I have a table of accounts and a table of transactions. In a report I need to show the following for each account:
First Purchase Date,
First Purchase Amount,
Last Purchase Date,
Last Purchase Amount,
# of Purchases,
Total of All Purchases.
The transaction table looks like this:
TX_UID
Card_Number
Post_Date
TX_Type
TX_Amount
Currently the query I've inherited has a sub-query for each of these elements. It seems to me that there's got to be a more efficient way. I'm able to use a stored procedure for this and not a single query.
A sample of a query to get all transactions for a single account would be:
select * from tx_table where card_number = '12345' and TX_Type = 'Purchase'
Any ideas?
try this:
select tt1.post_date as first_purchase_date,
tt1.tx_amount as first_purchase_amount,
tt2.post_date as last_purchase_date,
tt2.tx_amount as last_purchase_amount,
tg.pc as purchase_count,
tg.amount as Total
from (select Card_Number,min(post_date) as mipd, max(post_date) as mxpd, count(*) as pc, sum(TX_Amount) as Amount from tx_table where TX_Type = 'Purchase' group by card_number) tg
join tx_table tt1 on tg.card_number=tt1.card_number and tg.mipd=tt1.post_date
join tx_table tt2 on tg.card_number=tt2.card_number and tg.mxpd=tt2.post_date
where TX_Type = 'Purchase'
I added the count .. I didn't see it first time.
If you need also the summary on multiple TX_Types, you have to take it from the where clause and put it in the group and the inner selection join. But I guess you need only for purchases
;with cte as
(
select
Card_Number,
TX_Type,
Post_Date,
TX_Amount,
row_number() over(partition by TX_Type, Card_Number order by Post_Date asc) as FirstP,
row_number() over(partition by TX_Type, Card_Number order by Post_Date desc) as LastP
from tx_table
)
select
F.Post_Date as "First Purchase Date",
F.TX_Amount as "First Purchase Amount",
L.Post_Date as "Last Purchase Date",
L.TX_Amount as "Last Purchase Amount",
C.CC as "# of Purchases",
C.Amount as "Total of All Purchases"
from (select Card_Number, TX_Type, count(*) as CC, sum(TX_Amount) as Amount
from cte
group by Card_Number, TX_Type) as C
inner join cte as F
on C.Card_Number = F.Card_Number and
C.TX_Type = F.TX_Type and
F.FirstP = 1
inner join cte as L
on C.Card_Number = L.Card_Number and
C.TX_Type = L.TX_Type and
L.LastP = 1
Related
I can not execute this code on Oracle, the error shows:
"ORA-00979: not a GROUP BY expression"
However, I was able to run it successfully on MySQL.
How does this happen?
SELECT CONCAT(i.lname, i.fname) AS inst_name,
CONCAT(s.lname, s.fname) AS stu_name,
t.avg_grade AS stu_avg_grade
FROM(
SELECT instructor_id, student_id, AVG(grade) as avg_grade, RANK() OVER(PARTITION BY instructor_id ORDER BY grade DESC) AS rk
FROM grade
GROUP BY 1,2) t
JOIN instructor i
ON t.instructor_id = i.instructor_id
JOIN student s
ON s.student_id = t.student_id
WHERE t.rk = 1
ORDER BY 3 DESC
You can't use ordinals like GROUP BY 1,2 in Oracle. In addition, the ORDER BY grade clause inside your RANK() function has a problem. Keep in mind that analytic functions evaluate after the GROUP BY aggregation, so grade is no longer available. Here is a version which should work without error:
SELECT CONCAT(i.lname, i.fname) AS inst_name,
CONCAT(s.lname, s.fname) AS stu_name,
t.avg_grade AS stu_avg_grade
FROM
(
SELECT instructor_id, student_id, AVG(grade) AS avg_grade,
RANK() OVER (PARTITION BY instructor_id ORDER BY AVG(grade) DESC) AS rk
FROM grade
GROUP BY instructor_id, student_id
) t
INNER JOIN instructor i
ON t.instructor_id = i.instructor_id
INNER JOIN student s
ON s.student_id = t.student_id
WHERE t.rk = 1
ORDER BY t.avg_grade DESC;
So I have two tables salary and emp whose definition is shown as below
[
I am tring to create a query that Find the employee who draws the maximum salary and Display the employee details along with the nationality.
I created this query
select empcode,
max(basic) as "Highest Sal"
from salary
join emp on empcode;
Please help with this
Your query uses a simple aggregate max(basic) which would find the highest salary. Except you need to join to the EMP table to display other details. This means you can't use aggregation, because we need to GROUP BY the non-aggregated columns, which would make a nonsense of the query.
Fortunately we can solve the problem with an analytic function. The subquery selects all the relevant information and ranks each employee by salary, with a rank of 1 being the highest paid. We use rank() here because that will handle ties: two employees with the same basic will be in the same rank.
select empcode
, empname
, nationality
, "Highest Sal"
from (
select emp.empcode
, emp.empname
, emp.nationality
, salary.basic as "Highest Sal"
, rank() over (order by salary.basic desc ) as rnk
from salary join emp on emp.empcode = salary.empcode
)
where rnk = 1;
Find the employee who draws the maximum salary
An employee can have multiple salaries in your datamodel. An employee's (total) salary hence is the sum of these. You want to find the maximum salary per employee and show the employee(s) earning that much.
You can use MAX OVER to find the maximum sum:
select e.*, s.total_salary
from emp e
join
(
select
empcode,
sum(basic) as total_salary,
max(sum(basic)) over () as max_total_salary
from salary
) s on s.empcode = e.empcode and s.total_salary = s.max_total_salary
order by e.empcode;
Try this:
SELECT * FROM
(SELECT E.EmpCode, E.EmpName, E.DOB, E.DOJ, E.DeptCode, E.DesgCode, E.PhNo,
E.Qualification, E.Nationality, S.Basic, S.HRA, S.TA, S.UTA, S.OTRate
FROM EMP AS E JOIN SALARY AS S ON (E.EmpCode = S.EmpCode) order by S.Basic desc)
WHERE rownum = 1
I'm using this query:
SELECT *
FROM HISTORY
LEFT JOIN CUSTOMER ON CUSTOMER.CUST_NUMBER = HISTORY.CUST_NUMBER
LEFT JOIN (
Select LOAN_DATE, CUST_NUMBER, ACCOUNT_NUMBER, STOCK_NUMBER, LOC_SALE
From LOAN
WHERE ACCOUNT_NUMBER != 'DD'
ORDER BY LOAN_DATE DESC
) LOAN ON LOAN.CUST_NUMBER = HISTORY.CUST_NUMBER
order by DATE desc
But I want only the top result from the loan table to be joined (Most recent by Loan_date). For some reason, it's getting three records (one for each loan on the customer I'm looking at). I'm sure I'm missing something simple?
If you're after joining the latest loan row per cust_number, then this ought to do the trick:
select *
from history
left join customer on customer.cust_number = history.cust_number
left join (select loan_date,
cust_number,
account_number,
stock_number,
loc_sale
from (select loan_date,
cust_number,
account_number,
stock_number,
loc_sale,
row_number() over (partition by cust_number
order by loan_date desc) rn
from loan
where account_number != 'DD')
where rn = 1) loan on loan.cust_number = history.cust_number
order by date desc;
If there are two rows with the same loan_date per cust_number and you want to retrieve both, then change the row_number() analytic function for rank().
If you only want to retreive one row, then you'd have to make sure you add additional columns into the order by, to make sure that the tied rows always display in the same order, otherwise you could find that sometimes you get different rows returned on subsequent runs of the query.
I have query
SELECT ID, TIME, PRICE, QTY
FROM myTable
that returns:
ID TIME PRICE QTY
1295179228 1/29/2015 20:59:37 15.24 1112
1295179228 1/29/2015 20:59:37 15.23 2
1295179228 1/29/2015 20:59:38 15.28 22
1295179228 1/29/2015 20:59:38 15.27 1800
I am then using group by to return the min time, average price and sum of qty BUT I also want to return the first time
SELECT ID, t2.name, min(TIME) as MinTIME, avg(PRICE) as AVGPrice, sum( QTY) as SUMQTY
FROM myTable t inner join table2 t2 on t.id = t2.id
group by ID, t2.name
But how do I add a column in that group by query above that will also return the first PRICE. In that case that would be 15.24
I have been googling and I see oracle has FIRST() and FIRST_VALUE() functions but I could not get them to work.
Thank you.
Assuming that "first price" means "price with the earliest time"
min(price) keep (dense_rank first order by time)
WITH CTE AS
(SELECT ID,"TIME",PRICE,QTY,ROW_NUMBER() OVER (PARTITION BY id ORDER BY TIME ASC) as rn
FROM t)
SELECT ID, min("TIME") as MinTIME, avg(PRICE) as AVGPrice, sum( QTY) as SUMQTY,
MAX(CASE WHEN rn=1 THEN PRICE ELSE 0 END) as FirstPrice
FROM CTE
group by ID
For Oracle,
I have 2 tables; first is Store and another is Book whereas they are connected by store ID (PK FK).
I would like to lists the name of the store which has the highest numbers of books.
However, the result showed every store in orders but I just want the highest.
SELECT STORE.STORE_NAME
FROM STORE, BOOK
WHERE STORE.STORE_ID=BOOK.BOOK_STOREID
GROUP BY STORE.STORE_NAME
ORDER BY COUNT(BOOK.BOOK_STOREID) DESC;
the result is
Store:
D
E
F
B
A
C
It should be only 'D'. What should I do? Thank you.
Try
SELECT STORE_NAME
FROM
(SELECT STORE.STORE_NAME
FROM STORE, BOOK
WHERE STORE.STORE_ID=BOOK.BOOK_STOREID
GROUP BY STORE.STORE_NAME
ORDER BY COUNT(BOOK.BOOK_STOREID) DESC)
WHERE rownum = 1
Here is a sqlfiddle demo
BTW, you can also use row_number() function
select STORE_NAME
from
(SELECT STORE.STORE_NAME,
row_number() over( order by COUNT(BOOK.BOOK_STOREID)desc) rn
FROM STORE join BOOK on STORE.STORE_ID=BOOK.BOOK_STOREID
group by STORE.STORE_NAME)
where rn = 1;
UPDATE If you want to see all stores which have the max number of books you can use rank instead of row_number:
select STORE_NAME
from
(SELECT STORE.STORE_NAME,
rank() over( order by COUNT(BOOK.BOOK_STOREID)desc) rn
FROM STORE join BOOK on STORE.STORE_ID=BOOK.BOOK_STOREID
group by STORE.STORE_NAME)
where rn = 1;
Just for fun, here's another formulation:
with store_counts as (
select store_name,
count(*) books
from store join book on store_id=book_storeid
group by store_name)
select *
from store_counts
where books = (
select max(books)
from store_counts)