I have inherited a SQL Server table in the (abbreviated) form of (includes sample data set):
| SID | name | Invite_Date |
|-----|-------|-------------|
| 101 | foo | 2013-01-06 |
| 102 | bar | 2013-04-04 |
| 101 | fubar | 2013-03-06 |
I need to select all SID's and the Invite_date, but if there is a duplicate SID, then just get the latest entry (by date).
So the results from the above would look like:
101 | fubar | 2013-03-06
102 | bar | 2013-04-04
Any ideas please.
N.B the Invite_date column has been declared as a nvarchar, so to get it in a date format I am using CONVERT(DATE, Invite_date)
You can use a ranking function like ROW_NUMBER or DENSE_RANK in a CTE:
WITH CTE AS
(
SELECT SID, name, Invite_Date,
rn = Row_Number() OVER (PARTITION By SID
Order By Invite_Date DESC)
FROM dbo.TableName
)
SELECT SID, name, Invite_Date
FROM CTE
WHERE RN = 1
Demo
Use Row_Number if you want exactly one row per group and Dense_Rank if you want all last Invite_Date rows for each group in case of repeating max-Invite_Dates.
select t1.*
from your_table t1
inner join
(
select sid, max(CONVERT(DATE, Invite_date)) mdate
from your_table
group by sid
) t2 on t1.sid = t2.sid and CONVERT(DATE, t1.Invite_date) = t2.mdate
select
SID,name,MAX(Invite_date)
FROM
Table1
GROUP BY
SID
http://sqlfiddle.com/#!2/6b6f66/1
Related
I have the following table
| id | date | team |
|----|------------|------|
| 1 | 2019-01-05 | A |
| 2 | 2019-01-05 | A |
| 3 | 2019-01-01 | A |
| 4 | 2019-01-04 | B |
| 5 | 2019-01-01 | B |
How can I query the table to receive the most recent values for the teams?
For example, the result for the above table would be ids 1,2,4.
In this case, you can use window functions:
select t.*
from (select t.*, rank() over (partition by team order by date desc) as seqnum
from t
) t
where seqnum = 1;
In some databases a correlated subquery is faster with the right indexes (I haven't tested this with Postgres):
select t.*
from t
where t.date = (select max(t2.date) from t t2 where t2.team = t.team);
And if you wanted only one row per team, then the canonical answer is:
select distinct on (t.team) t.*
from t
order by t.team, t.date desc;
However, that doesn't work in this case because you want all rows from the most recent date.
If your dataset is large, consider the max analytic function in a subquery:
with cte as (
select
id, date, team,
max (date) over (partition by team) as max_date
from t
)
select id
from cte
where date = max_date
Notionally, max is O(n), so it should be pretty efficient. I don't pretend to know the actual implementation on PostgreSQL, but my guess is it's O(n).
One more possibility, generic:
select * from t join (select max(date) date,team from t
group by team) tt
using(date,team)
Window function is the best solution for you.
select id
from (
select team, id, rank() over (partition by team order by date desc) as row_num
from table
) t
where row_num = 1
That query will return this table:
| id |
|----|
| 1 |
| 2 |
| 4 |
If you to get it one row per team, you need to use array_agg function.
select team, array_agg(id) ids
from (
select team, id, rank() over (partition by team order by date desc) as row_num
from table
) t
where row_num = 1
group by team
That query will return this table:
| team | ids |
|------|--------|
| A | [1, 2] |
| B | [4] |
I don't really know how to describe it. I have a table:
ID | Name | Date
-------------------------
1 | Mike | 01.01.2016
1 | Michael | 02.03.2016
2 | Samuel | 23.12.2015
2 | Sam | 05.03.2015
3 | Tony | 02.04.2012
I want to select pairs of IDs and Names with latest dates in each pair. The result here should be:
ID | Name | Date
-------------------------
1 | Michael | 02.03.2016
2 | Samuel | 23.12.2015
3 | Tony | 02.04.2012
How do I achieve this?
Oracle Database 11g
You can do it using the ROW_NUMBER() analytic function:
SELECT id, name, "date"
FROM (
SELECT t.*,
ROW_NUMBER() OVER ( PARTITION BY id ORDER BY "date" DESC ) rn
FROM table_name t
)
WHERE rn = 1
This requires only a single table scan (it does not have a self-join or correlated sub-query - i.e. IN (...) or EXISTS(...)).
Have a sub-select that returns each id and it's max date:
select * from table
where (id, date) in (select id, max(date) from table group by id)
You can use NOT EXISTS() :
SELECT * FROM YourTable t
WHERE NOT EXISTS(SELECT 1 FROM YourTable s
WHERE t.id = s.id and s.date > t.date)
Possibly the most efficient method is:
select t.*
from table t
where t.date = (select max(date) from table t2 where t2.id = t.id);
along with an index on table(id, date).
This version should scan the table and look up the correct value in the index.
Or, if there are only three columns, you can use keep:
select id, max(date) as date,
max(name) keep (dense_rank first order by date desc) as name
from table
group by id;
I have found that this version works very well in Oracle.
I have a table
User | Phone | Value
Peter | 0 | 1
Peter | 456 | 2
Peter | 456 | 3
Paul | 456 | 7
Paul | 789 | 10
I want to select MAX value for every user, than it also lower than a tresshold
For tresshold 8, I want result to be
Peter | 456 | 3
Paul | 456 | 7
I have tried the GROUP BY with HAVING, but I am getting
column "phone" must appear in the GROUP BY clause or be used in an aggregate function
Similar query logic works in MySQL, but I am not quite sure how to operate with GROUP BY in PostgreSQL. I dont want to GROUP BY phone.
After I have results from "juergen d" solution, I came up with this which gives me the same results faster
SELECT DISTINCT ON(user) user, phone, value
FROM table
WHERE value < 8
ORDER BY user, value DESC;
select t1.*
from your_table t1
join
(
select user, max(value) as max_value
from your_table
where value < 8
group by user
) t2 on t1.user = t2.user and t1.value = t2.max_value
Alternatively, you could use a ranking function:
select * from
(
select *, RANK() OVER (partition by [user] ORDER BY t.value desc ) as value_rank from test_table as t
where t.value < 8
) as t1
where value_rank = 1
I have the following data structure
ID | REFID | NAME
1 | 100 | A
2 | 101 | B
3 | 101 | C
With
SELECT DISTINCT REFID, ID, NAME
FROM my_table
ORDER BY ID
I would like to have the following result:
1 | 100 | A
2 | 101 | B
Colum NAME and ID should contain the MIN or FIRST value.
But actually I get stuck at using MIN/FIRST here.
I welcome every tipps :-)
select id,
refid,
name
from (select id,
refid,
name,
row_number() over(partition by refid order by name) as rn
from my_table)
where rn = 1
order by id
You can use a subquery to do this.
WITH Q AS
( SELECT MIN(NAME) AS NAME, REFID FROM T GROUP BY REFID )
SELECT T.ID, T.REFID, T.NAME
FROM T
JOIN Q
ON (T.NAME = Q.NAME)
Also, note that SQL tables have no order. So there's no "First" value.
I have the following table:
| ID | ExecOrd | date |
| 1 | 1.0 | 3/4/2014|
| 1 | 2.0 | 7/7/2014|
| 1 | 3.0 | 8/8/2014|
| 2 | 1.0 | 8/4/2013|
| 2 | 2.0 |12/2/2013|
| 2 | 3.0 | 1/3/2014|
| 2 | 4.0 | |
I need to get the date of the top ExecOrd per ID of about 8000 records, and so far I can only do it for one ID:
SELECT TOP 1 date
FROM TABLE
WHERE DATE IS NOT NULL and ID = '1'
ORDER BY ExecOrd DESC
A little help would be appreciated. I have been trying to find a similar question to mine with no success.
There are several ways of doing this. A generic approach is to join the table back to itself using max():
select t.date
from yourtable t
join (select max(execord) execord, id
from yourtable
group by id
) t2 on t.id = t2.id and t.execord = t2.execord
If you're using 2005+, I prefer to use row_number():
select date
from (
select row_number() over (partition by id order by execord desc) rn, date
from yourtable
) t
where rn = 1;
SQL Fiddle Demo
Note: they will give different results if ties exist.
;with cte as (
SELECT id,row_number() over(partition by ID order byExecOrd DESC) r
FROM TABLE WHERE DATE IS NOT NULL )
select id from
cte where r=1