How can I query my SQL database and find all the duplicate orders where a customer ordered something more than once?
Not a super clear question, but I get the gist of it. I don't know what your database looks like, but your query would look something like this:
SELECT customer_id, count(*) FROM orders
GROUP BY customer_id
HAVING count(*) > 1
SELECT customerId, productId, count(productId)
FROM CustomerOrders
GROUP BY customerId
HAVING count(productId) > 1
A platform, version would help. So would a sample table. So here is a sample orders table:
SQL> select * from orders;
CUST_ID ORDER_ID ORDER_DAT
---------- ---------- ---------
1 1 25-FEB-09
1 2 24-FEB-09
1 3 23-FEB-09
2 4 24-FEB-09
2 5 23-FEB-09
2 6 22-FEB-09
3 7 23-FEB-09
9 8 22-FEB-09
9 9 21-FEB-09
4 10 22-FEB-09
4 11 21-FEB-09
4 12 20-FEB-09
5 13 21-FEB-09
5 14 20-FEB-09
5 15 19-FEB-09
6 16 20-FEB-09
11 17 19-FEB-09
10 18 18-FEB-09
7 19 19-FEB-09
7 20 18-FEB-09
7 21 17-FEB-09
8 22 18-FEB-09
8 23 17-FEB-09
8 24 16-FEB-09
24 rows selected.
The following select works on this table for a simple output:
1* select cust_id, count(*) from orders group by cust_id having count(*) > 1
SQL> /
CUST_ID COUNT(*)
---------- ----------
1 3
2 3
4 3
5 3
8 3
7 3
9 2
7 rows selected.
You can use a Select Distinct for it.
Related
Hi Experts I have a table like this
T1
Order_no
Qty
1
3
2
5
3
1
4
3
I need to generate a column 'serial no' having values based on 'qty'
Output needed
OrderNo
Qty
SerailNo
1
3
1
1
3
2
1
3
3
2
5
1
2
5
2
2
5
3
2
5
4
2
5
5
3
1
1
4
3
1
4
3
2
4
3
3
Any suggestions?
Thanks in advance!!
You don't mention the specific database so I'll assume you are using PostgreSQL, aren't you?
You can use a Recursive CTE to expand the rows. For example:
with recursive
n as (
select order_no, qty, 1 as serial_no from t1
union all
select order_no, qty, serial_no + 1
from n
where serial_no < qty
)
select * from n order by order_no, serial_no
Result:
order_no qty serial_no
--------- ---- ---------
1 3 1
1 3 2
1 3 3
2 5 1
2 5 2
2 5 3
2 5 4
2 5 5
3 1 1
4 3 1
4 3 2
4 3 3
See running example at DB Fiddle.
EDIT FOR ORACLE
If you are using Oracle the query changes a bit to:
with
n (order_no, qty, serial_no) as (
select order_no, qty, 1 from t1
union all
select order_no, qty, serial_no + 1
from n
where serial_no < qty
)
select * from n order by order_no, serial_no
Result:
ORDER_NO QTY SERIAL_NO
--------- ---- ---------
1 3 1
1 3 2
1 3 3
2 5 1
2 5 2
2 5 3
2 5 4
2 5 5
3 1 1
4 3 1
4 3 2
4 3 3
See running example at db<>fiddle.
You should first provide the database you're using. Whether it's oracle, Sql Server, PostGreSQL will determine which procedural language to use. It's very likely that you'll need to do this in two steps:
1st: Duplicate the number of rows based on the column Qty using a decreasing loop
2nd: You'll need to create a sequential partionned column based on the Qty column
I have the following table
Cash_table
ID Cash Rates Amount
1 50 3 16
2 100 4 25
3 130 10 7
3 130 10 6
4 13 7 1.8
5 30 8 2.5
5 30 10 1
6 10 5 2
What I want as a result is to cumulate all the entries that have a Count(id)>1 like this:
ID New_Cash New_Rates New_Amount
1 50 3 16
2 100 4 25
3 130 10+10 130/(10+10)
4 13 7 1.8
5 30 8+10 30/(8+10)
6 10 5 2
So I only want to change the rows where Count(id)>1 and leave the rest like it was.
For the rows with count(id)>1 I want to sum up the rates and take the cash and divide it by the sum of the rates. The Rates alone aren't a problem since I can sum them up and group by id and get the desired result.
The problem is with the New_Amount column:
I am trying to do it with a case statement but it isn't working:
select id,
cash as new_cash,
sum(rates) as new_rates,
(case count(id)
when 1 then amount
else cash/sum(nvl(rates,null))
end) as new_amount
from Cash_table
group by id
As the cash value is always the same for an ID, you can group by that as well:
select id,
cash as new_cash,
sum(rates) as new_rates,
case count(id)
when 1 then max(amount)
else cash/sum(rates)
end as new_amount
from cash_table
group by id, cash
order by id
ID NEW_CASH NEW_RATES NEW_AMOUNT
---------- ---------- ---------- ----------
1 50 3 16
2 100 4 25
3 130 20 6.5
4 13 7 1.8
5 30 18 1.66666667
6 10 5 2
The first branch of the case expression needs an aggregate because you aren't grouping by amount; and the sum(nvl(rates,null)) can just be sum(rates). If you're expecting any null rates then you need to decide how you want the amount to be handled, but nvl(rates,null) isn't doing anything.
You can do the same thing without a case expression if you prefer, manipulating all the values - which might be more expensive:
select id,
cash as new_cash,
sum(rates) as new_rates,
sum(amount * rates)/sum(rates) as new_amount
from cash_table
group by id, cash
order by id
I have data that looks like this, where there are multiple values for each ID that correspond to an ascending date variable:
ID LEVEL DATE
1 10 10/1/2000
1 10 11/20/2001
1 10 12/01/2001
1 30 02/15/2002
1 30 02/15/2002
1 20 05/17/2002
1 20 01/04/2003
1 30 07/20/2003
1 30 03/16/2004
1 30 04/15/2004
I want to acquire a count per each ID/LEVEL/DATE block that looks like this:
ID LEVEL COUNT
1 10 3
1 30 2
1 20 2
1 30 3
The problem is that if I use the count windows function and partition by level, it groups 30 together regardless of the temporal sequence. I want the count for level 30 both before and after 20 to be distinct. Does anyone know how to do that?
A standard gaps and islands solution using ROW_NUMBER(), if it's available on your particular DBMS...
WITH
ordered AS
(
SELECT
*,
ROW_NUMBER() OVER (PARTITION BY id ORDER BY date) AS set_ordinal,
ROW_NUMBER() OVER (PARTITION BY id, level ORDER BY date) AS grp_ordinal
FROM
yourData
)
SELECT
id,
level,
set_ordinal - grp_ordinal,
MIN(date),
COUNT(*)
FROM
ordered
GROUP BY
id,
level,
set_ordinal - grp_ordinal
ORDER BY
id,
MIN(date)
Visualising the effect of the two row numbers...
ID LEVEL DATE set_ordinal grp_ordinal set-grp GROUP
-- ----- ---------- ----------- ----------- ------- --------
1 10 10/01/2000 1 1 0 1,10,0
1 10 11/20/2001 2 2 0 1,10,0
1 10 12/01/2001 3 3 0 1,10,0
1 30 02/15/2002 4 1 3 1,30,3
1 30 02/15/2002 5 2 3 1,30,3
1 20 05/17/2002 6 1 5 1,20,5
1 20 01/04/2003 7 2 5 1,20,5
1 30 07/20/2003 8 3 5 1,30,5
1 30 03/16/2004 9 4 5 1,30,5
1 30 04/15/2004 10 5 5 1,30,5
I have table like this:
ID COUNT
-----------
7 2
7 2
8 3
8 3
9 4
9 4
And I want to summarise COUNT with same ID to have
ID COUNT
-----------
7 4
8 6
9 8
How to get it in oracle db?
EDIT:
I get my table from pipelined function with this type:
TYPE g_item IS RECORD (
g_id NUMBER,
g_count VARCHAR2(50),
g_who VARCHAR2(50)
);
Code to get correct table:
SQL> select g_id, sum(g_count)
from (SELECT * from TABLE(my_package.get_items_tab ()))
group by g_id;
G_ID SUM(G_COUNT)
----- ------------
1 9
6 7
2 7
4 7
5 7
3 7
7 7
7 rows selected.
EDIT2:
Oh.. it's my fault, I must group it by g_who, not by g_id
select id, sum(count)
from your_table
group by id
This is a simple group by operation:
select id,
sum(count) as total_count
from the_table
group by id
order by id;
Btw: it's a bad habit to use reserved words (count) as column names.
I have a table which rank the items which i have.
I need a queries which will pick up only the top 2 ranks for a given item, the rank may not be in sequential order.
I need to fetch the item with least two ranks, there will same rank for two items as well.
Here is the snap shot of my table.
Item Id Supp Id Rank
1 2 2
1 1 7
1 7 5
1 9 11
2 67 4
2 9 14
2 10 14
2 34 4
2 25 3
2 60 3
2 79 5
my requirement is if I enter 2 i should get the result as below
Item Id Supp_id Rank
2 25 3
2 60 3
2 67 4
2 34 4
I am using oracle 10g version.
As one of the approaches it can be done as follows. Here we are using dense_rank() over() analytic function to assign a rank for a row in a ordered by rank group of rows .
select t.item_id
, t.supp_id
, t.rank
from (select item_id
, supp_id
, rank
, dense_rank() over(partition by item_id
order by rank) as rn
from t1
where item_id = 2
) t
where t.rn <= 2
Result:
ITEM_ID SUPP_ID RANK
---------- ---------- ----------
2 25 3
2 60 3
2 67 4
2 34 4
SQLFiddle Demo