It's my first question here so I hope I can explain it well enough,
I want to order my data by amount of occurrences in the table.
My table is like this:
id Daynr
1 2
1 4
2 4
2 5
2 6
3 1
4 2
4 5
And I want it to sort it like this:
id Daynr
3 1
1 2
1 4
4 2
4 5
2 4
2 5
2 6
Player #3 has one day in the table, and Player #1 has 2.
My table is named "dayid"
Both id and Daynr are foreign keys, together making it a primary key
I hope this explains my problem enough, Please ask for more information it's my first time here.
Thanks in advance
You can do this by counting the number of times that things occur for each id. Most databases support window functions, so you can do this as:
select id, daynr
from (select t.*, count(*) over (partition by id) as cnt
from table t
) t
order by cnt, id;
You can also express this as a join:
select t.id, t.daynr
from table as t inner join
(select id, count(*) as cnt
from table
group by id
) as tg
on t.id = tg.id
order by tg.cnt, id;
Note that both of these include the id in the order by. That way, if two ids have the same count, all rows for the id will appear together.
Related
I've got this table ratings:
id
user_id
type
value
0
0
Rest
4
1
0
Bar
3
2
0
Cine
2
3
0
Cafe
1
4
1
Rest
4
5
1
Bar
3
6
1
Cine
2
7
1
Cafe
5
8
2
Rest
4
9
2
Bar
3
10
3
Cine
2
11
3
Cafe
5
I want to have a table with a row for every pair (user_id, type) for the top 3 rated types through all users (ranked by sum(value) across the whole table).
Desired result:
user_id
type
value
0
Rest
4
0
Cafe
1
0
Bar
3
1
Rest
4
1
Cafe
5
1
Bar
3
2
Rest
4
3
Cafe
5
2
Bar
3
I was able to do this with two queries, one to get the top 3 and then another to get the rows where the type matches the top 3 types.
Does someone know how to fit this into a single query?
Get rows per user for the 3 highest ranking types, where types are ranked by the total sum of their value across the whole table.
So it's not exactly about the top 3 types per user, but about the top 3 types overall. Not all users will have rows for the top 3 types, even if there would be 3 or more types for the user.
Strategy:
Aggregate to get summed values per type (type_rnk).
Take only the top 3. (Break ties ...)
Join back to main table, eliminating any other types.
Order result by user_id, type_rnk DESC
SELECT r.user_id, r.type, r.value
FROM ratings r
JOIN (
SELECT type, sum(value) AS type_rnk
FROM ratings
GROUP BY 1
ORDER BY type_rnk DESC, type -- tiebreaker
LIMIT 3 -- strictly the top 3
) v USING (type)
ORDER BY user_id, type_rnk DESC;
db<>fiddle here
Since multiple types can have the same ranking, I added type to the sort order to break ties alphabetically by their name (as you did not specify otherwise).
Turns out, we don't need window functions - the ones with OVER and, optionally, PARTITION for this. (Since you asked in a comment).
I think you just want row_number(). Based on your results, you seem to want three rows per type, with the highest value:
select t.*
from (select t.*,
row_number() over (partition by type order by value desc) as seqnum
from t
) t
where seqnum <= 3;
Your description suggests that you might just want this per user, which is a slight tweak:
select t.*
from (select t.*,
row_number() over (partition by user order by value desc) as seqnum
from t
) t
where seqnum <= 3;
There is a table ID in my PostgreSQL database.
select * from ID;
ID
1
2
3
4
5
6
7
8
I need to write a query that will give me this output:
id id
1 2
3 4
5 6
7 8
There are no such things as even and odd rows. SQL tables represent unordered sets (well technically multi-sets).
Assuming id is the ordering, you can split them using aggregation like this:
select min(id), max(id)
from (select t.*, row_number() over (order by id) - 1 as seqnum
from t
) t
group by floor(seqnum / 2)
I have an existing table to which I have added a new column which is supposed to hold the Id of a record in another (new) table.
Simplified structure is sort of like this:
Customer table
[CustomerId] [GroupId] [LicenceId] <-- new column
Licence table <-- new table
[LicenceId] [GroupId]
The Licence table has a certain number of licences per group than can be assigned to customers in that same group. There are multiple groups, and each group has a variable number of customers and licences.
So say there are 100 licences available for group 1 and there are 50 customers in group 1, so each can get a license. There are never more customers than there are licences.
Sample
Customer
[CustomerId] [GroupId] [LicenceId]
1 1 NULL
2 1 NULL
3 1 NULL
4 1 NULL
5 2 NULL
6 2 NULL
7 2 NULL
8 3 NULL
9 3 NULL
Licence
[LicenceId] [GroupId]
1 1
2 1
3 1
4 1
5 1
6 1
7 2
8 2
9 2
10 2
11 2
12 3
13 3
14 3
15 3
16 3
17 3
Desired outcome
Customer
[CustomerId] [GroupId] [LicenceId]
1 1 1
2 1 2
3 1 3
4 1 4
5 2 7
6 2 8
7 2 9
8 3 12
9 3 13
So now I have to do this one time update to give every customer a licence and I have no idea how to go about it.
I'm not allowed to use a cursor. I can't seem to do a MERGE UPDATE, because joining the Customer to the Licence table by GroupId will result in multiple hits.
How do I assign each customer the next available LicenceId within their group in one query?
Is this even possible?
You can use window functions:
with c as (
select c.*, row_number() over (partition by groupid order by newid()) as seqnum
from customers c
),
l as (
select l.*, row_number() over (partition by groupid order by newid()) as seqnum
from licenses c
)
update c
set c.licenceid = l.licenseid
from c join
l
on c.seqnum = l.seqnum and c.groupid = l.groupid;
This assigns the licenses randomly. That is really just for fun. The most efficient method is to use:
row_number() over (partition by groupid order by (select null)) as seqnum
SQL Server often avoids an additional sort operation in this case.
But you might want to order them by something else -- for instance by the ordering of the customer ids, or by some date column, or something else.
Gordon has put it very well in his answer.
Let me break it down into simpler steps for you.
Step 1. Use the ROW_NUMBER() function to assign a SeqNum to the Customers. Use PARTITION BY GroupId so that the number starts from 1 in every group. I would ORDER BY CustomerId
Step 2. Use the ROW_NUMBER() function to assign a SeqNum to the Licences. Use PARTITION BY GroupId so that the number starts from 1 in every group. ORDER BY LicenseId because your ask is to "assign each customer the next available LicenceId within their group".
Now use these 2 queries to update LicenseId in Customer table.
I create frequencies on one column in SQL in a standard way.
My code is
select id , count(*) as counts
from TABLE
group by id
order by counts desc
Suppose the output is as follows for six id
id counts
-- -----
1 3 two id have 3 counts per
2 3
---------
3 6 three id have 6 counts per
4 6
5 6
---------
6 2 one id has 2 counts
How can I produce the following?
nid counts
--- ------
1 2
2 3
3 6
I am writing in a hive environment, but that should be standard SQL.
Thanks in advance for answering.
You want two levels of aggregation:
select counts, count(*)
from (select id , count(*) as counts
from TABLE
group by id
) c
group by counts
order by counts;
I call this a "histogram-of-histograms" query. I usually include min(id) and max(id) in the outer select, so I have examples of ids with given frequencies.
I need to find the repeated rows from a table where the first three columns will make up the primary key. Then after finding out which one's are repeated, those repeated rows need to be removed from the query results as this example shows:
Given this table. The first 3 columns act as the primary key.
--------------1 2 3 4 5 6-------------------------------------------------------------------------------------------------------------------------------1 2 3 9 8 9-------------------------------------------------------------------------------------------------------------------------------1 4 3 9 8 9-------------------------------------------------------------------------------------------------------------------------------3 4 2 2 2 1-------------------------------------------------------------------------------------------------------------------------------2 3 4 1 1 3-------------------------------------------------------------------------------------------------------------------------------2 3 4 9 9 0--------
Since 1 2 3 is the composite primary key. The first 2 rows should be considered repeated and therefore eliminated from the results. Just as the two 2 3 4 rows.
The only rows in the result set should be:
1 4 3 9 8 9 and 3 4 2 2 2 1
Could you please help?
Thanks a lot in advance..
Your question doesn't fully make sense. A composite primary key would prevent duplicates in the table. So, these columns are not declared as a comosite primary key if the data contains duplicates.
If you just one one row for each group, you can use row_number() for this, something like this (the column names are obviously invalid):
select t.*
from (select t.*, row_number() over (partition by 1, 2, 3 order by 1) as seqnum
from table t
) t
where seqnum = 1;
If you want to delete extra rows, you can try:
delete from t
where rowid not in (select min(rowid) from table t group by 1, 2, 3);
EDIT:
If you want to remove cases where the rows are repeated, then you want count() instead of row_number():
select t.*
from (select t.*, count() over (partition by 1, 2, 3) as cnt
from table t
) t
where cnt > 1;