There are groups like this;
USER_ID SEQ_ID NAME
1 2 Armut
1 3 Elma
1 4 Kiraz
2 1 Nar
2 2 Uzum
4 3 Sheftali
4 4 Karpuz
4 5 Kavun
After select query I want to see only;
USER_ID SEQ_ID NAME
1 2 Armut
2 1 Nar
4 3 Karpuz
That is, I want the row with the least SEQ_ID for each USER_ID. What SQL query will give me this result?
Best regards
SELECT USER_ID, SEQ_ID, NAME
FROM table
WHERE NAME IN ('Armut', 'Nar', 'Karpuz')
ORDER BY USER_ID
If you have something else in mind, please clarify your question.
Looks to me like it should be:
SELECT USER_ID, MIN(SEQ_ID) AS SEQ_ID, NAME
FROM table
GROUP BY USER_ID, NAME
ORDER BY USER_ID;
Related
I have a table like this.
id
grade_1
grade_2
createdAt
1
1
1
20220304
2
1
1
20220301
3
4
2
20220228
I want to select the current row(in here, id=1) and a row where the grade's value is different with the row I selected.(in here, id=3)
Like This
id
grade_1
grade_2
createdAt
1
1
1
20220304
3
4
2
20220228
I tried to use subquery but it doesn't really worked for me. Is there any way to skip the duplicated value when selecting table?
You can just do it with group by and a max value to retieve the one you want
SELECT
grade_1,
grade_2,
Max(createdAt)
from
yourTable
Group by
grade_1,
grade_2
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.
I have table like this-
id object_id product_id
1 1 1
2 1 1
4 2 2
6 3 2
7 3 2
8 1 2
9 1 1
I want to delete all rows except these-
1 1 1
4 2 2
6 3 2
9 1 2
Basically there are duplicates and I want to remove them but keep one copy intact.
what would be the most efficient way for this?
If this is a one-off then you can simply identify the records you want to keep like so:
SELECT MIN(id) AS id
FROM yourtable
GROUP BY object_id, product_id;
You want to check that this works before you do the next thing and actually throw records out. To actually delete those duplicate records you do:
DELETE FROM yourtable WHERE id NOT IN (
SELECT MIN(id) AS id
FROM yourtable
GROUP BY object_id, product_id
);
The MIN(id) obviously always returns the record with the lowest id for a set of (object_id, product_id). Change as desired.
I have a table "Employee" as shown below
Id Name
1 John
2 Jaffer
3 Syam
4 Aish
5 Gidson
1 Aboo
2 Sindhu
3 Saravanan
I want to get two outputs like
Id
1
2
3
Id
4
5
Which transformation should i use?
Could you Please help on that?
You will have to write two queries.
SELECT Id
FROM Employee
GROUP BY Id
HAVING COUNT(Id)>1
The above query will give you first output
SELECT Id
FROM Employee
GROUP BY Id
HAVING COUNT(Id)=1
This will give you 2nd output.
I need to find if a record id is on the same group id more than once.
group id | record id | comments
--------------------------------
3 1
3 1
4 2
In this case, the record with id 1 is being associated with group id 3 twice.
Is there a query that can provide all records with this behaviour?
Thanks
This query should do the job:
select t.group_id, t.record_id
from [your_table] t
group by t.group_id, t.record_id
having count(*) > 1