How to get max value and using group by clause - sql

I have a query like this:
select transactions_id,
time_stamp,
clock
from times
group by transactions_id
having sum(distinct type) = 1
now, I would like to get max value depending on id.
I used below queries but not worked:
select max(id),
transactions_id,
time_stamp,
clock
from times
group by transactions_id
having sum(distinct type) = 1
or
select transactions_id,
time_stamp,
clock
from times
group by transactions_id
having sum(distinct type) = 1
and max(id)
for example:
I have three conditions:
type must be 1
group by transactions_id
max id

You can find aggregates in one query and join its result with the table to get the relevant rows.
select *
from times t1
join (
select transactions_id,
max(id) as id
from times
where type = 1
group by transactions_id
) t2 using (transactions_id, id);

If I understand correctly, you can use the ANSI standard row_number() function:
select t.*
from (select t.*,
row_number() over (partition by transactions_id order by id desc) as seqnum
from times t
) t
where seqnum = 1;
I am not sure what having sum(distinct type) = 1. That condition is not explained in the question.

Related

select value based on max of other column

I have a few questions about a table I'm trying to make in Postgres.
The following table is my input:
id
area
count
function
1
100
20
living
1
200
30
industry
2
400
10
living
2
400
10
industry
2
400
20
education
3
150
1
industry
3
150
1
education
I want to group by id and get the dominant function based on max area. With summing up the rows for area and count. When area is equal it should be based on max count, when area and count is equal it should be based on prior function (i still have to decide if education is prior to industry or vice versa). So the result should be:
id
area
count
function
1
300
50
industry
2
1200
40
education
3
300
2
industry
I tried a lot of things and maybe it's easy, but i don't get it. Can someone help to get the right SQL?
One method uses row_number() and conditional aggregation:
select id, sum(area), sum(count),
max(function) over (filter where seqnum = 1) as function
from (select t.*,
row_number() over (partition by id order by area desc) as seqnum
from t
) t
group by id;
Another method uses ``distinct on`:
select id, sum(area) over (partition by id) as area,
sum(count) over (partition by id) as count,
function
from t
order by id, area desc;
Use a scalar sub-query for "function".
select t.id, sum(t.area), sum(t.count),
(
select "function"
from the_table
where id = t.id
order by area desc, count desc, "function" desc
limit 1
) as "function"
from the_table as t
group by t.id order by t.id;
SQL Fiddle
you can use sum as window function:
select distinct on (t.id)
id,
sum(area) over (partition by id) as area,
sum(count) over (partition by id) as count,
( select function from tbl_test where tbl_test.id = t.id order by count desc limit 1 ) as function
from tbl_test t
This is how you get the function for each group based on id:
select id, function
from yourtable yt1
left join yourtable yt2
on yt1.id = yt2.id and yt1.area < yt2.area
where yt2.area.id is null;
(we ensure that no yt2 exists that would be of the same id but of higher areay)
This would work nicely, but you might have several max areas with different values. To cope with this isue, let's ensure that exactly one is chosen:
select id, max(function) as function
from yourtable yt1
left join yourtable yt2
on yt1.id = yt2.id and yt1.area < yt2.area
where yt2.area.id is null
group by id;
Now, let's join this to our main table;
select yourtable.id, sum(yourtable.area), sum(yourtable.count), t.function
from yourtable
join (
select id, max(function) as function
from yourtable yt1
left join yourtable yt2
on yt1.id = yt2.id and yt1.area < yt2.area
where yt2.area.id is null
group by id
) t
on yourtable.id = t.id
group by yourtable.id;

sql: Select count(*) - nth record from each group

I'm grouping by tenant_id. I want to select the count() - 1000th record (ordered by _updated time) from each GROUPBY group, for the groups where count() is greater than 1000. As follows:
select t1.tenant_id,
(select temp._updated
from trace temp
where temp.tenant_id = t1.tenant_id
order by _updated limit 1 offset
count(*) - 1000
) as timekey
from fgc.trace as t1
group by tenant_id
having count(*) > 1000;
But this is not allowed as count(*) cannot be used inside the subquery.
So I tried the following, which still doesn't work as I don't have access to t1 since this is not a join.
select t1.tenant_id,
(select temp._updated
from trace temp
where temp.tenant_id = t1.tenant_id
order by _updated limit 1 offset
(select count(*)-1000
from trace t2
group by tenant_id
having t2.tenant_id = t1.tenant_id)
) as timekey
from fgc.trace as t1
group by tenant_id
having count(*) > 1000;
So how can I get the following?
tenant_id | timekey
+-----------+----------------------------------+
n7ia6ryc | 2019-07-23 23:09:49.951406+00:00
You seem to want ROW_NUMBER(). Cockroach supports windows functions, so:
SELECT updated
FROM (
SELECT
tenant_id,
updated,
ROW_NUMBER() OVER(PARTITION BY tenant_id ORDER BY updated DESC) rn
FROM trace
) x WHERE rn = 1001
For each tenant_id, this will return the timestamp of the 1001th less recent record. If a given tenant has less than 1000 records, it will not appear in the results.
select x.tenant_id
from (
select t.tenant_id,
row_number() over (partition by t.tenant_id order by t.timekey) as tenant_number
from fgc.trace as t
) x
where x.tenant_number > 1000
group by x.tenant_id
just the one timestamp would look like this:
select min(x.timekey) as min_timestamp
from (
select t.tenant_id, t.timekey,
row_number() over (partition by t.tenant_id order by t.timekey) as tenant_number
from fgc.trace as t
) x
where x.tenant_number > 1000
note that grouping does not matter here because each row can only be in one group and you are only looking at one row.

Finding top count of a value in a table using SQL

I'm looking for a way to find the top count value of a column by SQL.
If for example this is my data
id type
----------
1 A
1 B
1 A
2 C
2 D
2 D
I would like the result to be:
1 A
2 D
I'm looking for a way to do it without groping by the column I count (type in the example)
Thanks
Statistically, this is called the "mode". You can calculate it using window functions:
select id, type, cnt
from (select id, type, count(*) as cnt,
row_number() over (partition by id order by count(*) desc) as seqnum
from t
group by id, type
) t
where seqnum = 1;
If there are ties, then an arbitrary value is chosen from among the ties.
You are looking for the statistic mode (the most often ocurring value):
select id, stats_mode(type)
from mytable
group by id
order by id;
Not all DBMS support this however. Check your docs, wheher this function or a similar one is available in your DBMS.
Just GROUP BY id, type and keep the rows with the maximum counter:
select id, type
from tablename
group by id, type
having count(*) = (
select count(*) from tablename group by id, type order by count(*) desc limit 1
)
See the demo
Or
select id, type
from tablename
group by id, type
having count(*) = (
select max(t.counter) from (select count(*) counter from tablename group by id, type) t
)
See the demo

Aggregate function like MAX for most common cell in column?

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;

SQL Server Group By with Max on Date field

I hope i can explain the issue i'm having and hopefully so can point me in the same direction.
I'm trying to do a group by (Email Address) on a subset of data, then i'm using a max() on a date field but because of different values in other fields its bring back more rows then require.
I would just like to return the max record per email address and return the fields that are on the same row that are on the max record.
Not sure how i can write this query?
This is a task for ROW_NUMBER:
select *
from
(
select t.*,
-- assign sequential number starting with 1 for the maximum date
row_number() over (partiton by email_address order by datecol desc) as rn
from tab
) as dt
where rn = 1 -- only return the latest row
You can write this query using row_number():
select t.*
from (select t.*,
row_number() over (partition by emailaddress order by date desc) as seqnum
from t
) t
where seqnum = 1;
How about something like this?
select a.*
from baseTable as a
inner join
(select Email,
Max(EmailDate) as EmailDate
from baseTable
group by Email) as b
on a.Email = b.Email
and a.EmailDate = b.EmailDate