I have built an SQL Query that returns me the top 10 customers which have the highest outstanding. The oustanding is on product level (each product has its own outstanding).
Untill now everything works fine, my only problem is that if a certain customer has more then 1 product then the second product or more should be categorized under the same customer_id like in the second picture (because the first product that has the highest outstanding contagions the second product that may have a lower outstanding that the other 9 clients of top 10).
How can I modify my query in order to do that? Is it possible in SQL Server 2012?
My query is:
select top 10 CUSTOMER_ID
,S90T01_GROSS_EXPOSURE_THSD_EUR
,S90T01_COGNOS_PROD_NAME
,S90T01_DPD_C
,PREVIOUS_BUCKET_DPD_REP
,S90T01_BUCKET_DPD_REP
from [dbo].[DM_07MONTHLY_DATA]
where S90T01_CLIENT_SEGMENT = 'PI'
and YYYY_MM = '2017_01'
group by CUSTOMER_ID
,S90T01_GROSS_EXPOSURE_THSD_EUR
,S90T01_COGNOS_PROD_NAME
,S90T01_DPD_C
,PREVIOUS_BUCKET_DPD_REP
,S90T01_BUCKET_DPD_REP
order by S90T01_GROSS_EXPOSURE_THSD_EUR desc;
You need to calculate the top Customers first, then pull out all their products. You can do this with a Common Table Expression.
As you haven't provided any test data this is untested, but I think it will work for you:
with top10 as
(
select top 10 CUSTOMER_ID
,sum(S90T01_GROSS_EXPOSURE_THSD_EUR) as TOTAL_S90T01_GROSS_EXPOSURE_THSD_EUR
from [dbo].[DM_07MONTHLY_DATA]
where S90T01_CLIENT_SEGMENT = 'PI'
and YYYY_MM = '2017_01'
group by CUSTOMER_ID
order by TOTAL_S90T01_GROSS_EXPOSURE_THSD_EUR desc
)
select m.CUSTOMER_ID
,m.S90T01_GROSS_EXPOSURE_THSD_EUR
,m.S90T01_COGNOS_PROD_NAME
,m.S90T01_DPD_C
,m.PREVIOUS_BUCKET_DPD_REP
,m.S90T01_BUCKET_DPD_REP
from [dbo].[DM_07MONTHLY_DATA] m
join top10 t
on m.CUSTOMER_ID = t.CUSTOMER_ID
order by t.TOTAL_S90T01_GROSS_EXPOSURE_THSD_EUR desc
,m.S90T01_GROSS_EXPOSURE_THSD_EUR;
Related
I am attempting to join a customer table with sales table where I show the list of all customers in database and any paid sale the customer might have in the sales tables. Now a customer can have multiple sales rows in the sales table.
This is an example sales record of one customer with multiple sales in the sale tables
while extracting this record I would like to get only the MAX (q_saledatetime) WHERE the q_paidamount is > 0.
as in show me the last time this customer made a payment to us. So in this case row 2 where they paid 8.90 is what I would like to get for that customer. If a customer has no record in the sales table, show their name/details on the list either way.
My failure at the moment is how to include the where clause of the paid amount + max date column.
ATTEMPT A
select DISTINCT ON (q_customer.q_code)
q_customer.q_code, q_customer.q_name, -- customer info
MAX(q_saleheader.q_saledatetime) AS latestDate, q_saleheader.q_paidamount -- saleheader info
FROM q_customer
LEFT JOIN q_saleheader ON (q_customer.q_code = q_saleheader.q_customercode)
group by q_customer.q_code, q_customer.q_name , q_saleheader.q_saledatetime, q_saleheader.q_paidamount
order by q_customer.q_code ASC
which results in
so for Fred Blogg is picking up details from row 4 instead of 2 (first image). As there's no rule for q_paidamount at this point
ATTEMPT B
SELECT
customer.q_code, customer.q_name, -- customer info
sale.q_saledatetime, sale.q_paidamount -- sale info
FROM q_customer customer
LEFT JOIN (SELECT * FROM q_saleheader WHERE q_saledatetime =
(SELECT MAX(q_saledatetime) FROM q_saleheader b1 where q_paidamount > 0 ))
sale ON sale.q_customercode = customer.q_code
which results in
This doesnt seem to be getting any information from the sale table at all.
Update:
After having a closer look at my first attempt I amended the statement and came up with this solution which achieves the same results as Michal's answer. I just curious to know is there any pitfalls or perfomance disadvantages with the following way.
select DISTINCT ON (q_customer.q_code)
q_customer.q_code, q_customer.q_name, -- customer info
q_saleheader.q_saledatetime, q_saleheader.q_paidamount -- saleheader info
FROM q_customer
LEFT JOIN q_saleheader ON (q_customer.q_code = q_saleheader.q_customercode AND
q_saleheader.q_paidamount > 0 )
group by q_customer.q_code, q_customer.q_name , q_saleheader.q_saledatetime,
q_saleheader.q_paidamount
order by q_customer.q_code ASC, q_saleheader.q_saledatetime DESC
main change was adding AND q_saleheader.q_paidamount > 0 on the join and q_saleheader.q_saledatetime DESC to make sure are getting the top row of that related data. As mentioned both Michal's answer and this solution achieve the same results. Just curious about pitfalls in either of the two ways.
Try this query:
SELECT c.q_code,
c.q_name,
CASE WHEN q_saledatetime <> '1900-01-01 00:00:00.000' THEN q_saledatetime END q_saledatetime,
q_paidamount
FROM (
SELECT c.q_code,
c.q_name,
coalesce(s.q_saledatetime, '1900-01-01 00:00:00.000') q_saledatetime, --it will indicate customer with no data
s.q_paidamount,
ROW_NUMBER() OVER (PARTITION BY c.q_code ORDER BY COALESCE(s.q_saledatetime, '1900-01-01') DESC) rn
FROM q_customer c
LEFT JOIN (SELECT q_saledatetime,
q_paidamount
FROM q_saleheader
WHERE q_paidamount > 0) s
ON c.q_code = s.q_customercode
) c WHERE rn = 1
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 want to write a query that allows me to only get the specific data I want and nothing more.
We will use TV's as an example. I have three brands of TVs and I want to see the top ten selling models of each brand. I only want to return 30 rows. One solution is unions, but that can get messy fast. Ideally there would be a WHERE ROWNUM grouping by situation.
SELECT
A.Brand
, A.Model
, A.Sales
FROM
( SELECT
TV.Brand
, TV.Model
, SUM(TV.SALES) AS SALES
FROM TV_TABLE as TV
ORDER BY
TV.Brand
, SALES DESC
) A
WHERE ROWNUM <10
In my code above I will get the top 10 total results from the inner query, but not 10 from each Grouping.
What I want to see is something like this:
Brand: Model: Sales
Sony: x10: 20
Sony: X20: 18
Sony: X30: 10
VISIO: A40: 40
VISIO: A20: 10
This is an oversimplified example, in practice I'll need to have 20-50 gropings and would like to avoid downloading all of the data and using a Pivot feature.
select Brand, Model, SALES
from(
select Brand, Model, SALES,row_number()over(partition by Brand order by SALES desc) rn
from (
SELECT TV.Brand, TV.Model,SUM(TV.SALES) AS SALES,
FROM TV_TABLE as TV
group BY TV.Brand,TV.Model
)a
)b
where rn <= 10
SELECT TV.Brand, TV.Model, SUM(TV.SALES) AS SALES
FROM TV_TABLE TV
group by TV.Brand, TV.Model
order by SUM(TV.SALES) desc, TV.Brand
limit 30
I would like to retrieve only the customers from an order table who have paid for all the orders. (Paid = 'Y').
The order table looks like this:
Cus # Order # Paid
111 1 Y
111 2 Y
222 3 Y
222 4 N
333 5 N
In this example the query should only return customer 111.
The query
Select * from order where Paid = 'Y';
returns customers that have paid and unpaid orders (ex. customer 222) in addition to customers who have paid for all of their orders (customer 111).
How do I structure the query to evaluate all the orders for a customer and only return information for a customer that has paid for all the orders?
Looking the problem a different way, you need only the customers who don't have any unpaid order.
sel cus from order group by Cus having min(Paid) = 'Y';
The above query also utilizes the fact that 'Y' > 'N'.
SQL Fiddle:
http://sqlfiddle.com/#!4/f6022/1
If you need to select all different orders for eligible customers, you may use OLAP functions:
select cus,order,paid from (select cus,order,paid,min(paid)
over (partition by cus)minz from order)dt where minz='Y';
With Oracle, you can also do select customer from your_table where paid='Y' minus select customer from your_table where paid='N', although I don't know if this is faster and, of course, you don't get the other fields in this case.
Don't know Oracle but often work in SQL Server, I would write query something like this
Select Cus from order group by Cus having count(*) = sum(case when Paid = ‘Y’ then 1 else 0);
Basically you retrieve customers where total order count equals to sum or paid orders.
Again, apologize for not giving proper Oracle syntax but hopefully that will point you to the right direction.
SELECT *
FROM orders oo
WHERE NOT EXISTS
(
SELECT 1
FROM orders io
WHERE oo.cus# = io.cus#
AND io.paid = 'N'
)
;