subtract 13 if you want today and 13 days before today.)
If you want it after Monday of last week then TRUNCate to the start of the ISO Week (which is always a Monday) and subtract 7 days:
SELECT visit_date
FROM patient
WHERE visit_date >= TRUNC(SYSDATE, 'IW') - INTERVAL '7' DAY
I ended up figuring it out based on an answer from another forum (linked below);
it appears that my original to_date() was incomplete without a second and operator in my where clause. This code is working perfectly
select
"User_Name",
"Visit_Date",
"Visit_Narrative",
"Visit_Narrative_Overflow"
from "SQLUser"."Visits"
where "SQLUser"."Visits"."Visit_Date" >= to_date('5/10/2021', 'MM/DD/YYYY')
and "SQLUser"."Visits"."Visit_Date" < to_date('5/24/2021', 'MM/DD/YYYY')
I'm looking to get a quick and efficient code to get the first day of current year and everyday after that and their week numbers. So far I have something like this to start me off:
SELECT
to_char(TRUNC(SysDate,'YEAR'), 'WW')Week_No,
to_char(TRUNC(SysDate,'YEAR'), 'DD/MM/YY')First_Day
FROM dual;
Which gets me week number and first day of current year. I also have this to get me the last calendar day of current year:
SELECT ADD_MONTHS(trunc(sysdate,'YEAR'),12)-1 from dual;
I am being extremely silly or maybe it's end of the day brain mush but I'm trying to get the bit in the middle now. The point of this is to get a list of each calendar day and week number.
If anyone can guide me here I'd appreciate this.
Thanks
Are you looking for something like this:
SELECT
ST_DT + LEVEL - 1,
TO_CHAR(ST_DT + LEVEL - 1, 'IW') AS DY
FROM
(
SELECT
TRUNC(SYSDATE, 'YEAR') ST_DT,
TRUNC(SYSDATE, 'YEAR') + INTERVAL '12' MONTH - INTERVAL '1' DAY END_DT
FROM DUAL
)
CONNECT BY LEVEL <= END_DT - ST_DT
ORDER BY LEVEL
Cheers!!
I have spent days trying to figure this out to no avail, so hopefully someone can help me. I have a queried date set which contains several fields including a column of dates. What I want to do is create a new field in my query that tells what the Monday and Friday is for the week of that row's particular date.
So for example; if the date in one of my rows is "1/16/18",
the new field should indicate "1/15/18 - 1/19/18".
So basically I need to be able to extract the Monday date (1/15/18) and the Friday date (1/19/18) of the week of 1/16/18 and then concatenate the two with a dash ( - ) in between. I need to do this for every row.
How on earth do I do this? I've been struggling just to figure out how to find the Monday or Friday of the given date...
Assuming that your column is of type date, you can use trunc to get the first day of the week (monday) and then add 4 days to get the friday.
For example:
with yourTable(d) as (select sysdate from dual)
select trunc(d, 'iw'), trunc(d, 'iw') + 4
from yourTable
To format the date as a string in the needed format, you can use to_char; for example:
with yourTable(d) as (select sysdate from dual)
select to_char(trunc(d, 'iw'), 'dd/mm/yy') ||'-'|| to_char(trunc(d, 'iw') + 4, 'dd/mm/yy')
from yourTable
gives
15/01/2018-19/01/18
There may be a simpler, canonical Oracle method to this but you can still reduce it to a simple calculation on your own either way. I'm going to assume you're dealing with only dates falling Monday through Friday. If you do need to deal with weekend dates then you might have to be more explicit about which logical week they should be attached to.
<date> - (to_char(<date>, 'D') - 2) -- Monday
<date> + (6 - to_char(<date>, 'D')) -- Friday
In principle all you need to do is add/subtract the appropriate number of days based on the current day of week (from 1 - 7). There are some implicit casts going on in there and it would probably be wise to handle those better. You might also want to check into NLS settings to make sure you can rely on to_char() using Sunday as the first day of week.
https://docs.oracle.com/cd/B19306_01/server.102/b14200/sql_elements004.htm
You can also use the NEXT_DAY function, as in:
SELECT TRUNC(NEXT_DAY(SYSDATE, 'MON')) - INTERVAL '7' DAY AS PREV_MONDAY,
TRUNC(NEXT_DAY(SYSDATE, 'FRI')) AS NEXT_FRIDAY
FROM DUAL;
Note that using the above, on weekends the Monday will be the Monday preceding the current date, and the Friday will be the Friday following the current date, i.e. there will be 11 days between the two days.
You can also use
SELECT TRUNC(NEXT_DAY(SYSDATE, 'MON')) - INTERVAL '7' DAY AS PREV_MONDAY,
TRUNC(NEXT_DAY(SYSDATE, 'MON')) - INTERVAL '3' DAY AS NEXT_FRIDAY
FROM DUAL;
in which case the Monday and Friday will always be from the same week, but if SYSDATE is on a weekend the Monday and Friday returned will be from the PREVIOUS week.
The following query returns
select to_char( trunc(sysdate) - numtoyminterval(level - 1, 'month'), 'mon-yy') as month from dual connect by level <= 12
last 12 months according to today's date(i.e. 2-Jan-18).
Say if today's date is 29-DEC-17 it gives oracle sql error:
ORA-01839: date not valid for month specified
(since on subtracting there would be a date in the result as '29-FEB-17' which is not possible). So on specific dates this error would pop-up. How do you suggest to overcome this?
I prefer ADD_MONTHS (TRUNC of a date literal is stupid, but I left it as you'd have SYSDATE anyway, wouldn't you?):
SQL> select to_char(add_months(trunc(date '2017-12-29'), -level), 'dd-mon-yy',
2 'nls_Date_language = english') as month
3 from dual
4 connect by level <= 12;
MONTH
------------------
29-nov-17
29-oct-17
29-sep-17
29-aug-17
29-jul-17
29-jun-17
29-may-17
29-apr-17
29-mar-17
28-feb-17
29-jan-17
29-dec-16
12 rows selected.
SQL>
This behavior of INTERVAL YEAR TO MONTH is as documented, see Datetime/Interval Arithmetic
You should consider function ADD_MONTHS:
If date is the last day of the month or if the resulting month has
fewer days than the day component of date, then the result is the last
day of the resulting month. Otherwise, the result has the same day
component as date.
It depends on your requirements what you consider as "right". In fact "one month" does not have a fixed duration.
Alright so I am trying to retrieve data a field we will call DATE_OF_ENTRY and the field is like this.
Example DATE_OF_ENTRY Data
28-NOV-15
So I need to use this field in a script that will be running twice a month to pull certain records. Basically when it's the 16th day of the current month I want all the records from the 1st-15th to be pulled up. When I run this script on the 1st of the next month I want all the records from the 16th-End of last month.
What I am using now
WHERE ROUND(DATE_OF_ENTRY,'MM') = ROUND(sysdate-1,'MM') AND DATE_OF_ENTRY < trunc(sysdate)
The problem with this statement is that it works on the 1st for the 16th to End of the last month, but on the 16th it gets data from the prior month still.
Any help is appreciated!
Using TRUNC() function with MONTH parameter will get the first day of the month.
Using TRUNC() function with DATE_OF_ENTRY will remove the TIME part.
Use + operator to add days to a DATE
SELECT TRUNC(sysdate, 'MONTH') firstDay,
TRUNC(sysdate, 'MONTH') + 15 Day15,
*
FROM yourTable
WHERE TRUNC(DATE_OF_ENTRY) >= TRUNC(sysdate, 'MONTH')
AND TRUNC(DATE_OF_ENTRY) <= TRUNC(sysdate, 'MONTH') + 15