Is there a way to change "LIMIT 1" and get the same output? I have to get client's name, surname and a quantity of books that has the most books
SELECT stud.skaitytojas.name, stud.skaitytojas.surname,
COUNT (stud.skaitytojas.nr) AS quantity
FROM stud.egzempliorius , stud.skaitytojas
WHERE stud.egzempliorius.client = stud.skaitytojas.nr
GROUP BY stud.skaitytojas.nr
ORDER BY quantity DESC
LIMIT 1
Postgres supports the ANSI standard FETCH FIRST 1 ROW ONLY, so you can do:
SELECT s.name, s.surname, COUNT(s.nr) AS quantity
FROM stud.egzempliorius e JOIN
stud.skaitytojas s
ON e.client = s.nr
GROUP BY s.name, s.surname
ORDER BY quantity DESC
FETCH FIRST 1 ROW ONLY;
Also notice the use of table aliases and proper JOIN syntax. I also prefer to list the columns in the SELECT in the GROUP BY, although that is optional if s.nr is unique.
You can select the row with the highest quantity using row_number()
SELECT * FROM (
SELECT * , row_number() over (order by quantity desc) rn FROM (
SELECT stud.skaitytojas.name, stud.skaitytojas.surname, COUNT (stud.skaitytojas.nr) AS quantity
FROM stud.egzempliorius , stud.skaitytojas
WHERE stud.egzempliorius.client = stud.skaitytojas.nr
GROUP BY stud.skaitytojas.name, stud.skaitytojas.surname
) t
) t where rn = 1
If you want to include ties for the highest quantity, then use rank() instead.
Related
I am using SQL Server and I have a table "a"
month segment_id price
-----------------------------
1 1 100
1 2 200
2 3 50
2 4 80
3 5 10
I want to make a query which presents the original columns where the price will be the max per month
The result should be:
month segment_id price
----------------------------
1 2 200
2 4 80
3 5 10
I tried to write SQL code:
Select
month, segment_id, max(price) as MaxPrice
from
a
but I got an error:
Column segment_id is invalid in the select list because it is not contained in either an aggregate function or the GROUP BY clause
I tried to fix it in many ways but didn't find how to fix it
Because you need a group by clause without segment_id
Select month, max(price) as MaxPrice
from a
Group By month
as you want results per each month, and segment_id is non-aggregated in your original select statement.
If you want to have segment_id with maximum price repeating per each month for each row, you need to use max() function as window analytic function without Group by clause
Select month, segment_id,
max(price) over ( partition by month order by segment_id ) as MaxPrice
from a
Edit (due to your lastly edited desired results) : you need one more window analytic function row_number() as #Gordon already mentioned:
Select month, segment_id, price From
(
Select a.*,
row_number() over ( partition by month order by price desc ) as Rn
from a
) q
Where rn = 1
I would recommend a correlated subquery:
select t.*
from t
where t.price = (select max(t2.price) from t t2 where t2.month = t.month);
The "canonical" solution is to use row_number():
select t.*
from (select t.*,
row_number() over (partition by month order by price desc) as seqnum
from t
) t
where seqnum = 1;
With the right indexes, the correlated subquery often performs better.
Only because it was not mentioned.
Yet another option is the WITH TIES clause.
To be clear, the approach by Gordon and Barbaros would be a nudge more performant, but this technique does not require or generate an extra column.
Select Top 1 with ties *
From YourTable
Order By row_number() over (partition by month order by price desc)
With not exists:
select t.*
from tablename t
where not exists (
select 1 from tablename
where month = t.month and price > t.price
)
or:
select t.*
from tablename inner join (
select month, max(price) as price
from tablename
group By month
) g on g.month = t.month and g.price = t.price
I'm looking for a way to let me select all orders that have multiple distinct names within the same order-number, it looks like this:
order - name
111-Paul
112-Paula
113-John
113-John
113-Jessica
114-Eric
114-Eric
114-John
115-Zack
115-Zack
115-Zack
etc.
so that i would get all the orders that have 2 or more distinct names in it:
113-John
113-Jessica
114-Eric
114-John
with which I could do further queries but I'm stuck. Can anyone give me some hints on how to tackle this problem please? I've tried it with count(*) which looked like this:
select order, name, count(name) from dbo.orders
group by order, name
having count(name) > 1
which gave me all the orders which had more than 1 name in it but I don't know how to let it only show orders with the distinct names.
Here's one approach using exists:
select distinct [order], name
from orders o
where exists (
select 1
from orders o2
where o.[order] = o2.[order] and o.name != o2.name)
Fiddle Demo
I would use windows functions for this
For example:
select distinct order
from
(select
order,
row_number() over(partition by order, name order by order asc) as rn
) as t1
where rn > 1
you can do the same with count
count(*) over(partition by order,name order by order asc) as cnt
Here's a straight forward implementation in Sql Server:
select distinct *
from table1
where [order] in (
select [order]
from (select distinct * from table1) iq
group by [order]
having count(*) > 1)
It's essentially breaking down the problem into:
Finding the orders that have more than one distinct value.
Finding the pairs of distinct order - name that belong to the list previously calculated.
When you use HAVING COUNT(name) > 1, it is counting all of the rows in those groups, including duplicate rows (rows 113-John and 113-John are 2 rows for order 113). I would query distinct rows from your table, and then select from that:
SELECT [order], [name] FROM (
SELECT DISTINCT [order], [name] FROM dbo.orders
) A
GROUP BY [order], [name]
HAVING COUNT([name]) > 1
As a note, if a [name] is null, then it will not be counted with COUNT(name). If you want nulls to be counted, use COUNT(*) instead.
You can use count(distinct name) to get the number of unique names for each order:
select [order], count(distinct name)
from orders
group by [order]
To just get the order for those you can use having:
select [order]
from orders
group by [order]
having count(distinct name) > 1
To get the details for those orders you can put that in the where clause to just return the rows with order in that list:
select *
from orders
where [order] in (
select [order]
from orders
group by [order]
having count(distinct name) > 1
)
sqlfiddle
I would use RANK (or DENSE_RANK) for this as shown below.
SELECT [Order]
FROM (SELECT
[Order],
RANK() OVER(PARTITION BY [Order] ORDER BY Name) AS NameRank
FROM [StackOverflow].[dbo].[OrderAndName]) ranked
WHERE ranked.NameRank > 1
GROUP BY [Order]
The sub-query ranks (gives a seeding) to the names in an order according to their value. Names with the same value would have the same rank i.e. when an order has one name multiple times (like 115) the rank of all names would be 1.
The partition is important here as otherwise you would get the rank for all names for all orders which wouldn't give you the result you'd like.
It is then just a case of pulling out the orders that have a RANK greater than 1 and grouping (could use distinct if that's a preference).
You can then join to this table to get get the orders and names as follows;
SELECT oan.[Order], [Name]
FROM [StackOverflow].[dbo].[OrderAndName] oan
INNER JOIN (SELECT [Order]
FROM (SELECT [Order],
RANK() OVER(PARTITION BY [Order] ORDER BY Name) AS NameRank
FROM [StackOverflow].[dbo].[OrderAndName]) ranked
WHERE ranked.NameRank > 1
GROUP BY [Order]) twoOrMore ON oan.[Order] = twoOrMore.[Order]
The results are below. I need to get the records (seller and purchaser) with the max count- grouped by purchaser (marked with yellow)
You can use window functions:
with q as (
<your query here>
)
select q.*
from (select q.*,
row_number() over (order by seller desc) as seqnum_s,
row_number() over (order by purchaser desc) as seqnum_p
from q
) q
where seqnum_s = 1 or seqnum_p = 1;
Try this:
SELECT COUNT,seller,purchaser FROM YourTable ORDER BY seller,purchaser DESC
SELECT T2.MaxCount,T2.purchaser,T1.Seller FROM <Yourtable> T1
Inner JOIN
(
Select Max(Count) as MaxCount, purchaser
FROM <Yourtable>
GROUP BY Purchaser
)T2
On T2.Purchaser=T1.Purchaser AND T2.MaxCount=T1.Count
First you select the Seller from which will give you a list of all 5 sellers. Then you write another query where you select only the Purchaser and the Max(count) grouped by Purchaser which will give you the two yellow-marked lines. Join the two queries on fields Purchaser and Max(Count) and add the columns from the joined table to your first query.
I can't think of a faster way but this works pretty fast even with rather large queries. You can further-by order the fields as needed.
Group by the highest Number in a column worked great with MAX(), but what if I would like to get the cell that is at most common.
As example:
ID
100
250
250
300
200
250
So I would like to group by ID and instead of get the lowest (MIN) or highest (MAX) number, I would like to get the most common one (that would be 250, because there 3x).
Is there an easy way in SQL Server 2012 or am I forced to add a second SELECT where I COUNT(DISTINCT ID) and add that somehow to my first SELECT statement?
You can use dense_rank to return all the id's with the highest counts. This would handle cases when there are ties for the highest counts as well.
select id from
(select id, dense_rank() over(order by count(*) desc) as rnk from tablename group by id) t
where rnk = 1
A simple way to do what you want uses top and order by:
SELECT top 1 id
FROM t
GROUP BY id
ORDER BY COUNT(*) DESC;
This is a statistic called the mode. Getting the mode and max is a bit challenging in SQL Server. I would approach it as:
WITH cte AS (
SELECT t.id, COUNT(*) AS cnt,
row_number() OVER (ORDER BY COUNT(*) DESC) AS seqnum
FROM t
GROUP BY id
)
SELECT MAX(id) AS themax, MAX(CASE WHEN seqnum = 1 THEN id END) AS MODE
FROM cte;
I have a simple assignment but I am stuck, I have a table and need to print out the ID with the maximum total of sales. I have managed to print a sorted list of the IDs based on the sum of each one's sales:
SELECT "COMPANY"."ID", SUM("COMPANY"."PRICE") As PriceSum
FROM "COMPANY"
WHERE "COMPANY"."DATEOFSALE" >= DATE '2016-01-01'
GROUP BY "COMPANY"."ID"
ORDER BY PriceSum DESC;
I just want to show the ID and the total sales of the top selling company.
TIA
This is in Oracle, so I can't be cheap and use LIMIT 1.
You can use a subquery instead:
SELECT c.*
FROM (SELECT "COMPANY"."ID", SUM("COMPANY"."PRICE") As PriceSum
FROM "COMPANY"
WHERE "COMPANY"."DATEOFSALE" >= DATE '2016-01-01'
GROUP BY "COMPANY"."ID"
ORDER BY PriceSum DESC
) c
WHERE rownum = 1;
In Oracle 12c+, you can use FETCH FIRST 1 ROW ONLY without the subquery. This is the ANSI standard equivalent of LIMIT.
EDIT:
If you want all companies with the maximum, use rank() or dense_rank():
SELECT c.*
FROM (SELECT "COMPANY"."ID", SUM("COMPANY"."PRICE") As PriceSum,
RANK() OVER (ORDER BY SUM("COMPANY"."PRICE") DESC) as seqnum
FROM "COMPANY"
WHERE "COMPANY"."DATEOFSALE" >= DATE '2016-01-01'
GROUP BY "COMPANY"."ID"
ORDER BY PriceSum DESC
) c
WHERE seqnum = 1;
You can replace RANK() with ROW_NUMBER() and get the previous result as well.