Having an issue in SQL script where I’m trying to achieve filter criteria of rolling 12 months in the day column which stored data as a text in server.
Goal is to count sizes for product at retail store location over the last 12 months from the current day. Currently, in my query I'm using the criteria of year 2019 which only counts the sizes for that year but not for rolling 12 months from current date.
CALENDARDAY column is in text field in the data set and data stores in yyyymmdd format.
When trying to run below script in Tableau with GETDATE and DATEADD function it is giving me a functional error. I am trying to access SAP HANA server with below query.
Any help would be appreciated
Select
SKU, STYLE_ID, Base_Style_ID, COLOR, SIZEKEY, STORE, Year,
count(SIZEKEY)over(partition by STYLE_ID,COLOR,STORE,Year) as SZ_CNT
from
(
select
a."RAW" As SKU,
a."STYLENUM" As STYLE_ID,
mat."BASENUM" AS Base_Style_ID,
a."COLORNUM" AS COLOR,
a."SIZE" AS SIZEKEY,
a."STORENUM" AS STORE,
substring(a."CALENDARDAY",1,4) As year
from PRTRPT_XRE as a
JOIN ZAT_SKU As mat On a."RAW" = mat."SKU"
where a."ORGANIZATION" = 'M20'
and a."COLORNUM" is not null
and substring(a."CALENDARDAY",1,4) = '2019'
Group BY
a."RAW",
a."STYLENUM",
mat."BASENUM",
a."ZCOLORCD",
a."SIZE",
a."STORENUM",
substring(a."CALENDARDAY",1,4)
)
I have never worked on that DB / Server, so I don't have a way to test this.
But hopefully this will work (expecting exact 12 months before today's date)
AND ADD_MONTHS (TO_DATE (a."CALENDARDAY", 'YYYY-MM-DD'), 12) > CURRENT_DATE
or
AND ADD_MONTHS (a."CALENDARDAY", 12) > CURRENT_DATE
Below condition from one of our CALENDAR table also worked same way as ADD_MONTHS mentioned in above response
select distinct CALENDARDAY
from
(
select FISCALWEEK, CALENDARDAY, CNST, row_number()over(partition by CNST order by FISCALWEEK desc) as rnum
from
(
select distinct FISCALWEEK, CALENDARDAY, 'A' as CNST
from CALENDARTABLE
where CALENDARDAY < current_date
order by 1,2
)
) where rnum < 366
Related
I am doing a query where the query will append the data from previous date as the outcome in BigQuery.
So, the result data for today will be higher than yesterdays as the data is appending by days.
So far, what I only managed to get the outcome is the data by days (where you can see the number of ID declining and is not appending from previous day) as this result:
What should I do to add appending function in the query so each day will get the result of data from the previous day in bigquery?
code:
WITH
table1 AS (
SELECT
ID,
...
FROM t
WHERE DATE_SUB('2020-01-31', INTERVAL 31 DAY) and '2020-01-31'
),
table2 AS (
SELECT
ID,
COUNTIF((rating < 7) as bad,
COUNTIF((rating >= 7 AND SAFE_CAST(NPS_Rating as INT64) < 9) as intermediate,
COUNTIF((rating as good
FROM
t
WHERE DATE_SUB('2020-01-31', INTERVAL 31 DAY) and '2020-01-31'
)
SELECT
DATE_SUB('2020-01-31', INTERVAL 31 DAY) as date,
*
FROM table1
FULL OUTER JOIN table2 USING (ID)
If you have counts that you want to accumulate, then you want a cumulative sum. The query would look something like this:
select datecol, count(*), sum(count(*)) over (order by datecol)
from t
group by datecol
order by datecol;
i have the following table RENTAL(book_date, copy_id, member_id, title_id, act_ret_date, exp_ret_date). Where book_date shows the day the book was booked. I need to write a query that for every day of the month(so from 1-30 or from 1-29 or from 1-31 depending on month) it shows me the number of books booked.
i currently know how to show the number of books rented in the days that are in the table
select count(book_date), to_char(book_date,'DD')
from rental
group by to_char(book_date,'DD');
my questions are:
How do i show the rest of the days(if let's say for some reason in my database i have no books rented on 20th or 19th or multiple days) and put the number 0 there?
How do i show the number of days only of the current month so(28,29,30,31 all these 4 are possible depending on month or year)... i am lost . This must be done using only SQL query no pl/SQL or other stuff.
The following query would give you all days in the current month, in your case you can replace SYSDATE with your date column and join with this query to know how many for a given month
SELECT DT
FROM(
SELECT TRUNC (last_day(SYSDATE) - ROWNUM) dt
FROM DUAL CONNECT BY ROWNUM < 32
)
where DT >= trunc(sysdate,'mm')
The answer is to create a table like this:
table yearsmonthsdays (year varchar(4), month varchar(2), day varchar(2));
use any language you wish, e.g. iterate in java with Calendar.getInstance().getActualMaximum(Calendar.DAY_OF_MONTH) to get the last day of the month for as many years and months as you like, and fill that table with the year, month and days from 1 to last day of month of your result.
you'd get something like:
insert into yearsmonthsdays ('1995','02','01');
insert into yearsmonthsdays ('1995','02','02');
...
insert into yearsmonthsdays ('1995','02','28'); /* non-leap year */
...
insert into yearsmonthsdays ('1996','02','01');
insert into yearsmonthsdays ('1996','02','02');
...
insert into yearsmonthsdays ('1996','02','28');
insert into yearsmonthsdays ('1996','02','29'); /* leap year */
...
and so on.
Once you have this table done, your work is almost finished. Make an outer left join between your table and this table, joining year, month and day together, and when no lines appear, the count will be zero as you wish. Without using programming, this is your best bet.
In oracle, you can query from dual and use the conncect by level syntax to generate a series of rows - in your case, dates. From there on, it's just a matter of deciding what dates you want to display (in my example I used all the dates from 2014) and joining on your table:
SELECT all_date, COALESCE (cnt, 0)
FROM (SELECT to_date('01/01/2014', 'dd/mm/yyyy') + rownum - 1 AS all_date
FROM dual
CONNECT BY LEVEL <= 365) d
LEFT JOIN (SELECT TRUNC(book_date), COUNT(book_date) AS cnt
FROM rental
GROUP BY book_date) r ON d.all_date = TRUNC(r.book_date)
There's no need to get ROWNUM involved ... you can just use LEVEL in the CONNECT BY:
WITH d1 AS (
SELECT TRUNC(SYSDATE, 'MONTH') - 1 + LEVEL AS book_date
FROM dual
CONNECT BY TRUNC(SYSDATE, 'MONTH') - 1 + LEVEL <= LAST_DAY(SYSDATE)
)
SELECT TRUNC(d1.book_date), COUNT(r.book_date)
FROM d1 LEFT JOIN rental r
ON TRUNC(d1.book_date) = TRUNC(r.book_date)
GROUP BY TRUNC(d1.book_date);
Simply replace SYSDATE with a date in the month you're targeting for results.
All days of the month based on current date
select trunc(sysdate) - (to_number(to_char(sysdate,'DD')) - 1)+level-1 x from dual connect by level <= TO_CHAR(LAST_DAY(sysdate),'DD')
It did works to me:
SELECT DT
FROM (SELECT TRUNC(LAST_DAY(SYSDATE) - (CASE WHEN ROWNUM=1 THEN 0 ELSE ROWNUM-1 END)) DT
FROM DUAL
CONNECT BY ROWNUM <= 32)
WHERE DT >= TRUNC(SYSDATE, 'MM')
In Oracle SQL the query must look like this to not miss the last day of month:
SELECT DT
FROM(
SELECT trunc(add_months(sysdate, 1),'MM')- ROWNUM dt
FROM DUAL CONNECT BY ROWNUM < 32
)
where DT >= trunc(sysdate,'mm')
I have a Postgres table that I'm trying to analyze based on some date columns.
I'm basically trying to count the number of rows in my table that fulfill this requirement, and then group them by month and year. Instead of my query looking like this:
SELECT * FROM $TABLE WHERE date1::date <= '2012-05-31'
and date2::date > '2012-05-31';
it should be able to display this for the months available in my data so that I don't have to change the months manually every time I add new data, and so I can get everything with one query.
In the case above I'd like it to group the sum of rows which fit the criteria into the year 2012 and month 05. Similarly, if my WHERE clause looked like this:
date1::date <= '2012-06-31' and date2::date > '2012-06-31'
I'd like it to group this sum into the year 2012 and month 06.
This isn't entirely clear to me:
I'd like it to group the sum of rows
I'll interpret it this way: you want to list all rows "per month" matching the criteria:
WITH x AS (
SELECT date_trunc('month', min(date1)) AS start
,date_trunc('month', max(date2)) + interval '1 month' AS stop
FROM tbl
)
SELECT to_char(y.mon, 'YYYY-MM') AS mon, t.*
FROM (
SELECT generate_series(x.start, x.stop, '1 month') AS mon
FROM x
) y
LEFT JOIN tbl t ON t.date1::date <= y.mon
AND t.date2::date > y.mon -- why the explicit cast to date?
ORDER BY y.mon, t.date1, t.date2;
Assuming date2 >= date1.
Compute lower and upper border of time period and truncate to month (adding 1 to upper border to include the last row, too.
Use generate_series() to create the set of months in question
LEFT JOIN rows from your table with the declared criteria and sort by month.
You could also GROUP BY at this stage to calculate aggregates ..
Here is the reasoning. First, create a list of all possible dates. Then get the cumulative number of date1 up to a given date. Then get the cumulative number of date2 after the date and subtract the results. The following query does this using correlated subqueries (not my favorite construct, but handy in this case):
select thedate,
(select count(*) from t where date1::date <= d.thedate) -
(select count(*) from t where date2::date > d.thedate)
from (select distinct thedate
from ((select date1::date as thedate from t) union all
(select date2::date as thedate from t)
) d
) d
This is assuming that date2 occurs after date1. My model is start and stop dates of customers. If this isn't the case, the query might not work.
It sounds like you could benefit from the DATEPART T-SQL method. If I understand you correctly, you could do something like this:
SELECT DATEPART(year, date1) Year, DATEPART(month, date1) Month, SUM(value_col)
FROM $Table
-- WHERE CLAUSE ?
GROUP BY DATEPART(year, date1),
DATEPART(month, date1)
I have a table that contains multiple records for each day of the month, over a number of years. Can someone help me out in writing a query that will only return the last day of each month.
SQL Server (other DBMS will work the same or very similarly):
SELECT
*
FROM
YourTable
WHERE
DateField IN (
SELECT MAX(DateField)
FROM YourTable
GROUP BY MONTH(DateField), YEAR(DateField)
)
An index on DateField is helpful here.
PS: If your DateField contains time values, the above will give you the very last record of every month, not the last day's worth of records. In this case use a method to reduce a datetime to its date value before doing the comparison, for example this one.
The easiest way I could find to identify if a date field in the table is the end of the month, is simply adding one day and checking if that day is 1.
where DAY(DATEADD(day, 1, AsOfDate)) = 1
If you use that as your condition (assuming AsOfDate is the date field you are looking for), then it will only returns records where AsOfDate is the last day of the month.
Use the EOMONTH() function if it's available to you (E.g. SQL Server). It returns the last date in a month given a date.
select distinct
Date
from DateTable
Where Date = EOMONTH(Date)
Or, you can use some date math.
select distinct
Date
from DateTable
where Date = DATEADD(MONTH, DATEDIFF(MONTH, -1, Date)-1, -1)
In SQL Server, this is how I usually get to the last day of the month relative to an arbitrary point in time:
select dateadd(day,-day(dateadd(month,1,current_timestamp)) , dateadd(month,1,current_timestamp) )
In a nutshell:
From your reference point-in-time,
Add 1 month,
Then, from the resulting value, subtract its day-of-the-month in days.
Voila! You've the the last day of the month containing your reference point in time.
Getting the 1st day of the month is simpler:
select dateadd(day,-(day(current_timestamp)-1),current_timestamp)
From your reference point-in-time,
subtract (in days), 1 less than the current day-of-the-month component.
Stripping off/normalizing the extraneous time component is left as an exercise for the reader.
A simple way to get the last day of month is to get the first day of the next month and subtract 1.
This should work on Oracle DB
select distinct last_day(trunc(sysdate - rownum)) dt
from dual
connect by rownum < 430
order by 1
I did the following and it worked out great. I also wanted the Maximum Date for the Current Month. Here is what I my output is. Notice the last date for July which is 24th. I pulled it on 7/24/2017, hence the result
Year Month KPI_Date
2017 4 2017-04-28
2017 5 2017-05-31
2017 6 2017-06-30
2017 7 2017-07-24
SELECT B.Year ,
B.Month ,
MAX(DateField) KPI_Date
FROM Table A
INNER JOIN ( SELECT DISTINCT
YEAR(EOMONTH(DateField)) year ,
MONTH(EOMONTH(DateField)) month
FROM Table
) B ON YEAR(A.DateField) = B.year
AND MONTH(A.DateField) = B.Month
GROUP BY B.Year ,
B.Month
SELECT * FROM YourTableName WHERE anyfilter
AND "DATE" IN (SELECT MAX(NameofDATE_Column) FROM YourTableName WHERE
anyfilter GROUP BY
TO_CHAR(NameofDATE_Column,'MONTH'),TO_CHAR(NameofDATE_Column,'YYYY'));
Note: this answer does apply for Oracle DB
Here's how I just solved this. day_date is the date field, calendar is the table that holds the dates.
SELECT cast(datepart(year, day_date) AS VARCHAR)
+ '-'
+ cast(datepart(month, day_date) AS VARCHAR)
+ '-'
+ cast(max(DATEPART(day, day_date)) AS VARCHAR) 'DATE'
FROM calendar
GROUP BY datepart(year, day_date)
,datepart(month, day_date)
ORDER BY 1
I would appreciate a little expert help please.
in an SQL SELECT statement I am trying to get the last day with data per month for the last year.
Example, I am easily able to get the last day of each month and join that to my data table, but the problem is, if the last day of the month does not have data, then there is no returned data. What I need is for the SELECT to return the last day with data for the month.
This is probably easy to do, but to be honest, my brain fart is starting to hurt.
I've attached the select below that works for returning the data for only the last day of the month for the last 12 months.
Thanks in advance for your help!
SELECT fd.cust_id,fd.server_name,fd.instance_name,
TRUNC(fd.coll_date) AS coll_date,fd.column_name
FROM super_table fd,
(SELECT TRUNC(daterange,'MM')-1 first_of_month
FROM (
select TRUNC(sysdate-365,'MM') + level as DateRange
from dual
connect by level<=365)
GROUP BY TRUNC(daterange,'MM')) fom
WHERE fd.cust_id = :CUST_ID
AND fd.coll_date > SYSDATE-400
AND TRUNC(fd.coll_date) = fom.first_of_month
GROUP BY fd.cust_id,fd.server_name,fd.instance_name,
TRUNC(fd.coll_date),fd.column_name
ORDER BY fd.server_name,fd.instance_name,TRUNC(fd.coll_date)
You probably need to group your data so that each month's data is in the group, and then within the group select the maximum date present. The sub-query might be:
SELECT MAX(coll_date) AS last_day_of_month
FROM Super_Table AS fd
GROUP BY YEAR(coll_date) * 100 + MONTH(coll_date);
This presumes that the functions YEAR() and MONTH() exist to extract the year and month from a date as an integer value. Clearly, this doesn't constrain the range of dates - you can do that, too. If you don't have the functions in Oracle, then you do some sort of manipulation to get the equivalent result.
Using information from Rhose (thanks):
SELECT MAX(coll_date) AS last_day_of_month
FROM Super_Table AS fd
GROUP BY TO_CHAR(coll_date, 'YYYYMM');
This achieves the same net result, putting all dates from the same calendar month into a group and then determining the maximum value present within that group.
Here's another approach, if ANSI row_number() is supported:
with RevDayRanked(itemDate,rn) as (
select
cast(coll_date as date),
row_number() over (
partition by datediff(month,coll_date,'2000-01-01') -- rewrite datediff as needed for your platform
order by coll_date desc
)
from super_table
)
select itemDate
from RevDayRanked
where rn = 1;
Rows numbered 1 will be nondeterministically chosen among rows on the last active date of the month, so you don't need distinct. If you want information out of the table for all rows on these dates, use rank() over days instead of row_number() over coll_date values, so a value of 1 appears for any row on the last active date of the month, and select the additional columns you need:
with RevDayRanked(cust_id, server_name, coll_date, rk) as (
select
cust_id, server_name, coll_date,
rank() over (
partition by datediff(month,coll_date,'2000-01-01')
order by cast(coll_date as date) desc
)
from super_table
)
select cust_id, server_name, coll_date
from RevDayRanked
where rk = 1;
If row_number() and rank() aren't supported, another approach is this (for the second query above). Select all rows from your table for which there's no row in the table from a later day in the same month.
select
cust_id, server_name, coll_date
from super_table as ST1
where not exists (
select *
from super_table as ST2
where datediff(month,ST1.coll_date,ST2.coll_date) = 0
and cast(ST2.coll_date as date) > cast(ST1.coll_date as date)
)
If you have to do this kind of thing a lot, see if you can create an index over computed columns that hold cast(coll_date as date) and a month indicator like datediff(month,'2001-01-01',coll_date). That'll make more of the predicates SARGs.
Putting the above pieces together, would something like this work for you?
SELECT fd.cust_id,
fd.server_name,
fd.instance_name,
TRUNC(fd.coll_date) AS coll_date,
fd.column_name
FROM super_table fd,
WHERE fd.cust_id = :CUST_ID
AND TRUNC(fd.coll_date) IN (
SELECT MAX(TRUNC(coll_date))
FROM super_table
WHERE coll_date > SYSDATE - 400
AND cust_id = :CUST_ID
GROUP BY TO_CHAR(coll_date,'YYYYMM')
)
GROUP BY fd.cust_id,fd.server_name,fd.instance_name,TRUNC(fd.coll_date),fd.column_name
ORDER BY fd.server_name,fd.instance_name,TRUNC(fd.coll_date)