I have this table:
id sale_id price
-------------------
1 1 100
2 1 200
3 2 50
4 3 50
I want this result:
id sale_id price sum(price by sale_id)
------------------------------------------
1 1 100 300
2 1 200 300
3 2 50 50
4 3 50 50
I tried this:
SELECT id, sale_id, price,
(SELECT sum(price) FROM sale_lines GROUP BY sale_id)
FROM sale_lines
But get the error that subquery returns different number of rows.
How can I do it?
I want all the rows of sale_lines table selecting all fields and adding the sum(price) grouped by sale_id.
You can use window function :
sum(price) over (partition by sale_id) as sum
If you want sub-query then you need to correlate them :
SELECT sl.id, sl.sale_id, sl.price,
(SELECT sum(sll.price)
FROM sale_lines sll
WHERE sl.sale_id = sll.sale_id
)
FROM sale_lines sl;
Don't use GROUP BY in the sub-query, make it a co-related sub-query:
SELECT sl1.id, sl1.sale_id, sl1.price,
(SELECT sum(sl2.price) FROM sale_lines sl2 where sl2.sale_id = sl.sale_id) as total
FROM sale_lines sl1
In addition to other approaches, You can use CROSS APPLY and get the sum.
SELECT id, sale_id,price, Price_Sum
FROM YourTable AS ot
CROSS APPLY
(SELECT SUM(price) AS Price_Sum
FROM YourTable
WHERE sale_id = ot.sale_id);
SELECT t1.*,
total_price
FROM `sale_lines` AS t1
JOIN(SELECT Sum(price) AS total_price,
sale_id
FROM sale_lines
GROUP BY sale_id) AS t2
ON t1.sale_id = t2.sale_id
Related
Given a table products
pid
name
123
Milk
456
Tea
789
Cake
...
...
and a table sales
stamp
pid
units
14:54
123
3
15:02
123
9
15:09
456
1
15:14
456
1
15:39
456
2
15:48
789
12
...
...
...
How would I be able to get the product(s) with the most sold units?
My goal is to run a SELECT statement that results in, for this example,
pid
name
123
Milk
789
Cake
because the sum of sold units of both those products is 12, the maximum value (greater than 4 for Tea, despite there being more sales for Tea).
I have the following query:
SELECT DISTINCT products.pid, products.name
FROM sales
INNER JOIN products ON sale.pid = products.pid
INNER JOIN (
SELECT pid, SUM(units) as sum_units
FROM sales
GROUP BY pid
) AS total_units ON total_units.pid = sales.pid
WHERE total_units.sum_units IN (
SELECT MAX(sum_units) as max_units
FROM (
SELECT pid, SUM(units) as sum_units
FROM sales
GROUP BY pid
) AS total_units
);
However, this seems very long, confusing, and inefficient, even repeating the sub-query to obtain total_units, so I was wondering if there was a better way to accomplish this.
How can I simplify this? Note that I can't use ORDER BY SUM(units) LIMIT 1 in case there are multiple (i.e., >1) products with the most units sold.
Thank you in advance.
Since Postgres 13 it has supported with ties so your query can be simply this:
select p.pId, p.name
from sales s
join products p on p.pid = s.pid
group by p.pId, p.name
order by Sum(units) desc
fetch first 1 rows with ties;
See demo Fiddle
Solution for your problem:
WITH cte1 AS
(
SELECT s.pid, p.name,
SUM(units) as total_units
FROM sales s
INNER JOIN products p
ON s.pid = p.pid
GROUP BY s.pid, p.name
),
cte2 AS
(
SELECT *,
DENSE_RANK() OVER(ORDER BY total_units DESC) as rn
FROM cte1
)
SELECT pid,name
FROM cte2
WHERE rn = 1
ORDER BY pid;
Working example: db_fiddle link
I have the following table named customerOrders.
ID user order
1 1 2
2 1 3
3 1 1
4 2 1
5 1 5
6 2 4
7 3 1
8 6 2
9 2 2
10 2 3
I want to return to users with most orders. Currently, I have the following QUERY:
SELECT user, COUNT(user) AS UsersWithMostOrders
FROM customerOrders
GROUP BY user
ORDER BY UsersWithMostOrders DESC;
This returns me all the values grouped by total orders like.
user UsersWithMostOrders
1 4
2 4
3 1
6 1
I only want to return the users with most orders. In my case that would be user 1 and 2 since both of them have 4 orders. If I use TOP 1 or LIMIT, it will only return the first user. If I use TOP 2, it will only work in this scenario, it will return invalid data when top two users have different count of orders.
Required Result
user UsersWithMostOrders
1 4
2 4
You can use TOP 1 WITH TIES:
SELECT TOP 1 WITH TIES
[user], COUNT(*) AS UsersWithMostOrders
FROM customerOrders
GROUP BY [user]
ORDER BY UsersWithMostOrders DESC;
See the demo.
Results:
> user | UsersWithMostOrders
> ---: | ------------------:
> 1 | 4
> 2 | 4
Option 1
Should work with most versions of SQL.
select *
from (
select *,
rank() over(order by numOrders desc) as rrank
from (
select `user`, count(*) as numOrders
from customerOrders
group by `user`
) summed
) ranked
where rrank = 1
Play around with the code here
Option 2
If your version of SQL allows window functions (with), here is a much more readable solution which does the same thing
with summed as (
select `user`, count(*) as numOrders
from customerOrders
group by `user`
),
ranked as (
select *,
rank() over(order by numOrders desc) as rrank
from summed
)
select *
from ranked
where rrank = 1
Play around with the code here
You can use a CTE to attain this Req:
;WITH CTE AS(
SELECT [user], COUNT(user) AS UsersWithMostOrders
FROM #T
GROUP BY [user])
SELECT M.* from CTE M
INNER JOIN ( SELECT
MAX(UsersWithMostOrders) AS MaximumOrders FROM CTE) S ON
M.UsersWithMostOrders=S.MaximumOrders
Below Oracle Query can help:
WITH test_table AS
(
SELECT user, COUNT(order) AS total_order , DENSE_RANK() OVER (ORDER BY
total_order desc) AS rank_orders FROM customerOrders
GROUP BY user
)
select * from test_table where rank_orders = 1
ID Name Price
--------------------------------------
1 item1 10
2 item2 40
3 item3 10
4 item4 20
5 item5 50
6 item6 20
Say you had this table above and wanted to retrieve the following:
ID Name Price
5 item5 50
But you wanted to retrieve the above by using the highest price. I am currently using the below code.
SELECT
MAX(price) AS Price,
Name,
ID
FROM
ExampleTable
GROUP BY
Name, ID;
In SQL Server we can try:
SELECT TOP 1 *
FROM yourTable
ORDER BY Price DESC;
If there could be more than one record tied for the highest price, and you also wanted to report all ties, then we could use WITH TIES:
SELECT TOP 1 WITH TIES *
FROM yourTable
ORDER BY Price DESC;
If you want to use TOP to select only certain columns, then just list those columns out, e.g.
SELECT TOP 1 ID, Price
to select only the ID and Price columns.
use top as your DBMS is sql server
select top 1 * from your_table
order by Price desc
You could also use window function
with t1 as
(
select * , row_number() over(order by Price desc) as rn from your_table
) select ID ,Name ,Price from t1 where rn=1
I am trying to build a query results with SQL. Here is my table:
CUST_ID ORDER_ID STORE_FREQUENCY
---------- ----------- ---------------
100 20122 500
100 20100 500
100 20100 737
200 20119 287
300 20130 434
300 20150 434
300 20130 434
300 20120 120
The expected output is:
CUST_ID UNIQUE_ORDERS TOP_STORE
--------- ----------------- ---------
100 2 737
200 1 287
300 3 434
The requirement for the output is:
TOP_STORE = Per CUST_ID, sort the STORE_FREQUENCY column by DESC and get the greatest store frequency
UNIQUE_ORDERS = Per CUST_ID, the number of unique ORDER_IDs in the column
I have started this SELECT statement, but having difficulties completing it to include the 2 columns correctly:
Select cust_id, Count(order_id) as unique_orders
From ORDERS_TABLE
Group By Order_ID
Can you help me complete the 2 columns?
Use aggregate functions such as COUNT(DISTINCT ...) and MAX()
SELECT CUST_ID, COUNT(DISTINCT ORDER_ID), MAX(STORE_FREQUENCY )
FROM TableName
GROUP BY CUST_ID
Here's a DEMO.
It seems to be that the top store should be the store with the greatest number of orders. If so, then CUST_ID 100 should have store 500 as the top store, not 737. In other words, I would expect the following output:
This requirement changes the query strategy, because we no longer can just do a single simple aggregation over the entire table. One approach is to do a separate calculation to find the top store for each customer, then join that result to a query similar to the other answers.
WITH cte AS (
SELECT CUST_ID, STORE_FREQUENCY, cnt,
ROW_NUMBER() OVER (PARTITION BY CUST_ID ORDER BY cnt DESC) rn
FROM
(
SELECT CUST_ID, STORE_FREQUENCY,
COUNT(*) OVER (PARTITION BY CUST_ID, STORE_FREQUENCY) cnt
FROM yourTable
) t
)
SELECT
t1.CUST_ID,
t1.UNIQUE_ORDERS,
t2.TOP_STORE
FROM
(
SELECT CUST_ID, COUNT(DISTINCT ORDER_ID) AS UNIQUE_ORDERS
FROM yourTable
GROUP BY CUST_ID
) t1
INNER JOIN
(
SELECT CUST_ID, STORE_FREQUENCY AS TOP_STORE
FROM cte
WHERE rn = 1
) t2
ON t1.CUST_ID = t2.CUST_ID;
Demo
i have 3 tables
Buyer
buyer_id | name
50 |Joe
60 |Astor
70 |Cloe
Item
item_id | description
1 | iphone
2 | ipod
3 | imac
Item_Sold
buyer_id | item_id
50 | 1
50 | 2
60 | 1
60 | 3
70 | 1
70 | 2
70 | 3
I want to find out the description of the best-selling item, in this case:
Best-Selling
iphone
SELECT description AS Best_Selling
FROM item
WHERE item_id = (SELECT item_id FROM( SELECT item_id ,COUNT(*) as num_items
FROM Item_Sold
GROUP BY item_id
ORDER BY num_items DESC
LIMIT 1
)z
)
See SQL FIDDLE
This answer is not totally correct . If two items have same sale amount then it will return only one of them.
This query will give all item id decription whose sale is maximum i.e. when two or more item id have equal amount of sale....
;WITH CTE1(Id,Counts) as
(
SelectItem_Id,COUNT(buyer_id ) AS C FROM T GROUP BY ID
)
Select Item.Description from CTE1 A inner join
(Select MAX(Counts) AS MaxCount FROM CTE1 ) b on a.Counts=b.MaxCount
inner join
Item on Item.Item_Id=a.Item_Id
If Common table Expression Not Work you Can Try Like this....
Select Item.Description from (Select Item_Id,COUNT(buyer_id ) AS Counts FROM item_sold GROUP BY Item_Id) A inner join
(Select MAX(Counts) AS MaxCount FROM
(
Select Item_Id,COUNT(buyer_id) AS Counts
FROM item_sold GROUP BY Item_Id) v
) b
on a.Counts=b.MaxCount
inner join
Item on Item.Item_Id=a.Item_Id
SQL Fiddle Demo
Here Is the Liknk of Fiddle the case i m talknig about....it give all description who have maximun sale....
Case Sql Fiddle Demo
select description as "Best-Selling"
from (select a.item_id, b.description, count(*) count
from Item_Sold a,Items b
where a.item_id = b.item_id
group by a.item_id ) temp
where count = (select max(count)
from (select a.item_id, count(*) count
from Item_Sold a,Items b
where a.item_id = b.item_id
group by a.item_id ) temp1)
pl-sql:
select description as "Best-Selling"
from item
where item_id in (
select item_id from (
select item_id, count(item_id) as item_count
from item_sold
group by item_id)
where item_count = (
select max(item_count) from (
select item_id, count(item_id) as item_count
from item_sold
group by item_id)
)
)