I have a table PAYTB with transaction information. The table contains ACAUREQ_AUREQ_ENV_M_CMONNM which is the Common Merchant Name.
Now I want an output like this:
orders ACAUREQ_AUREQ_ENV_M_CMONNM
---------+---------+----------------
100 Antique Shop
30 Airleisure
23 Books
12 ....
How can I construct the "orders" column which is the count of all transaction with a certain common merchant name?
You need to group on ACAUREQ_AUREQ_ENV_M_CMONNM and find count of rows for each, then order by that count in descending order.
SELECT COUNT(*) orders,
ACAUREQ_AUREQ_ENV_M_CMONNM
FROM PAYTB
GROUP BY ACAUREQ_AUREQ_ENV_M_CMONNM
ORDER BY orders desc;
SELECT count(*) as orders , ACAUREQ_AUREQ_ENV_M_CMONNM
FROM PAYTB
GROUP BY ACAUREQ_AUREQ_ENV_M_CMONNM
ORDER BY 1 desc;
you want to use a group by:
select count(*) as orders, ACAUREQ_AUREQ_ENV_M_CMONNM
from PAYTB
group by ACAUREQ_AUREQ_ENV_M_CMONNM;
Order by count(*) desc to show descending counts in a group by expression.
SELECT count(*) as orders , ACAUREQ_AUREQ_ENV_M_CMONNM
FROM PAYTB
GROUP BY ACAUREQ_AUREQ_ENV_M_CMONNM
ORDER BY count(*) desc;
Related
Hello I am trying to group multiple customer orders into buckets in SQL, the output should look something like it does below. Do I have to use a case statement to group them?
Table1 looks like:
CustomerID
Order_date
1
somedate
2
somedate
3
somedate
2
somedate
Edit: # of customers meaning if CustomerID 2 had 2 orders he/she would be of the in the bucket of #of orders of 2.
Output should be something like this?
# of Customers
# of Orders
2
1
1
2
My code so far is:
select count(*) CustomerID
FROM Table1
GROUP BY CustomerID;
Use a double aggregation:
SELECT COUNT(*) AS num_customers, cnt AS num_orders
FROM
(
SELECT CustomerID, COUNT(*) AS cnt
FROM Table1
GROUP BY CustomerID
) t
GROUP BY cnt;
The inner subquery finds the number of orders for each customer. The outer query then aggregates by number of orders and finds out the number of customers having each number of orders.
If you want to sort your tables and your users depending on the number of orders they made, this query should work:
SELECT CustomerID, COUNT(CustomerID) as NbOrder
FROM Table1
GROUP BY(NbOrder)
I believe what you want to do is get the count of orders by customer, first, via aggregation. Then get the count of customers by order count from that query.
SELECT count(*) as count_of_customers, count_of_orders
FROM
(
SELECT customerid, count(*) as count_of_orders
FROM your_table
GROUP BY customerid
) sub
GROUP BY count_of_orders
ORDER BY count_of_orders
i want to list customer id with highest sum of price of his orders. Please see graph oí orders down.
SQL DEMO
SELECT *
FROM (
SELECT "customerid", SUM("price")
FROM Orders
GROUP BY "customerid"
ORDER BY SUM("price") DESC
) T
WHERE ROWNUM <= 1
You need to group by customer_id to get all prices of each customers.
Then sum these prices filter it with max( sum(price))or get first row by descending order of sum(price).
--for Oracle
select * from (Select c.name,c.id,sum(o.price) from Customer c
inner join order o on o.customer_id=c.id
group by c.name,c.id
order by sum(o.price)desc
)where rownum =1
--For sql server and mysql
Select top 1 c.name,c.id,sum(o.price) from Customer c
inner join order o on o.customer_id=c.id
group by c.name,c.id
order by sum(o.price)desc
I am wanting to list names and the number of times they have done a certain action. I then want to order the names by the most amount of times.
I have the below code so far but I keep getting errors:
select name, count(*) as NoOfTimes
from CustName
group by count(*);
order by count(*) asc;
I should note that if you want the most times at the beginning of the result set, the you want a descending sort:
select name, count(*) as NoOfTimes
from CustName
group by name
order by count(*) desc;
In order to show count by name, you must group by name
select name, count(*) as NoOfTimes
from CustName
group by name
order by NoOfTimes desc
Order by index also a good idea:
select name, count(*) as NoOfTimes
from CustName
group by name
order by 2 DESC
Assume I have these 3 tables :
The first 2 tables define customers of different types ,i.e second table has other columns which are not included in table 1 i just left them the same to save complexity.
The third table defines orders for both types of customers . Each customer has more than one orders
I want to select the last order for every customer, i.e the order with order_id 4 for customer 1 which was created on 23/12/2016 and the order with order_id 5 for customer 2 which was created on 26/12/2016
I tried something like this :
select *
from customertype1
left join order on order.customer_id = customertype1.customer_id
order by order_id desc;
But this gives me multiple records for every customer, as I have stated above I want only the last order for every customertype1.
If you want the last order for each customer, then you only need the orders table:
select o.*
from (select o.*,
row_number() over (partition by customer_id order by datecreated desc) as seqnum
from orders o
) o
where seqnum = 1;
If you want to include all customers, then you need to combine the two tables. Assuming they are mutually exclusive:
with c as (
select customer_id from customers1 union all
select customer_id from customers2
)
select o.*
from c left join
(select o.*,
row_number() over (partition by customer_id order by datecreated desc) as seqnum
from orders o
) o
on c.customer_id = o.customer_id and seqnum = 1;
A note about your data structure: You should have one table for all customers. You can then define a foreign key constraint between orders and customers. For the additional columns, you can have additional tables for the different types of customers.
Use ROW_NUMBER() and PARTITION BY.
ROW_NUMBER(): it will give sequence no to your each row
PARTITION BY: it will group your data by given column
When you use ROW_NUMBER() and PARTITION BY both together then first partition by group your records and then row_number give then sequence no by each group, so for each group you have start sequence from 1
Help Link: Example of ROW_NUMBER() and PARTITION BY
This is the general idea. You can work out the details.
with customers as
(select customer_id, customer_name
from table1
union
select customer_id, customer_name
from table2)
, lastOrder as
(select customer_id, max(order_id) maxOrderId
from orders
group by customer_id)
select *
from lastOrder join customers on lastOrder.Customer_id = customers.customer_id
join orders on order_id = maxOrderId
I have a set of records and I want to sort these records on the basis of the number of items in a group.
I want to arrange the records in such a way that Products with maximum number of items are at the top i.e. the required order is- Product_ID 3 (with 6 items), then Product_ID 1 (with 5 items) and the last one would be Product_ID 2(with 3 items).
The following query returns the count of the items with same Product_ID, however, I want Item_Name, Item_Description and Item_Number to be arranged as well.
Select Product_ID, Count(*) from Product group by Product_ID order by Count(*) DESC
I have tried another query as follows, but I know I am wrong somewhere that it is not giving the desired results and I can't think of a possible solution:
Select Product_ID, Item_Name, Item_Description, Item_Number from Product
group by Product_ID,item_name,item_description,item_number
order by COUNT(product_ID)
Thanks in advance for your help!!
Select Product_ID, Item_Name, Item_Description, Item_Number
from Product
order by COUNT(1) over (partition by Product_ID) desc
I assume you want to group by the ID only but you want to list all other fields, you don't need to group by at all if you just want to order by:
SELECT product_id,
item_name,
item_description,
item_number
FROM product p1
ORDER BY (SELECT Count(product_id)
FROM product p2
WHERE p1.product_id = p2.product_id) DESC
Try using an alias:
Select Product_ID, Count(*) AS num_products from Product group by Product_ID order by num_products DESC;