I have made a query to bring me the number of products that have not been in stock (I know that by looking at the orders which the manufacturer returned with some status code), by product, date and storage, that looks like this:
SELECT count(*) as out_of_stock,
prod.id as product_id,
ped.data_envio::date as date,
opl.id as storage_id
from sub_produtos_pedidos spp
left join cad_produtos prod ON spp.ean_produto = prod.cod_ean
left join sub_pedidos sp ON spp.id_pedido = sp.id
left join pedidos ped ON sp.id_pedido = ped.id
left join op_logisticos opl ON sp.id_op_logistico = opl.id
where spp.motivo = '201' -- this is the code that means 'not in inventory'
group by storage_id,product_id,date
That produces an answer like this:
out_of_stock | product_id | date | storage_id
--------------|------------|-------------|-------------
1 | 5 | 2012-10-16 | 1
5 | 4 | 2012-10-16 | 2
Now I need to get the number of occurrences, by product and storage, of products that have been out of stock for 2 or more days, 5 or more days and so on.
So I guess I need to do a new count on the first query, aggregating the resultant rows in some defined day intervals.
I tried looking at the datetime functions in Postgres (http://www.postgresql.org/docs/7.3/static/functions-datetime.html), but couldn't find what I need.
May be I didn't get correctly you question, but it looks you need leverage sub-query.
Now I need to get the number of occurrences, by product and storage, of products that have been out of stock for 2 or more days
So:
SELECT COUNT(*), date, product_id FROM ( YOUR BIG QUERY IS THERE ) a
WHERE a.date < (CURRENT_DATE - interval '2' day)
GROUP BY date, product_id
Since you seem to want every row in the result individually, you cannot aggregate. Use a window function instead to get the count per day. The well known aggregate function count() can also serve as window aggregate function:
SELECT current_date - ped.data_envio::date AS days_out_of_stock
,count(*) OVER (PARTITION BY ped.data_envio::date)
AS count_per_days_out_of_stock
,ped.data_envio::date AS date
,p.id AS product_id
,opl.id AS storage_id
FROM sub_produtos_pedidos spp
LEFT JOIN cad_produtos p ON p.cod_ean = spp.ean_produto
LEFT JOIN sub_pedidos sp ON sp.id = spp.id_pedido
LEFT JOIN op_logisticos opl ON opl.id = sp.id_op_logistico
LEFT JOIN pedidos ped ON ped.id = sp.id_pedido
WHERE spp.motivo = '201' -- code for 'not in inventory'
ORDER BY ped.data_envio::date, p.id, opl.id
Sort order: Products having been out of stock for the longest time first.
Note, you can just subtract dates to get an integer in Postgres.
If you want a running count in the sense of "n rows have been out of stock for this number of days or more", use:
count(*) OVER (ORDER BY ped.data_envio::date) -- ascending order!
AS running_count_per_days_out_of_stock
You get the same count for the same day, peers are lumped together.
Related
I have two tables which name shoes_type and shoes_list. The shoes_type table includes shoes_id, shoes_size, shoes_type, date, project_id. Meanwhile, on the shoes_list table, I have shoes_quantity, shoes_id, shoes_color, date, project_id.
I need to get the sum of shoes_quantity based on the shoes_type, shoes_size, date, and also project_id.
I get how to sum the shoes_quantity based on color by doing:
select shoes_color, sum(shoes_quantity)
from shoes_list group by shoes_color
Basically what I want to see is the total quantity of shoes based on the type, size, date and project_id. The type and size information are available on shoes_type table, while the quantity is coming from the shoes_list table. I expect to see something like:
shoes_type shoes_size total quantity date project_id
heels 5 3 19/10/02 1
sneakers 5 3 19/10/02 1
sneakers 6 1 19/10/05 1
heels 7 5 19/10/03 1
While for the desired result, I have tried:
select shoes_type, shoes_size, date, project_id, sum(shoes_quantity)
from shoes_type st
join shoes_list sl
on st.project_id = sl.project_id
and st.shoes_id = sl.shoes_id
and st.date = sl.date
group by shoes_type, shoes_size, date, project_id
Unfortunately, I got an error that says that the column reference "date" is ambiguous.
How should I fix this?
Thank you.
The date column exists in both tables, so you have to specify where to select it from. Replace date with shoes_type.date or shoes_list.date
Qualify all column references to remove the "ambiguous" column error:
select st.shoes_type, st.shoes_size, st.date, st.project_id, sum(slshoes_quantity)
from shoes_type st join
shoes_list sl
on st.project_id = sl.project_id and
st.shoes_id = sl.shoes_id and
st.date = sl.date
group by st.shoes_type, st.shoes_size, st.date, st.project_id;
If you want all columns from shoes_type, you might find that a correlated subquery is faster:
select st.*,
(select sum(slshoes_quantity)
from shoes_list sl
where st.project_id = sl.project_id and
st.shoes_id = sl.shoes_id and
st.date = sl.date
)
from shoes_type st;
I´m having issues with the following query. I have two tables; Table Orderheader and table Bought. The first query I execute gives me, for example, two dates. Based on these two dates, I need to find Production data AND, based on the production data, I need to find the Bought data, and combine those data together. Lets say I do the following:
Select Lotdate From Orderheader where orhsysid = 1
This results in two rows: '2019-02-05' and '2019-02-04'. Now I need to do two things: I need two run two queries using this result set. The first one is easy; use the dates returned and get a sum of column A like this:
Select date, SUM(Amount) from Orderheader where date = Sales.date() [use the two dates here]
The second one is slighty more complicated, I need to find the last day where something has been bought based on the two dates. Production is everyday so Productiondate=Sales.date()-1. But Bought is derived from Productionday and is not everyday so for every Productionday it needs to find the last Boughtday. So I can't say where date = Orderheader.date. I need to do something like:
Select date, SUM(Amount)
FROM Bought
WHERE date = (
SELECT top 1 date
FROM Bought
WHERE date < Orderheader.date)
But twice, for both the dates I got.
This needs to result in 1 table giving me:
Bought.date, Bought.SUM(AMOUNT), Orderheader.date, Orderheader.SUM(AMOUNT)
All based on the, possible multiple, Lotdate(s) I got from the first query from Sales table.
I've been struggling with this for a moment now, using joins and nested queries but I can't seem to figure it out!
Example sample:
SELECT CONVERT(date,ORF.orfDate) as Productiedatum, SUM(orlQuantityRegistered) as 'Aantal'
FROM OrderHeader ORH
LEFT JOIN OrderFrame ORF ON ORH.orhFrameSysID = ORF.orfSysID
LEFT JOIN OrderLine ORL ON ORL.orhSysID = ORH.orhSysID
LEFT JOIN Item ON Item.itmSysID = ORL.orlitmSysID
where CONVERT(date,ORF.orfDate) IN
(
SELECT DISTINCT(CONVERT(date, Lot.lotproductiondate)) as Productiedatum
FROM OrderHeader ORH
LEFT JOIN Registration reg ON reg.regorhSysID = ORH.orhSysID
LEFT JOIN StockRegistration stcreg ON stcreg.stcregRegistrationSysID = reg.regSysID
LEFT JOIN Lot ON Lot.lotSysID = stcregSrclotSysID
WHERE ORH.orhSysID = 514955
AND regRevokeRegSysID IS NULL
AND stcregSrcitmSysID = 5103
)
AND ORL.orlitmSysID = 5103
AND orldirSysID = 2
AND NOT orlQuantityRegistered IS NULL
GROUP BY Orf.orfDate
Sample output:
Productiedatum Aantal
2019-02-05 20
2019-02-06 20
Here I used a nested subquery to get the results from 'Production' (orderheader) because I just can use date = date. I'm struggling with the Sales part where I need to find the last date(s) and use those dates in the Sales table to get the sum of that date.
Expected output:
Productiedatum Aantal Boughtdate Aantal
2019-02-04 20 2019-02-01 55
2019-02-05 20 2019-02-04 60
Try this.
IF OBJECT_ID('tempdb..#Production') IS NOT NULL DROP TABLE #Production
IF OBJECT_ID('tempdb..#Bought') IS NOT NULL DROP TABLE #Bought
CREATE table #Production(R_NO int,ProductionDate datetime,ProductionAmount float)
CREATE table #Bought(R_NO int,Boughtdate datetime,Boughtamount float)
insert into #Production(ProductionDate,ProductionAmount,R_NO)
select p.date ProductionDate,sum(Amount) ProductionAmount,row_number()over (order by p.date) R_NO
from Production P
join Sales s on p.date=S.date-1
where orhsysid=1
group by p.date
declare #loop int,#ProdDate datetime
select #loop =max(R_NO) from #Production
while (1<=#loop)
begin
select #ProdDate=ProductionDate from #Production where r_no=#loop
insert into #Bought(Boughtdate,Boughtamount,R_NO)
select Date,Sum(Amount),#loop R_NO from Bought where date=(
select max(date) from bought B
where B.Date<#ProdDate)
group by Date
set #loop=#loop-1
end
select ProductionDate,ProductionAmount,Boughtdate,Boughtamount from #Bought B
join #Production p on B.R_NO=P.R_NO
I am using Terdata SQL Assistant connected to an enterprise DW. I have written the query below to show an inventory of outstanding items as of a specific point in time. The table referenced loads and stores new records as changes are made to their state by load date (and does not delete historical records). The output of my query is 1 row for the specified date. Can I create a stored procedure or recursive query of some sort to build a history of these summary rows (with 1 new row per day)? I have not used such functions in the past; links to pertinent previously answered questions or suggestions on how I could get on the right track in researching other possible solutions are totally fine if applicable; just trying to bridge this gap in my knowledge.
SELECT
'2017-10-02' as Dt
,COUNT(DISTINCT A.RECORD_NBR) as Pending_Records
,SUM(A.PAY_AMT) AS Total_Pending_Payments
FROM DB.RECORD_HISTORY A
INNER JOIN
(SELECT MAX(LOAD_DT) AS LOAD_DT
,RECORD_NBR
FROM DB.RECORD_HISTORY
WHERE LOAD_DT <= '2017-10-02'
GROUP BY RECORD_NBR
) B
ON A.RECORD_NBR = B.RECORD_NBR
AND A.LOAD_DT = B.LOAD_DT
WHERE
A.RECORD_ORDER =1 AND Final_DT Is Null
GROUP BY Dt
ORDER BY 1 desc
Here is my interpretation of your query:
For the most recent load_dt (up until 2017-10-02) for record_order #1,
return
1) the number of different pending records
2) the total amount of pending payments
Is this correct? If you're looking for this info, but one row for each "Load_Dt", you just need to remove that INNER JOIN:
SELECT
load_Dt,
COUNT(DISTINCT record_nbr) AS Pending_Records,
SUM(pay_amt) AS Total_Pending_Payments
FROM DB.record_history
WHERE record_order = 1
AND final_Dt IS NULL
GROUP BY load_Dt
ORDER BY 1 DESC
If you want to get the summary info per record_order, just add record_order as a grouping column:
SELECT
load_Dt,
record_order,
COUNT(DISTINCT record_nbr) AS Pending_Records,
SUM(pay_amt) AS Total_Pending_Payments
FROM DB.record_history
WHERE final_Dt IS NULL
GROUP BY load_Dt, record_order
ORDER BY 1,2 DESC
If you want to get one row per day (if there are calendar days with no corresponding "load_dt" days), then you can SELECT from the sys_calendar.calendar view and LEFT JOIN the query above on the "load_dt" field:
SELECT cal.calendar_date, src.Pending_Records, src.Total_Pending_Payments
FROM sys_calendar.calendar cal
LEFT JOIN (
SELECT
load_Dt,
COUNT(DISTINCT record_nbr) AS Pending_Records,
SUM(pay_amt) AS Total_Pending_Payments
FROM DB.record_history
WHERE record_order = 1
AND final_Dt IS NULL
GROUP BY load_Dt
) src ON cal.calendar_date = src.load_Dt
WHERE cal.calendar_date BETWEEN <start_date> AND <end_date>
ORDER BY 1 DESC
I don't have access to a TD system, so you may get syntax errors. Let me know if that works or you're looking for something else.
I am still new to SQL and getting my head around the whole sub-query aggregation to display some results and was looking for some advice:
The tables might look something like:
Customer: (custID, name, address)
Account: (accountID, reward_balance)
Shop: (shopID, name, address)
Relational tables:
Holds (custID*, accountID*)
With (accountID*, shopID*)
How can I find the store that has the least reward_balance?
(The customer info is not required at this point)
I tried:
SELECT accountID AS ACCOUNT_ID, shopID AS SHOP_ID, MIN(reward_balance) AS LOWEST_BALANCE
FROM Account, Shop, With
WHERE With.accountID = Account.accountID
AND With.shopID=Shop.shopID
GROUP BY
Account.accountID,
Shop.shopID
ORDER BY MIN(reward_balance);
This works in a way that is not intended:
ACCOUNT_ID | SHOP_ID | LOWEST_BALANCE
1 | 1 | 10
2 | 2 | 40
3 | 3 | 100
4 | 4 | 1000
5 | 4 | 5000
As you can see Shop_ID 4 actually has a balance of 6000 (1000+5000) as there are two customers registered with it. I think I need to SUM the lowest balance of the shops based on their balance and display it from low-high.
I have been trying to aggregate the data prior to display but this is where I come unstuck:
SELECT shopID AS SHOP_ID, MIN(reward_balance) AS LOWEST_BALANCE
FROM (SELECT accountID, shopID, SUM(reward_balance)
FROM Account, Shop, With
WHERE
With.accountID = Account.accountID
AND With.shopID=Shop.shopID
GROUP BY
Account.accountID,
Shop.shopID;
When I run something like this statement I get an invalid identifier error.
Error at Command Line : 1 Column : 24
Error report -
SQL Error: ORA-00904: "REWARD_BALANCE": invalid identifier
00904. 00000 - "%s: invalid identifier"
So I figured I might have my joining condition incorrect and the aggregate sorting incorrect, and would really appreciate any general advice.
Thanks for the lengthy read!
Approach this problem one step at time.
We're going to assume (and we should probably check this) that by least reward_balance, that refers to the total of all reward_balance associated with a shop. And we're not just looking for the shop that has the lowest individual reward balance.
First, get all of the individual "reward_balance" for each shop. Looks like the query would need to involve three tables...
SELECT s.shop_id
, a.reward_balance
FROM `shop` s
LEFT
JOIN `with` w
ON w.shop_id = s.shop_id
LEFT
JOIN `account` a
ON a.account_id = w.account_id
That will get us the detail rows, every shop along with the individual reward_balance amounts associated with the shop, if there are any. (We're using outer joins for this query, because we don't see any guarantee that a shops is going to be related to at least one account. Even if it's true for this use case, that's not always true in the more general case.)
Once we have the individual amounts, the next step is to total them for each shop. We can do that using a GROUP BY clause and a SUM() aggregate.
SELECT s.shop_id
, SUM(a.reward_balance) AS tot_reward_balance
FROM `shop` s
LEFT
JOIN `with` w
ON w.shop_id = s.shop_id
LEFT
JOIN `account` a
ON a.account_id = w.account_id
GROUP BY s.shop_id
At this point, with MySQL we could add an ORDER BY clause to arrange the rows in ascending order of tot_reward_balance, and add a LIMIT 1 clause if we only want to return a single row. We can also handle the case when tot_reward_balance is NULL, assigning a zero in place of the NULL.
SELECT s.shop_id
, IFNULL(SUM(a.reward_balance),0) AS tot_reward_balance
FROM `shop` s
LEFT
JOIN `with` w
ON w.shop_id = s.shop_id
LEFT
JOIN `account` a
ON a.account_id = w.account_id
GROUP BY s.shop_id
ORDER BY tot_reward_amount ASC, s.shop_id ASC
LIMIT 1
If there are two (or more) shops with the same least value of tot_reward_amount, this query returns only one of those shops.
Oracle doesn't have the LIMIT clause like MySQL, but we can get equivalent result using analytic function (which is not available in MySQL). We also replace the MySQL IFNULL() function with the Oracle equivalent NVL() function...
SELECT v.shop_id
, v.tot_reward_balance
, ROW_NUMBER() OVER (ORDER BY v.tot_reward_balance ASC, v.shop_id ASC) AS rn
FROM (
SELECT s.shop_id
, NVL(SUM(a.reward_balance),0) AS tot_reward_balance
FROM shop s
LEFT
JOIN with w
ON w.shop_id = s.shop_id
LEFT
JOIN account a
ON a.account_id = w.account_id
GROUP BY s.shop_id
) v
HAVING rn = 1
Like the MySQL query, this returns at most one row, even when two or more shops have the same "least" total of reward_balance.
If we want to return all of the shops that have the lowest tot_reward_balance, we need to take a slightly different approach.
The best approach to building queries is step wise refinement; in this case, start by getting all of the individual reward_amount for each shop. Next step is to aggregate the individual reward_amount into a total. The next steps is to pickout the row(s) with the lowest total reward_amount.
In SQL Server, You can try using a CTE:
;with cte_minvalue as
(
select rank() over (order by Sum_Balance) as RowRank,
ShopId,
Sum_Balance
from (SELECT Shop.shopID, SUM(reward_balance) AS Sum_Balance
FROM
With
JOIN Shop ON With.ShopId = Shop.ShopId
JOIN Account ON With.AccountId = Account.AccountId
GROUP BY
Shop.shopID)ShopSum
)
select ShopId, Sum_Balance from cte_minvalue where RowRank = 1
I have a select statement that divides the count of sales by country, priceBanding (see example below)
The select statement looks like follows:
SELECT p.[Price Band]
,t.[Country]
,o.COUNT([Order]) as [Order Count]
FROM #price p (temp table)
INNER JOIN country t ON p.CountryCode = t.countryCode
INNER JOIN sales o ON o.salesValue >= p.startPrice and s.salesValue < p.endPrice
What i want to be able to do is based on this result i want to get an avg of the unit count i.e. For all orders that are under 20 what is the avg unit counts and the same for all others. How can i do this?
Its most likely simple but I cant think through it.
What I am after:
So as you can see, in the price band <20 in UK the order count is 50, and the avg Units of that is 2. As i mentioned earlier, I want the Avg Units of all orders that are under 20 (which is 50 in the picture).
Is that clearer?
Thanks in advance.
EDIT:
The first table: assume it to be the source
And the second table gets the avg, that's what I am after.
Wouldn't you just use avg()?
SELECT p.[Price Band], t.[Country],
o.COUNT(*) as [Order Count],
AVG(Items)
FROM #price p INNER JOIN
country t
ON p.CountryCode = t.countryCode INNER JOIN
sales o
ON o.salesValue >= p.startPrice and s.salesValue < p.endPrice
GROUP BY p.[Price Band], t.[Country]
ORDER BY t.[Country], p.[Price Band]
Note: SQL Server does integer division of integers (so 3/2 = 1 not 1.5) and similarly for AVG(). It is more accurate to use a decimal point number. An easy way is to use AVG(items * 1.0).