How can I find the sum of the lowest five points in the Point column
and group by ID
Table
The desired results should be;
Results
No idea where to start
Thanks
select a.ID, SUM(a.points) from(select ID , points,row_number() over
(partition by ID order by POINTS) as rownum_returned from your_table) a where
a.rownum_returned<6 group by a.ID;
Read about row_number() function here
If I've to do that I would solve it with a subquery.
In SQL server I will do subquery that retrieve the 5 lower point
Select Top 5 id, point from table
Order by point asc
Note: the keyword TOP that limit the result to the first 5
Note 2: order by point asc will order the result putting in top the lowest value
Now I use the query as subquery to complete the activity
Select id, sum (point) from
(Select top 5 id,point from table order by point asc) group by id
This should work
Related
I'm trying to make an SQL query that returns the greatest number from a column and its respective id.
For more information I have two columns ID and NUMBER. Both of them have 2 entries and I want to get the highest number with the ID next to it. This is what I tried but didn't success.
SELECT ID, MAX(NUMBER) AS MAXNUMB
FROM TABLE1
GROUP BY ID, MAXNUMB;
The problem I'm experiencing is that it just shows ALL the entries and if I add a "where" expression it just shows the same (all entries [ids+numbers]).
Pd.: Yes, I got what I wanted but only with one column (number) if I add another column (ID) to select it "brokes".
Try:
SELECT
ID,
A_NUMBER
FROM TABLE1
WHERE A_NUMBER = (
SELECT MAX(A_NUMBER)
FROM TABLE1);
Presuming you want the IDs* of the row with the highest number (and not, instead, the highest number for each ID -- if IDs were not unique in your table, for example).
* there may be more than one ID returned if there are two or more IDs with equal maximum numbers
you can try this
Select ID,maxNumber
From
(
SELECT
ID,
(Select Max(NUMBER) from Tmp where Id = t.Id) maxNumber
FROM
Tmp t
)T1
Group By ID,maxNumber
The query you posted has an illegal column name (number) and is group by the alias for the max value, which is illegal and also doesn't make sense; and you can't include the unaliased max() within the group-by either. So it's likely you're actually doing something like:
select id, max(numb) as maxnumb
from table1
group by id;
which will give one row per ID, with the maximum numb (which is the new name I've made up for your numeric column) for each ID. Or as you said you get "ALL the entries" you might have group by id, numb, which would show all rows from the table (unless there are duplicate combinations).
To get the maximum numb and the corresponding id you could group by id only, order by descending maxnumb, and then return the first row only:
select id, max(numb) as maxnumb
from table1
group by id
order by maxnumb desc
fetch first 1 row only
If there are two ID with the same maxnumb then you would only get one of them - and which one is indeterminate unless you modify the order by - but in that case you might prefer to use first 1 row with ties to see them all.
You could achieve the same thing with a subquery and analytic function to generating a ranking, and have the outer query return the highest-ranking row(s):
select id, numb as maxnumb
from (
select id, numb, dense_rank() over (order by numb desc) as rnk
from table1
)
where rnk = 1
You could also use keep to get the same result as first 1 row only:
select max(id) keep (dense_rank last order by numb) as id, max(numb) as maxnumb
from table1
fiddle
I'm trying to select the nth row of a table in MS Access (Office 365). I've seen both of the following solutions:
https://stackoverflow.com/a/45031166/1907765
https://stackoverflow.com/a/44891583/1907765
And neither of them have worked for me. When I wrote a query based on these answers, the query returned the last n rows in a table, and then selected the first result in that. E.g. if I was looking to select the 3rd row, it would select the 3rd-to-last row. Here's my query:
SELECT TOP 1 Sense.SenseID
FROM
(
SELECT TOP 3 Sense.SenseID
FROM Sense
ORDER BY Sense.SenseID DESC
)
ORDER BY Sense.SenseID ASC
Any idea what I'm doing wrong, and how to generate the correct result?
The order bys should be reversed:
SELECT TOP 1
s.SenseID
FROM
(SELECT TOP 3 s.SenseID
FROM Sense AS
ORDER BY s.SenseID Asc) AS s
ORDER BY
s.SenseID Desc;
You need a table alias so the syntax is correct. Try this:
SELECT TOP 1 s.SenseID
FROM (SELECT TOP 3 s.SenseID
FROM Sense as s
ORDER BY s.SenseID DESC
) as s
ORDER BY s.SenseID ASC;
This assumes that Sense.SenseID is unique -- but that seems like a reasonable assumption.
I want to use a query, showing the top two best Quantity. If the table is like the picture, how can the desired result be produced
You can use DENSE_RANK(). For example:
select
id, name, quantity
from (
select
id, name, quantity,
dense_rank() over(order by quantity desc) as rk
from t
) x
where rk <= 2
DENSE_RANK() computes a number for each row according to an ordering of your choosing. Identical values get the same number, and no numbers are skipped. See SQL Fiddle.
You can use the TOP/LIMIT functions in query. Which would allow you to select a specific number of rows.
using number as 5 you can get the desired result from:
SELECT columnname FROM tablename WHERE condition LIMIT number;
or
SELECT TOP (number)/(percent) columnname FROM tablename WHERE condition;
I have table as below:
I want write a sql query to get output as below:
the query should select all the records from the table but, when multiple records have same Id column value then it should take only one record having latest Date.
E.g., Here Rudolf id 1211 is present three times in input---in output only one Rudolf record having date 06-12-2010 is selected. same thing with James.
I tried to write a query but it was not succssful. So, please help me to form a query string in sql.
Thanks in advance
You can partition your data over Date Desc and get the first row of each partition
SELECT A.Id, A.Name, A.Place, A.Date FROM (
SELECT
*,
ROW_NUMBER() OVER (PARTITION BY Id ORDER BY Date DESC) AS rn
FROM [Table]
) A WHERE A.rn = 1
you can use WITH TIES
select top 1 PERCENT WITH TIES * from t
order by (row_number() over(partition by id order by date desc))
https://dbfiddle.uk/?rdbms=sqlserver_2017&fiddle=280b7412b5c0c04c208f2914b44c7ce3
As i can see from your example, duplicate rows differ only in Date. If it's a case, then simple GROUP BY with MAX aggregate function will do the job for you.
SELECT Id, Name, Place, MAX(Date)
FROM [TABLE_NAME]
GROUP BY Id, Name, Place
Here is working example: http://sqlfiddle.com/#!18/7025e/2
I have this query currently, which selects the top "number of pickups" in descending order. I need to filter only the top 10 rows/highest numbers though. How can I do this?
I have tried adding 'WHERE ROWNUM <= 10' at the bottom, to no avail.
SELECT customer.company_name, COUNT (item.pickup_reference) as "Number of Pickups"
FROM customer
JOIN item ON (customer.reference_no=item.pickup_reference)
GROUP BY customer.company_name, item.pickup_reference
ORDER BY COUNT (customer.company_name) DESC;
Thanks for any help!
You need to subquery it for the rownum to work.
SELECT *
FROM
(
SELECT customer.company_name, COUNT (item.pickup_reference) as "Number of Pickups"
FROM customer
JOIN item ON (customer.reference_no=item.pickup_reference)
GROUP BY customer.company_name, item.pickup_reference
ORDER BY COUNT (customer.company_name) DESC
)
WHERE rownum <= 10
You could alternatively use ranking functions, but given the relative simplicity of this, I'm not sure whether I would.
The solution by using the rank is something like this :
select customer.company_name, COUNT (item.pickup_reference) from (
select distinct customer.company_name, COUNT (item.pickup_reference) ,
rank() over ( order by count(item.pickup_reference) desc) rnk
from customer
JOIN item ON (customer.reference_no=item.pickup_reference)
group by customer.company_name, item.pickup_reference
order by COUNT (customer.company_name) )
where rnk < 10
Using the 'rownum' to get the top result doesn't give the expected result, because it get the 10 first rows which are not ordred, and then order them (Please notify this on a comment on Andrew's response, I don't have the right to add the comment) .