I have a table like this:
--------------------------------------
RecID|name |date
--------------------------------------
1 |John | 05/09/2016
2 |John | 05/02/2016
3 |Mary | 05/09/2016
4 |Mary | 05/08/2016
5 |Mary | 03/02/2016
and I want to get the count for name for each instance in which that name has appeared on or before that date in the row. So I want the output to look like this:
--------------------------------------
RecID|name |date |count
--------------------------------------
1 |John | 05/09/2016 | 2
2 |John | 05/02/2016 | 1
3 |Mary | 05/09/2016 | 3
4 |Mary | 05/08/2016 | 2
5 |Mary | 03/02/2016 | 1
Any ideas on how I should go about doing this?
You can use the count function with a window specification.
select t.*, count(*) over(partition by name order by date) as cnt
from tablename t
This will produce incorrect results if there are mutliple rows on a given date for a name. One way to avoid this is using a correlated sub-query.
select t.*,
(select count(distinct t2.date)
from tablename t2
where t2.name=t.name and t2.date<=t.date) as cnt
from tablename t
Or use row_number.
select t.*, row_number() over(partition by name order by date) as cnt
from tablename t
Or use dense_rank if there can be multiple rows for the same name on a given date.
select t.*, dense_rank() over(partition by name order by date) as cnt
from tablename t
The easiest solution of all would be to use dense_rank.
use
count(*) count
and
group by date
if your date is already a string (i.e. without hour/minute information)
Related
Given the following table P_PROV
+----+-----------+-----------+
| id | date | person_id |
+----+-----------+-----------+
| 1 |19/06/2019 | 1 |
| 2 |18/07/2010 | 2 |
| 3 |19/06/2020 | 1 |
| 4 |17/06/2020 | 2 |
| 5 |28/06/2020 | 3 |
+----+-----------+-----------+
I want this output
+----+-----------+-----------+
| id | date | person_id |
+----+-----------+-----------+
| 3 |19/06/2020 | 1 |
| 4 |17/06/2020 | 2 |
| 5 |28/06/2020 | 3 |
+----+-----------+-----------+
Putting this in words, I want to return per person the maximum date. I tried something like this
SELECT DISTINCT pp.date, pp.id FROM P_PROV pp
WHERE (SELECT MAX(aa.date)
FROM P_PROV aa) = pp.date;
This one is only returning one row (of course, because the MAX will return the maximum date only), but I really don't know how to approach this issue, any kind of help would be appreciated
ROW_NUMBER provides one way to handle this:
SELECT id, date, person_id
FROM
(
SELECT t.*, ROW_NUMBER() OVER (PARTITION BY person_id ORDER BY date DESC) rn
FROM yourTable t
) t
WHERE rn = 1;
Oracle has a fun way to do this using aggregation:
select max(id) keep (dense_rank first order by date desc) as id,
max(date) as date, person_id
from P_PROV
group by person_id;
Given that your ids are increasing, this probably also does what you want:
select max(id) as id, max(date) as date, person_id
from P_PROV
group by person_id;
My table looks like this, what I'm trying to achieve is to pull out all the records for one user for the product that have the earliest date
product |type_id| user | Date |Desired ROW_NUMBER as output |
-------+--------+------+-------+---------------------
1 | 1 | A | 0101 | 1
1 | 1 | A | 0102 | 1
2 | 3 | A | 0105 | 2
2 | 5 | A | 0105 | 2
3 | 7 | B | 0101 | 1
3 | 8 | B | 0104 | 1
So I want to pull all the records with "1" in the desired row_num column, but I haven't figured out hot to get this without doing another group by. Any helps would be appreciated.
You can use window functions:
select t.*
from (select t.*,
rank() over (partition by user order by min_date) as seqnum
from (select t.*,
min(date) over (partition by user, product) as min_date
from t
) t
) t
where seqnum = 1;
Or, with only one subquery:
select t.*
from (select t.*,
min(date) over (partition by user, product) as min_date_up,
min(date) over (partition by user) as min_date_u
from t
) t
where min_date_u = min_date_up;
You can interpret this as "return all rows where the product has the minimum date for the user".
Here is a db<>fiddle.
SELECT * FROM [tableName] WHERE Desired ROW_NUMBER = 1 ORDER BY Date[DESC, ASC]
Pass the Desired ROW_NUMBER value dynamically as a parameter.
Please help me to the SQL Oracle with the showing only 1 row result in each ID group, Order by DATE DESC below:
From this list
Table: Employees
____________________________
ID | Employees | Date
___+___________+____________
1 | A | 2017-08-01
1 | A | 2017-08-08
2 | B | 2017-07-01
2 | B | 2017-07-10
2 | B | 2017-07-05
Result
____________________________
ID | Employees | Date
___+___________+____________
1 | A | 2017-08-08
2 | B | 2017-07-10
There are several different ways of achieving this. It really depends on what business rules you want to use to choose the one date per employee. You haven't specified anything, but your desired output suggests you need the most recent date. If so you can use the max() aggregate function with a GROUP BY clause.
select id,
employees,
max(date) as date
from employee
group by id,
employees
You can use the analytic function :
https://docs.oracle.com/cloud/latest/db112/SQLRF/functions004.htm#SQLRF06174
select *, RANK() OVER (PARTITION BY Employees ORDER BY date desc) from Employees
then put an outer select to select rank 1
select id, emplyoees, max(date) as date1
from employees
group by id, emplyoees;
i have a table in a postgres DB which has the following structure:
id | date | groupme1 | groupme2 | value
----------------------------------------
1 |
2 |
3 |
Now i want to achieve the following:
Grouping the table after groupme1 and groupme2
Get the value for every group
But only the last entry for each group-compination (odered after date)
Example:
id | date | groupme1 | groupme2 | value
---------------------------------------
| | A | 1 | 4
| | A | 2 | 7
| | A | 3 | 3
| | B | 1 | 9
My current approach looks like this:
SELECT a.*
FROM table AS a
JOIN (SELECT max(id) AS id
FROM table
GROUP BY groupme1, groupme2) AS b
ON a.id = b.id
The Problems of this approach:
it asumes that higher dates have a higher id
it takes long
Is there a faster and better way of doing this? Can windowing function help with this?
I think you just want window functions:
select t.*
from (select t.*,
row_number() over (partition by groupme1, groupme2 order by date desc) as seqnum
from t
) t
where seqnum = 1;
Or, a better way to do this in Postgres uses distinct on:
select distinct on (groupme1, groupme2) t.*
from t
order by groupme1, groupme2, date desc;
I'm partitioning by some non unique identifier, but I'm only concerned in the partitions with at least two results. What would be the way to get out all the instances where there's exactly one of the specified identifier?
Query I'm using:
SELECT ROW_NUMBER() OVER
(PARTITION BY nonUniqueId ORDER BY nonUniqueId, aTimeStamp) as row
,nonUniqueId
,aTimeStamp
FROM myTable
What I'm getting:
row | nonUniqueId | aTimeStamp
---------------------------------
1 | 1234 | 2014-10-08...
2 | 1234 | 2014-10-09...
1 | 1235 | 2014-10-08...
1 | 1236 | 2014-10-08...
2 | 1236 | 2014-10-09...
What I want:
row | nonUniqueId | aTimeStamp
---------------------------------
1 | 1234 | 2014-10-08...
2 | 1234 | 2014-10-09...
1 | 1236 | 2014-10-08...
2 | 1236 | 2014-10-09...
Thanks for any direction :)
Based on syntax, I'm assuming this is SQL Server 2005 or higher. My answer will be meant for that.
You have a couple options.
One, use a CTE:
;WITH CTE AS (
SELECT ROW_NUMBER() OVER
(PARTITION BY nonUniqueId ORDER BY nonUniqueId, aTimeStamp) as row
,nonUniqueId
,aTimeStamp
FROM myTable
)
SELECT *
FROM CTE t
WHERE EXISTS (SELECT 1 FROM CTE WHERE row = 2 and nonUniqueId = t.nonUniqueId);
Or, you can use subqueries:
SELECT ROW_NUMBER() OVER
(PARTITION BY nonUniqueId ORDER BY nonUniqueId, aTimeStamp) as row
,nonUniqueId
,aTimeStamp
FROM myTable t
WHERE EXISTS (SELECT 1 FROM myTable
WHERE nonUniqueId = t.nonUniqueId GROUP BY nonUniqueId, aTimeStamp HAVING COUNT(*) >= 2);