I’m trying to get a date range from today’s date writing a sql query, today’s date but last 45 days from today so I want the dates in between 45days and today not only the date 45 days ago
To create a series of values, e.g. of dates, HANA provides the SERIES_GENERATE functions.
For the requirement to create all dates of the past 45 days, one could follow this approach
SELECT
GENERATED_PERIOD_START as DATE_OF_DAY
FROM
SERIES_GENERATE_DATE('INTERVAL 1 DAY'
, ADD_DAYS(current_date, -45)
, current_date);
You could use the ADD_DAYS function with a minus value as part of your range, ADD_DAYS(CURRENT_DATE, -45) or e.g. SELECT ADD_DAYS (TO_DATE ('2021-03-03', 'YYYY-MM-DD'), -45) FROM DUMMY
You can use M_TIME_DIMENSION table to get dates with other attributes (like month, year, date in ABAP format (dats) etc) with no calculation.
select date_sql
from _sys_bi.m_time_dimension
where date_sql between add_days(current_date, -45) and current_date
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 am new to SQL, here is my problem.
I have a table with daily dates:
Date:
20190101
20190102
20190103
.
**20190131**
20190201
20190202
20190203
.
**20190228**
20190301
20190302
20190303
.
**20190331**
I want to select only the month-end dates, what would be the code to do that?
thanks
I am using MS SQL Studio.
One method in standard SQL would be:
select t.*
from t
where extract(month from date + interval '1' day) <> extract(month from date);
Date/time functions vary significantly by database, so the exact functions might not match your database. However, the idea is simple: add one day and see if the month changes.
In standard SQL, you could do:
select date
from mytable
where date = date_trunc('month', date) + interval '1' month - interval '1' day
Edit
In SQL Server, you can just use eomonth(). Given a date, this functions returns the corresponding end of month, which you can compare against the date. So:
select date
from mytable
where date = eomonth(date)
Need help figuring out how to determine if the date is the same 'day' as today in teradata. IE, today 12/1/15 Tuesday, same day last year was actually 12/2/2014 Tuesday.
I tried using current_date - INTERVAL'1'Year but it returns 12/1/2014.
You can do this with a bit of math if you can convert your current date's "Day of the week" to a number, and the previous year's "Day of the week" to a number.
In order to do this in Teradata your best bet is to utilize the sys_calendar.calendar table. Specifically the day_of_week column. Although there are other ways to do it.
Furthermore, instead of using CURRENT_DATE - INTERVAL '1' YEAR, it's a good idea to use ADD_MONTHS(CURRENT_DATE, -12) since INTERVAL arithmetic will fail on 2012-02-29 and other Feb 29th leap year dates.
So, putting it together you get what you need with:
SELECT
ADD_MONTHS(CURRENT_DATE, -12)
+
(
(SELECT day_of_week FROM sys_calendar.calendar WHERE calendar_date = CURRENT_DATE)
-
(SELECT day_of_week FROM sys_calendar.calendar WHERE calendar_date = ADD_MONTHS(CURRENT_DATE, -12))
)
This is basically saying: Take the current dates day of week number (3) and subtract from it last years day of week number (2) to get 1. Add that to last year's date and you'll have the same day of the week as current date.
I tested this for all dates between 01/01/2010 and CURRENT_DATE and it worked as expected.
Why don't you simply subtract 52 weeks?
current_date - 364
The SQL below will get you to the abbreviated name for the day of week, it's cumbersome but it works across versions of Teradata.
SELECT CAST(CAST(ADD_MONTHS(CURRENT_DATE, -12) AS DATE FORMAT 'E3') AS CHAR(3)) AS LY_DayOfWeek
, CAST(CAST(CURRENT_DATE) AS DATE FORMAT 'E3') AS CHAR(3)) AS CY_DayOfWeek
Dates are internally represented at integers in Teradata as (Year-1900) * 100000 + (MONTH * 100) + DAY. You may be able to do some creative arithmetic to figure out that 12/1/2015 Tuesday was 12/2/2014 Tuesday last year.
I am working on a homework problem, I'm close but need some help with a data conversion I think. Or sysdate - start_date calculation
The question is:
Using the EX schema, write a SELECT statement that retrieves the date_id and start_date from the Date_Sample table (format below), followed by a column named Years_and_Months_Since_Start that uses an interval function to retrieve the number of years and months that have elapsed between the start_date and the sysdate. (Your values will vary based on the date you do this lab.) Display only the records with start dates having the month and day equal to Feb 28 (of any year).
DATE_ID START_DATE YEARS_AND_MONTHS_SINCE_START
2 Sunday , February 28, 1999 13-8
4 Monday , February 28, 2005 7-8
5 Tuesday , February 28, 2006 6-8
Our EX schema that refers to this question is simply a Date_Sample Table with two columns:
DATE_ID NUMBER NOT Null
START_DATE DATE
I Have written this code:
SELECT date_id, TO_CHAR(start_date, 'Day, MONTH DD, YYYY') AS start_date ,
NUMTOYMINTERVAL((SYSDATE - start_date), 'YEAR') AS years_and_months_since_start
FROM date_sample
WHERE TO_CHAR(start_date, 'MM/DD') = '02/28';
But my Years and months since start column is not working properly. It's getting very high numbers for years and months when the date calculated is from 1999-ish. ie, it should be 13-8 and I'm getting 5027-2 so I know it's not correct. I used NUMTOYMINTERVAL, which should be correct, but don't think the sysdate-start_date is working. Data Type for start_date is simply date. I tried ROUND but maybe need some help to get it right.
Something is wrong with my calculation and trying to figure out how to get the correct interval there. Not sure if I have provided enough information to everyone but I will let you know if I figure it out before you do.
It's a question from Murach's Oracle and SQL/PL book, chapter 17 if anyone else is trying to learn that chapter. Page 559.
you'll want MONTHS_BETWEEN in that numtoyminterval as the product of subtracting two date variables gives the answer in days which isn't usable to you and the reason its so high is you've told Oracle the answer was in years! Also use the fm modifier on the to_char to prevent excess whitespace.
select date_id,
to_char(start_date, 'fmDay, Month DD, YYYY') as start_date,
extract(year from numtoyminterval(months_between(trunc(sysdate), start_date), 'month') )
|| '-' ||
extract(month from numtoyminterval(months_between(trunc(sysdate), start_date), 'month') )
as years_and_months_since_start
from your_table
where to_char(start_date, 'MM/DD') = '02/28';
You can simplify the answer like this
SELECT date_id, start_date, numtoyminterval(months_between(sysdate, start_date), 'month') as "Years and Months Since Start"
FROM date_sample
WHERE EXTRACT (MONTH FROM start_date) = 2 AND EXTRACT (DAY FROM start_date) = 28;
This should be doable, but how can I extract the day of the week from a field containing data in date format with Netezza SQL? I can write the following query:
SELECT date_part('day',a.report_dt) as report_dt
FROM table as a
but that gives me the day of the month.
thanks for any help
The below queries give day numbers for any week,month,year for a particular date.
--Day of Week
SELECT EXTRACT(dow FROM report_dt) FROM table;
--Day of Month
SELECT DATE_PART('day', report_dt) FROM table;
--Day of Year
SELECT EXTRACT(doy FROM report_dt) FROM table;
Netezza is just ANSI SQL, originally derived from PostgreSQL. I'd expect this to work.
select extract(dow from a.report_dt) as report_dt
from table as a
Returns values should range from 0 to 6; 0 is Sunday. You might expect that to be an integer, but in PostgreSQL at least, the returned value is a double-precision floating point.
If you want to extract directly the day name :
Select to_char(date, 'Day') as Day_Name From table;
In Netezza SQL, SELECT EXTRACT(dow FROM report_dt) would return values 1 to 7. 1 is Sunday, 7 is Saturday.