PK Date ID
=== =========== ===
1 07/04/2017 22
2 07/05/2017 22
3 07/07/2017 03
4 07/08/2017 04
5 07/09/2017 22
6 07/09/2017 22
7 07/10/2017 05
8 07/11/2017 03
9 07/11/2017 03
10 07/11/2017 03
I want to count the number of ID occurred in a given week/month, something like this.
ID Count
22 3 --> count as 1 only in the same date occurred twice one 07/09/2017
03 2 --> same as above, increment only one regardless how many times it occurred in a same date
04 1
05 1
I'm trying to implement this in a perl file, to output/print it in a csv file, I have no idea on what query will I execute.
Seems like a simple case of count distinct and group by:
SELECT Id, COUNT(DISTINCT [Date]) As [Count]
FROM TableName
WHERE [Date] >= #StartDate
AND [Date] <= #EndDate
GROUP BY Id
ORDER BY [Count] DESC
You can use COUNT with DISTINCT e.g.:
SELECT ID, COUNT(DISTINCT Date)
FROM table
GROUP BY ID;
You can read more abot how to get month from a date in get month from a date (it also works for year).
Your query will be :
select DATEPART(mm,Date) AS month, COUNT(ID) AS count from table group by month
Hope that helped you.
Related
I have following data in my table:
eb |anz
05.03.2020 | 2
06.03.2020 | 3
07.03.2020 | 1
08.03.2020 | 9
09.03.2020 | 10
10.03.2020 | 2
11.03.2020 | 20
12.03.2020 | 25
Now I need to sum the values in specific range for each date.
For example "12.03.2020": I want to sum the value of the 12th, 11th, 10th and 9th of march for the date "12.03.2020". Additionally I want to sum the other four values before 9th of march and divide the summary 1 by summary 2 by select.
So my calculation would be: (25+20+2+10)/(9+1+3+2) = 3.8
I would like to output the date and the calculated value for each date in table.
I tried to sum the first group for each date (in example 9th to 12th march) but the output is the same as the data in the table.
select
eb,
sum(anz)
from (select eb, count(*) as anz from myTable where eb != '' group by eb) tmp
where
convert(date, eb, 104) >= dateadd(day,-3,convert(datetime, eb, 104))
and convert(date, eb, 104) <= convert(date, eb, 104)
group by eb
order by convert(date, eb, 104)
It looks like the condition is being ignored. Do you have any advice for me?
Thanks a lot
Let me assume that data is stored correctly as a date then you can use window functions:
select t.*,
(sum(anz) over (order by eb rows between 3 preceding and current row) /
sum(anz) over (order by eb rows between 8 preceding and 4 preceding)
)
from t;
Note that if value is an integer, then use * 1.0 / to avoid integer division.
Also, this assumes that you have data on each date.
So I have a table like:
UNIQUE_ID MONTH
abc 01
93j 01
acc 01
7as 01
oks 02
ais 02
asi 03
asd 04
etc
I query:
select count(unique_id) as amount, month
from table
group by month
now everything looks great:
AMOUNT MONTH
4 01
2 02
1 03
etc
is there a way to get oracle to split the amounts by weeks?
the way that the result look something like:
AMOUNT WEEK
1 01
1 02
1 03
1 04
etc
Assuming you know the year - lets say we go with 2014 then you need to generate all the weeks a year
select rownum as week_no
from all_objects
where rownum<53) weeks
then state which months contain the weeks (for 2014)
select week_no, to_char(to_date('01-JAN-2014','DD-MON-YYYY')+7*(week_no-1),'MM') month_no
from
(select rownum as week_no
from all_objects
where rownum<53) weeks
Then join in your data
select week_no,month_no, test.unique_id from (
select week_no, to_char(to_date('01-JAN-2014','DD-MON-YYYY')+7*(week_no-1),'MM') month_no
from
(select rownum as week_no
from all_objects
where rownum<53) weeks) wm
join test on wm.month_no = test.tmonth
This gives your data for the each week as you described above. You can redo your query and count by week instead of month.
This question already has answers here:
Closed 11 years ago.
Possible Duplicate:
Running total by grouped records in table
I am trying to put together an SQL statement that returns the SUM of a value by month, but on a year to date basis. In other words, for the month of March, I am looking to get the sum of a value for the months of January, February, and March.
I can easily do a group by to get a total for each month by itself, and potentially calculate the year to date value I need in my application from this data by looping through the results set. However, I was hoping to have some of this work handled with my SQL statement.
Has anyone ever tackled this type of problem with an SQL statement, and if so, what is the trick that I am missing?
My current sql statement for monthly data is similar to the following:
Select month, year, sum(value) from mytable group by month, year
If I include a where clause on the month, and only group by the year, I can get the result for a single month that I am looking for:
select year, sum(value) from mytable where month <= selectedMonth group by year
However, this requires me to have a particular month pre-selected or to utilize 12 different SQL statements to generate one clean result set.
Any guidance that can be provided would be greatly appreciated!
Update: The data is stored on an IBM iSeries.
declare #Q as table
(
mmonth INT,
value int
)
insert into #Q
values
(1,10),
(1,12),
(2,45),
(3,23)
select sum(January) as UpToJanuary,
sum(February)as UpToFebruary,
sum(March) as UpToMarch from (
select
case when mmonth<=1 then sum(value) end as [January] ,
case when mmonth<=2 then sum(value) end as [February],
case when mmonth<=3 then sum(value) end as [March]
from #Q
group by mmonth
) t
Produces:
UpToJanuary UpToFebruary UpToMarch
22 67 90
You get the idea, right?
NOTE: This could be done easier with PIVOT tables but I don't know if you are using SQL Server or not.
As far as I know DB2 does support windowing functions although I don't know if this is also supported on the iSeries version.
If windowing functions are supported (I believe IBM calls them OLAP functions) then the following should return what you want (provided I understood your question correctly)
select month,
year,
value,
sum(value) over (partition by year order by month asc) as sum_to_date
from mytable
order by year, month
create table mon
(
[y] int not null,
[m] int not null,
[value] int not null,
primary key (y,m))
select a.y, a.m, a.value, sum(b.value)
from mon a, mon b
where a.y = b.y and a.m >= b.m
group by a.y, a.m, a.value
2011 1 120 120
2011 2 130 250
2011 3 500 750
2011 4 10 760
2011 5 140 900
2011 6 100 1000
2011 7 110 1110
2011 8 90 1200
2011 9 70 1270
2011 10 150 1420
2011 11 170 1590
2011 12 600 2190
You should try to join the table to itself by month-behind-a-month condition and generate a synthetic month-group code to group by as follows:
select
sum(value),
year,
up_to_month
from (
select a.value,
a.year,
b.month as up_to_month
from table as a join table as b on a.year = b.year and b.month => a.month
)
group by up_to_month, year
gives that:
db2 => select * from my.rep
VALUE YEAR MONTH
----------- ----------- -----------
100 2011 1
200 2011 2
300 2011 3
400 2011 4
db2 -t -f rep.sql
1 YEAR UP_TO_MONTH
----------- ----------- -----------
100 2011 1
300 2011 2
600 2011 3
1000 2011 4
Being primarily a C# developer, I'm scratching my head when trying to create a pure T-SQL based solution to a problem involving summarizing days/month given a set of date ranges.
I have a set of data looking something like this:
UserID Department StartDate EndDate
====== ========== ========== ==========
1 A 2011-01-02 2011-01-05
1 A 2011-01-20 2011-01-25
1 A 2011-02-25 2011-03-05
1 B 2011-01-21 2011-01-22
2 A 2011-01-01 2011-01-20
3 C 2011-01-01 2011-02-03
The date ranges are non-overlapping, may span several months, there may exist several ranges for a specific user and department within a single month.
What I would like to do is to summarize the number of days (inclusive) per user, department, year and month, like this (with reservations for any math errors in my example...):
UserID Department Year Month Days
====== ========== ==== ===== ====
1 A 2011 01 10
1 A 2011 02 4
1 A 2011 03 5
1 B 2011 01 2
2 A 2011 01 20
3 C 2011 01 31
3 C 2011 02 3
This data is going into a new table used by reporting tools.
I hope the problem description is clear enough, this is my first posting here, be gentle :-)
Thanks in advance!
Working sample
-- sample data in a temp table
declare #t table (UserID int, Department char(1), StartDate datetime, EndDate datetime)
insert #t select
1 ,'A', '2011-01-02','2011-01-05'union all select
1 ,'A', '2011-01-20','2011-01-25'union all select
1 ,'A', '2011-02-25','2011-03-05'union all select
1 ,'B', '2011-01-21','2011-01-22'union all select
2 ,'A', '2011-01-01','2011-01-20'union all select
3 ,'C', '2011-01-01','2011-02-03'
-- the query you need is below this line
select UserID, Department,
YEAR(StartDate+v.number) Year,
MONTH(StartDate+v.number) Month, COUNT(*) Days
from #t t
inner join master..spt_values v
on v.type='P' and v.number <= DATEDIFF(d, startdate, enddate)
group by UserID, Department, YEAR(StartDate+v.number), MONTH(StartDate+v.number)
order by UserID, Department, Year, Month
If data is in the following format:
SID TID Tdatetime QID QTotal
----------------------------------------
100 1 01/12/97 9:00AM 66 110
100 1 01/12/97 9:00AM 66 110
100 1 01/12/97 10:00AM 67 110
100 2 01/19/97 9:00AM 66 .
100 2 01/19/97 9:00AM 66 110
100 2 01/19/97 10:00AM 66 110
100 3 01/26/97 9:00AM 68 120
100 3 01/26/97 9:00AM 68 120
110 1 02/03/97 10:00AM 68 110
110 3 02/12/97 9:00AM 64 115
110 3 02/12/97 9:00AM 64 115
120 1 04/05/97 9:00AM 66 105
120 1 04/05/97 10:00AM 66 105
I would like to be able to write a query to sum the QTotal column for all rows and find the count of duplicate rows for the Tdatetime column.
The output would look like:
Year Total Count
97 | 1340 | 4
The third column in the result does not include the count of distinct rows in the table. And the output is grouped by the year in the TDateTime column.
The following query may help:
SELECT
'YEAR ' + CAST(sub.theYear AS VARCHAR(4)),
COUNT(sub.C),
(SELECT SUM(QTotal) FROM MyTable WHERE YEAR(Tdatetime) = sub.theYear) AS total
FROM
(SELECT
YEAR(Tdatetime) AS theYear,
COUNT(Tdatetime) AS C
FROM MyTable
GROUP BY Tdatetime, YEAR(Tdatetime)
HAVING COUNT(Tdatetime) >= 2) AS sub
This will work if you really want to group by the tDateTime column:
SELECT DISTINCT tDateTime, SUM(QTotal), Count(distinct tDateTime)
FROM Table
GROUP BY tDateTime
HAVING Count(distinct tDateTime) > 1
But your results look like you want to group by the Year in the tDateTime column. Is this correct?
If so try this:
SELECT DISTINCT YEAR (tDateTime), SUM(QTotal), Count(distinct tDateTime)
FROM Table
GROUP BY YEAR (tDateTime)
HAVING Count(distinct tDateTime) > 1
You must do SELECT from this table GROUPing by QTotal, using COUNT(subSELECT from this table WHERE QTotal is the same). If I only I had time I would write you SQL statement, but it'll take some minutes.
Something like:
select Year(Tdatetime) ,sum(QTotal), count(1) from table group by year(Tdatetime )
or full date
select Tdatetime ,sum(QTotal), count(1) from table group by year(Tdatetime)
Or your ugly syntax ( : ) )
select 'Year ' + cast(Year(tdatetime) as varchar(4))
+ '|' + cast(sum(QTotal) as varchar(31))
+ '|' + cast(count(1) as varchar(31))
from table group by year(Tdatetime )
Or do you want just the year? Sum all columns? Or just by year?
SELECT
YEar + year(Tdatetime),
SUM ( QTotal ),
(SELECT COUNT(*) FROM (
SELECT Tdatetime FROM tDateTime GROUP BY Tdatetime
HAVING COUNT(QID) > 1) C
FROM
Tdatetime t
GROUP BY
YEar + year(Tdatetime)
This is the first time I have asked a question on stackoverflow. It looks like I have lost my original ID info. I had to register to login and add comments to the question I posted.
To answer OMG Ponies question, this is a SQL Server 2008 database.
#Abe Miessler , the row with SID 120 does not contain duplicates. the first row for SID 120 shows 9:00AM in the datetime column , and the second row shows 10:00AM.
#Zafer, your query is the accepted answer. I made a few minor tweaks to get it to work. Thanks.
Thanks due to Abe Miessler and the others for your help.