Sum across two tables returns unwanted Sum from one table multiplied by the number of rows in the other
I have 1 table with Actual results recorded by date and the other tables contains planned results recorded by month.
Table 1(Actual)
Date Location Amount
01/01/2019 Loc1 1000
01/02/2019 Loc1 700
01/01/2019 Loc2 7500
01/02/2019 Loc2 1000
02/01/2019 Loc1 500
Table 2(Plan)
Year Month Location Amount
2019 1 Loc1 1500
2019 1 Loc2 8000
2019 2 Loc1 800
I have tried various differed Joins using YEAR(Table1.date) and Month(table1.date) and grouping by
Month(Table1.Date) but I keep running into the same problem where the PlanAmount is multiplied by however many rows in the Actual table...
in the example of loc1 for Month 1 below I get
Year Month Location PlanAmount ActualAmount
2019 1 Loc1 3000 1700
I would like to return the below
Year Month Location PlanAmount ActualAmount
2019 1 Loc1 1500 1700
2019 1 Loc2 8000 8500
2019 2 Loc1 800 500
Thanks in advance for any help
D
You can do this with a full join or union all/group by:
select yyyy, mm, location,
sum(actual_amount) as actual_amount,
sum(plan_amount) as plan_amount
from ((select year(date) as yyyy, month(date) as mm, location,
amount as actual_amount, 0 as plan_amount
from actual
group by year(date) as yyyy, month(date) as mm, location
) union all
(select year, month, location,
0 as actual_amount, amount as plan_amount
from actual
group by year, month, location
)
) ap
group by yyyy, mm, location;
This ensures that you have rows, even when there are no matches in the other table.
To get the required results you need to group the first table on year of date, month of date and location and need to select the columns year, month, location and sum of amount from group after that you need to join that resultant r
SELECT
plans.year,
plans.month,
plans.location,
plans.plan_amount,
grouped_results.actual_amount
FROM plans
INNER JOIN (
SELECT
datepart(year, date) AS year,
datepart(month, date) AS month,
location,
SUM(amount) AS actual_amount
FROM actuals
GROUP BY datepart(year, date), datepart(month, date), location
) as grouped_results
ON
grouped_results.year = plans.year AND
grouped_results.month = plans.month AND
grouped_results.location = plans.location
I think the problem is that you are using sum(PlanTable.Amount) when grouping. Try using max(PlanTable.Amount) instead.
select
p.Year,
p.Month,
p.Location,
sum(a.Amount) as actual_amount,
max(p.Amount) as plan_amount
from
[Plan] p left join Actual a
on year(a.date) = p.year
and month(a.date) = p.Month
and a.Location = p.Location
group by
p.year,
p.month,
p.Location
SQL Fiddle
get year and month from date and use them in join , most dbms has year and month functions you can use according to your DBMS
select year(t1.date) yr,month(t1.date) as monthofyr ,t1.Location,
sum(t1.amount) as actual_amoun,
sum(t2.amount) as planamount
from table1 t1 left join table2 t2 on
month(t1.date)= t2.Month and t1.Location=t2.Location
and year(t1.date)=t2.year
group by year(t1.date) ,month(t1.date),Location
Related
I have a table with the columns Sales_Date and Sales. I am looking for a solution to get Sales for the last year from the Sales_Date Column. Sales_Date column has values from the year 2015 onwards.
For example:
Sales_Date
Sales
1/1/2016
$25
1/8/2016
$57
1/1/2015
$125
1/8/2015
$21
I am looking for the below result set:
Sales_Date
Sales
LYear_Sales_Date
LYear_Sales
1/1/2016
$25
1/1/2015
$125
1/8/2016
$57
1/8/2015
$21
Filter all data to this year (WHERE YEAR(Sales.Sales_Date) = 2016).
LEFT JOIN to the same table, combining each date with the same date one year prior (Sales LEFT JOIN Sales AS Sales_LastYear ON Sales_LastYear.Sales_Date = DATEADD(year, -1, Sales.Sales_Date)).
SELECT the fields that you want (SELECT Sales.Sales_Date, Sales_LastYear.Sales_Date AS LYear_Sales_Date, ...).
Replace the LEFT JOIN with an INNER JOIN, if you want only those records that have a matching last-year record.
Seems like LAG would work here. Assuming you are always wanting the for the same (day and) month:
WITH CTE AS(
SELECT Sales_Date,
Sales,
LAG(Sales_Date) OVER (PARTITION BY DAY(Sales_Date), MONTH(Sales_Date) ORDER BY YEAR(Sales_Date)) AS LYear_Sales_Date,
LAG(Sales) OVER (PARTITION BY DAY(Sales_Date), MONTH(Sales_Date) ORDER BY YEAR(Sales_Date)) AS LYear_Sales
FROM dbo.YourTable)
SELECT Sales_Date,
Sales,
LYear_Sales_Date,
LYear_Sales
FROM CTE
WHERE Sales_Date >= '20160101'
AND Sales_Date < '20170101';
I have a table where I have values by month and I want to spread these values by week, taking into account that weeks that spread into two month need to take part of the value of each of the month and weight on the number of days that correspond to each month.
For example I have the table with a different price of steel by month
Product Month Price
------------------------------------
Steel 1/Jan/2014 100
Steel 1/Feb/2014 200
Steel 1/Mar/2014 300
I need to convert it into weeks as follows
Product Week Price
-------------------------------------------
Steel 06-Jan-14 100
Steel 13-Jan-14 100
Steel 20-Jan-14 100
Steel 27-Jan-14 128.57
Steel 03-Feb-14 200
Steel 10-Feb-14 200
Steel 17-Feb-14 200
As you see above, the week that overlaps between Jan and Feb needs to be calculated as follows
(100*5/7)+(200*2/7)
This takes into account tha the week of the 27th has 5 days that fall into Jan and 2 into Feb.
Is there any possible way to create a query in SQL that would achieve this?
I tried the following
First attempt:
select
WD.week,
PM.PRICE,
DATEADD(m,1,PM.Month),
SUM(PM.PRICE/7) * COUNT(*)
from
( select '2014-1-1' as Month, 100 as PRICE
union
select '2014-2-1' as Month, 200 as PRICE
)PM
join
( select '2014-1-20' as week
union
select '2014-1-27' as week
union
select '2014-2-3' as week
)WD
ON WD.week>=PM.Month
AND WD.week < DATEADD(m,1,PM.Month)
group by
WD.week,PM.PRICE, DATEADD(m,1,PM.Month)
This gives me the following
week PRICE
2014-1-20 100 2014-02-01 00:00:00.000 14
2014-1-27 100 2014-02-01 00:00:00.000 14
2014-2-3 200 2014-03-01 00:00:00.000 28
I tried also the following
;with x as (
select price,
datepart(week,dateadd(day, n.n-2, t1.month)) wk,
dateadd(day, n.n-1, t1.month) dt
from
(select '2014-1-1' as Month, 100 as PRICE
union
select '2014-2-1' as Month, 200 as PRICE) t1
cross apply (
select datediff(day, t.month, dateadd(month, 1, t.month)) nd
from
(select '2014-1-1' as Month, 100 as PRICE
union
select '2014-2-1' as Month, 200 as PRICE)
t
where t1.month = t.month) ndm
inner join
(SELECT (a.Number * 256) + b.Number AS N FROM
(SELECT number FROM master..spt_values WHERE type = 'P' AND number <= 255) a (Number),
(SELECT number FROM master..spt_values WHERE type = 'P' AND number <= 255) b (Number)) n --numbers
on n.n <= ndm.nd
)
select min(dt) as week, cast(sum(price)/count(*) as decimal(9,2)) as price
from x
group by wk
having count(*) = 7
order by wk
This gimes me the following
week price
2014-01-07 00:00:00.000 100.00
2014-01-14 00:00:00.000 100.00
2014-01-21 00:00:00.000 100.00
2014-02-04 00:00:00.000 200.00
2014-02-11 00:00:00.000 200.00
2014-02-18 00:00:00.000 200.00
Thanks
If you have a calendar table it's a simple join:
SELECT
product,
calendar_date - (day_of_week-1) AS week,
SUM(price/7) * COUNT(*)
FROM prices AS p
JOIN calendar AS c
ON c.calendar_date >= month
AND c.calendar_date < DATEADD(m,1,month)
GROUP BY product,
calendar_date - (day_of_week-1)
This could be further simplified to join only to mondays and then do some more date arithmetic in a CASE to get 7 or less days.
Edit:
Your last query returned jan 31st two times, you need to remove the =from on n.n < ndm.nd. And as you seem to work with ISO weeks you better change the DATEPART to avoid problems with different DATEFIRST settings.
Based on your last query I created a fiddle.
;with x as (
select price,
datepart(isowk,dateadd(day, n.n, t1.month)) wk,
dateadd(day, n.n-1, t1.month) dt
from
(select '2014-1-1' as Month, 100.00 as PRICE
union
select '2014-2-1' as Month, 200.00 as PRICE) t1
cross apply (
select datediff(day, t.month, dateadd(month, 1, t.month)) nd
from
(select '2014-1-1' as Month, 100.00 as PRICE
union
select '2014-2-1' as Month, 200.00 as PRICE)
t
where t1.month = t.month) ndm
inner join
(SELECT (a.Number * 256) + b.Number AS N FROM
(SELECT number FROM master..spt_values WHERE type = 'P' AND number <= 255) a (Number),
(SELECT number FROM master..spt_values WHERE type = 'P' AND number <= 255) b (Number)) n --numbers
on n.n < ndm.nd
) select min(dt) as week, cast(sum(price)/count(*) as decimal(9,2)) as price
from x
group by wk
having count(*) = 7
order by wk
Of course the dates might be from multiple years, so you need to GROUP BY by the year, too.
Actually, you need to spred it over days, and then get the averages by week. To get the days we'll use the Numbers table.
;with x as (
select product, price,
datepart(week,dateadd(day, n.n-2, t1.month)) wk,
dateadd(day, n.n-1, t1.month) dt
from #t t1
cross apply (
select datediff(day, t.month, dateadd(month, 1, t.month)) nd
from #t t
where t1.month = t.month and t1.product = t.product) ndm
inner join numbers n on n.n <= ndm.nd
)
select product, min(dt) as week, cast(sum(price)/count(*) as decimal(9,2)) as price
from x
group by product, wk
having count(*) = 7
order by product, wk
The result of datepart(week,dateadd(day, n.n-2, t1.month)) expression depends on SET DATEFIRST so you might need to adjust accordingly.
I have a table with sales for products.
The sales are per day. like
product date sales
1 '2013-11-01' 100
1 '2013-11-02' 423
1 '2013-11-03' 700
1 '2013-11-04' 233
2 '2013-11-01' 623
2 '2013-11-02' 451
2 '2013-11-03' 9000
I want to get a query which will show me the week over week sum of sales
So something like:
product week ending sales
1 '2013-11-01' 10000
1 '2013-11-08' 15000
2 '2013-11-01' 4900
2 '2013-11-08' 30000
I'm not sure how I get this weekly groups when summing up.
I'm using teradata
If you are using Teradata 14 you can leverage the DayNumber_Of_Week() function in the database TD_SYSFNLIB:
SELECT s.Product
, s.Date + (7-DayNumber_Of_Week(s.date)) AS WeekEndingDate /* Saturday */
, SUM(s.Sales) AS Sales
FROM sales AS S
GROUP BY 1,2;
This should work in Teradata 13.10 as well.
Using Sys_Calendar:
SELECT s.Product
, s.DATE + (7-c.Day_Of_Week) AS WeekEndingDate /* Saturday */
, SUM(s.Sales) AS Sales
FROM sales AS S
INNER JOIN
Sys_Calendar.Calendar c
ON S.date = c.calendar_date
GROUP BY 1,2;
I know very little about TERADATA, but I believe you can leverage the sys_calendar.calendar table, something like:
SELECT s.Product, c.week_of_year, SUM(s.sales) AS Sales
FROM sales AS s
JOIN sys_calendar.calendar as C
ON s.date = c.date
You'd need the Year in there as well, so as to not group up week 1 of 2013 with week 1 of 2012.
my working table, Table name: sales
Here Is MY TABLE, [sl_no is primary key] table structure:
CREATE TABLE SALES
( SL_NO NUMBER PRIMARY KEY, REGION VARCHAR2(10) NOT NULL,
MONTH VARCHAR2(20) NOT NULL, YEAR NUMBER NOT NULL,
SALES_AMOUNT NUMBER NOT NULL )
and here is table data:
SQL> select * from sales;
SL_NO REGION MONTH YEAR SALES_AMOUNT
---------- ---------- -------------------- ---------- ------------
1 east december 2011 750000
2 east august 2011 800000
3 west january 2012 640000
5 east march 2012 1200000
6 west february 2011 580000
4 west april 2011 555000
6 rows selected.
I have tried this query to view total sales amount of those[2011,2012] year;
SELECT year, SUM(sales_amount) FROM sales GROUP BY year;
YEAR SUM(SALES_AMOUNT)
---------- -----------------
2011 2685000
2012 1840000
MY GOAL:> I want to find out the year of maximum sales amount.
I tried this,and work perfectly...but when i want to display that year also, it gives an Error.
SQL> select max(sum(sales_amount)) from sales group by year;
MAX(SUM(SALES_AMOUNT))
----------------------
2685000
SQL> select year, max(sum(sales_amount)) from sales group by year;
select year, max(sum(sales_amount)) from sales group by year
*
ERROR at line 1:
ORA-00937: not a single-group group function
Extra addition: if multiple rows have same value means....when sales amount of both year[2011,2012] remain same, Then....
plZ help me to Solve this problem.
This should work.
with yr_agg as (
select year, sum(sales_amount) as total
from sales
group by year
)
select year, total as max_total
from yr_agg
where total = (select max(total)
from yr_agg);
I think the simplest way is to order the results and take the first row:
select year, sales_amount
from (SELECT year, SUM(sales_amount) as sales_amount
FROM sales
GROUP BY year
order by sum(sales_amount) desc
) t
where rownum = 1;
EDIT:
If you need to display all the matching rows (which isn't mentioned in the question), I would suggest using the dense_rank() analytic function:
select year, sales_amount
from (SELECT year, SUM(sales_amount) as sales_amount,
dense_rank(over order by SUM(sales_amount) desc) as seqnum
FROM sales
GROUP BY year
order by sum(sales_amount) desc
) t
where seqnum = 1;
Or, you might like the max() version instead:
select year, sales_amount
from (SELECT year, SUM(sales_amount) as sales_amount,
max(sum(sales_amount)) over () as maxsa
FROM sales
GROUP BY year
order by sum(sales_amount) desc
) t
where sales_amount = maxsa;
Following select should do what you need (untested, do not have Oracle at home):
select year, total
from (
select year, sum(sales_amount) total
from sales
group by year
)
where total = (select max(total_amount)
from (
select year, sum(sales_amount) total_amount
from sales
group by year
))
Take in account, though, that it might give you different years in each execution if two of them have exactly the same total amount. You might want to include some more conditions to avoid this.
Here is my Query where multiple row can select
SELECT year,MAX(total_sale) as max_total
FROM
(SELECT year,SUM(sales_amount) AS total_sale FROM sales GROUP BY year)
GROUP BY
year HAVING MAX(total_sale) =
(SELECT MAX(total_sale) FROM (SELECT SUM(sales_amount) AS total_sale FROM sales GROUP BY year));
ACCOUNT Amount DATE
1 50 01-2010
1 100 03-2010
1 100 02-2011
2 100 01-2011
2 50 05-2011
2 50 09-2011
3 100 03-2012
3 100 03-2013
Is there a query structure that will allow me to count distinct accounts that has spanned current and past year? For example, account 1 has amounts in 2011 and 2010 so it should be counted once under 2011. Account 2 only has amounts in 2011 so it doesn't get counted while account 3 has amounts in 2013 and 2012, so it gets counted as 1 under 2013:
2010 2011 2012 2013
0 1 0 1
First, you need to know the years where you have data for an account:
select account, year(date) as yr
from t
group by account, year(date)
Next, you need to see if two years are in sequence. You can do this in 2012 with lag/lead. Instead, we'll just use a self join:
with ay as (
select account, year(date) as yr
from t
group by account, year(date)
)
select ay.account, ay.yr
from ay join
ay ayprev
on ay.account = ayprev.account and
ay.yr = ayprev.yr + 1
Next, if you want to count the number of accounts by year, just put this into an aggregation:
with ay as (
select account, year(date) as yr
from t
group by account, year(date)
)
select yr, count(*) as numaccounts
from (select ay.account, ay.yr
from ay join
ay ayprev
on ay.account = ayprev.account and
ay.yr = ayprev.yr + 1
) ayy
group by yr
Assuming you have a record id (call this ID)
SELECT COUNT(*),Year FROM Table t3
INNER JOIN (
SELECT record_id, Year(t1.Date) as Year FROM Table t1
INNER JOIN Table t2
WHERE Year(t1.Date)-1=Year(t2.Date) AND t1.Account == t2.Account
) x ON x.record_id = t3.record_id
GROUP BY Year
Use Below Query :
SELECT YEAR(T1.Date) AS D, COUNT(*) AS C
FROM YourTable AS T1
INNER JOIN YourTable T2 ON T2.Account = T1.Account AND YEAR(T2)=YEAR(T1)+1
GROUP BY T1.Account, YEAR(T1.Date)