I have a table with news which I then echo into a slider using foreach loop.
I would need to adjust the sort order of the query, so the last one added (max id) is always displayed first, with others sorted by id ASC.
So example: News 1, 2, 3,4. I need them echoed like 4,1,2,3.
SELECT * FROM news ORDER BY sort ASC
How add another condition in the form of max(id) first then sort ASC?
Thanks.
You can try this query:
SELECT *
FROM (
SELECT *,
(SELECT MAX(Id) FROM news) AS max_id
FROM news) AS t
ORDER BY CASE WHEN id = max_id THEN 0 ELSE 1 END asc,
id ASC
or, using CROSS JOIN:
SELECT n.*
FROM news AS n
CROSS JOIN (SELECT MAX(Id) AS max_id FROM news) AS m
ORDER BY CASE WHEN n.id = m.max_id THEN 0 ELSE 1 END asc,
n.id ASC
or use window function
SELECT *
FROM (
SELECT *, ROW_NUMBER() OVER (ORDER BY id DESC) AS RN
FROM news) AS T
ORDER BY CASE WHEN RN = 1 THEN 0 ELSE 1 END, id
SELECT *
FROM (
SELECT *, MAX(id) OVER () AS MaxID
FROM news) AS T
ORDER BY CASE WHEN MaxID = id THEN 0 ELSE 1 END, id
Related
How to put item in a specific position?
I'd like to put item on a 5th position:
`id`='randomId243'
So far my sorting looks like that:
ORDER BY
CASE
WHEN `id` = 'randomId123' THEN 0
WHEN `id` = 'randomId098' THEN 1
ELSE 2
END, `name` ASC
I don't know yet which id will be in position 2,3,4. I'd prefer to avoid run another query/subquery to get ids of items from position 2-4
So final order should be like this:
randomId123
randomId098
just item alphabetical
just item alphabetical
just item alphabetical
randomId243
just item alphabetical
Use ROW_NUMBER() window function for your current ordering to rank the rows initially.
Then create 3 groups of rows: the rows on top, the row with 'item5' and the rows below 'item5' which will be assembled with UNION ALL and sorted by group and row number:
WITH cte AS (
SELECT *,
ROW_NUMBER() OVER (ORDER BY CASE id WHEN 'item0' THEN 0 WHEN 'item1' THEN 1 ELSE 2 END, name) AS rn
FROM tablename
)
SELECT id, name
FROM (
SELECT * FROM (SELECT *, 1 grp FROM cte WHERE id <> 'item5' ORDER BY rn LIMIT 5)
UNION ALL
SELECT *, 2 FROM cte WHERE id = 'item5'
UNION ALL
SELECT * FROM (SELECT *, 3 FROM cte WHERE id <> 'item5' ORDER BY rn LIMIT -1 OFFSET 5) -- negative limit means there is no limit
)
ORDER BY grp, rn;
Note that instead of:
ORDER BY CASE id WHEN 'item0' THEN 0 WHEN 'item1' THEN 1 ELSE 2 END, name
you could use:
ORDER BY id = 'item0' DESC, id = 'item1' DESC, name
See a simplified demo.
SELECT * FROM
(
SELECT DISTINCT(TRUNC(receipt_dstamp))
FROM inventory
WHERE substr(location_id,1,3) = 'GI-'
ORDER BY 1 ASC
)
WHERE ROWNUM <= 5
Output:
Hi all, i've got this subeqery and in this case my oldest date is in row 1, i want to retrive only second from the last(from the top in this case) which is gonna be 01-SEP-21.
I was trying to play with ROWNUM and OVER but without any results, im getting blank output.
Thank you.
Full query:
SELECT TRUNC(receipt_dstamp) as old_putaway_date, COUNT(tag_id) as tag_old_putaway
FROM inventory
WHERE substr(location_id,1,3) = 'GI-'
AND TRUNC(receipt_dstamp) IN (
SELECT * FROM
(
SELECT DISTINCT(TRUNC(receipt_dstamp))
FROM inventory
WHERE substr(location_id,1,3) = 'GI-'
ORDER BY 1 ASC
)
WHERE ROWNUM = 1
)
GROUP BY TRUNC(receipt_dstamp);
You should be able to simplify the entire query to:
SELECT old_putaway_date,
COUNT(tag_id) as tag_old_putaway
FROM (
SELECT TRUNC(receipt_dstamp) as old_putaway_date,
tag_id,
DENSE_RANK() OVER (ORDER BY TRUNC(receipt_dstamp)) AS rnk
FROM inventory
WHERE substr(location_id,1,3) = 'GI-'
)
WHERE rnk = 3
GROUP BY
old_putaway_date;
You can use dense_rank() :
SELECT * FROM (
SELECT L.*,DENSE_RANK()
OVER (PARTITION BY L.TAG_OLD_PUTAWAY ORDER BY L.OLD_PUTAWAY_DATE DESC) RNK
FROM
(
SELECT TRUNC(receipt_dstamp) as old_putaway_date, COUNT(tag_id) as tag_old_putaway
FROM inventory
WHERE substr(location_id,1,3) = 'GI-'
AND TRUNC(receipt_dstamp) IN (
SELECT * FROM
(
SELECT DISTINCT(TRUNC(receipt_dstamp))
FROM inventory
WHERE substr(location_id,1,3) = 'GI-'
ORDER BY 1 ASC
)
WHERE ROWNUM = 1
)
GROUP BY TRUNC(receipt_dstamp)
) L
) WHERE RNK = 2
You are using an old Oracle syntax that is not standard compliant in the regard that it relies on a subquery result order. (Sub)query results are unordered data sets by definition, but Oracle lets this pass in order to make their ROWNUM work with it.
Oracle now supports the standard SQL FETCH clause, which you should use instead.
SELECT DISTINCT TRUNC(receipt_dstamp) AS receipt_date
FROM inventory
WHERE SUBSTR(location_id, 1, 3) = 'GI-'
ORDER BY receipt_date
OFFSET 2 ROWS
FETCH NEXT 1 ROW ONLY;
https://docs.oracle.com/en/database/oracle/oracle-database/19/sqlrf/SELECT.html#GUID-CFA006CA-6FF1-4972-821E-6996142A51C6
I have table that looks like the following
I have to select every second record per PatientID that would give the following result (my last query returns this result)
I then have to select the record with the oldest date which would be the following (this is the end result I want)
What I have done so far: I have a CTE that gets all the data I need
WITH cte
AS
(
SELECT visit.PatientTreatmentVisitID, mat.PatientMatchID,pat.PatientID,visit.RegimenDate AS VisitDate,
ROW_NUMBER() OVER(PARTITION BY mat.PatientMatchID, pat.PatientID ORDER BY visit.VisitDate ASC) AS RowNumber
FROM tblPatient pat INNER JOIN tblPatientMatch mat ON mat.PatientID = pat.PatientID
LEFT JOIN tblPatientTreatmentVisit visit ON visit.PatientID = pat.PatientID
)
I then write a query against the CTE but so far I can only return the second row for each patientID
SELECT *
FROM
(
SELECT PatientTreatmentVisitID,PatientMatchID,PatientID, VisitDate, RowNumber FROM cte
) as X
WHERE RowNumber = 2
How do I return the record with the oldest date only? Is there perhaps a MIN() function that I could be including somewhere?
If I follow you correctly, you can just order your existing resultset and retain the top row only.
In standard SQL, you would write this using a FETCH clause:
SELECT *
FROM (
SELECT
visit.PatientTreatmentVisitID,
mat.PatientMatchID,
pat.PatientID,
visit.RegimenDate AS VisitDate,
ROW_NUMBER() OVER(PARTITION BY mat.PatientMatchID, pat.PatientID ORDER BY visit.VisitDate ASC) AS rn
FROM tblPatient pat
INNER JOIN tblPatientMatch mat ON mat.PatientID = pat.PatientID
LEFT JOIN tblPatientTreatmentVisit visit ON visit.PatientID = pat.PatientID
) t
WHERE rn = 2
ORDER BY VisitDate
OFFSET 0 ROWS FETCH FIRST 1 ROW ONLY
This syntax is supported in Postgres, Oracle, SQL Server (and possibly other databases).
If you need to get oldest date from all selected dates (every second row for each patient ID) then you can try window function Min:
SELECT * FROM
(
SELECT *, MIN(VisitDate) OVER (Order By VisitDate) MinDate
FROM
(
SELECT PatientTreatmentVisitID,PatientMatchID,PatientID, VisitDate,
RowNumber FROM cte
) as X
WHERE RowNumber = 2
) Y
WHERE VisitDate=MinDate
Or you can use SELECT TOP statement. The SELECT TOP clause allows you to limit the number of rows returned in a query result set:
SELECT TOP 1 PatientTreatmentVisitID,PatientMatchID,PatientID, VisitDate FROM
(
SELECT *
FROM
(
SELECT PatientTreatmentVisitID,PatientMatchID,PatientID, VisitDate,
RowNumber FROM cte
) as X
WHERE RowNumber = 2
) Y
ORDER BY VisitDate
For simplicity add order desc on date column and use TOP to get the first row only
SELECT TOP 1 *
FROM
(
SELECT PatientTreatmentVisitID,PatientMatchID,PatientID, VisitDate, RowNumber FROM cte
) as X
WHERE RowNumber = 2
order by VisitDate desc
I have a table which has below values:
If Sum of values = 0 with same ID I want to delete them from the table. So result should look like this:
The code I have:
DELETE FROM tmp_table
WHERE ID in
(SELECT ID
FROM tmp_table WITH(NOLOCK)
GROUP BY ID
HAVING SUM(value) = 0)
Only deletes rows with ID = 2.
UPD: Including additional example:
Rows in yellow needs to be deleted
Your query is working correctly because the only group to total zero is id 2, the others have sub-groups which total zero (such as the first two with id 1) but the total for all those records is -3.
What you're wanting is a much more complex algorithm to do "bin packing" in order to remove the sub groups which sum to zero.
You can do what you want using window functions -- by enumerating the values for each id. Taking your approach using a subquery:
with t as (
select t.*,
row_number() over (partition by id, value order by id) as seqnum
from tmp_table t
)
delete from t
where exists (select 1
from t t2
where t2.id = t.id and t2.value = - t.value and t2.seqnum = t.seqnum
);
You can also do this with a second layer of window functions:
with t as (
select t.*,
row_number() over (partition by id, value order by id) as seqnum
from tmp_table t
),
tt as (
select t.*, count(*) over (partition by id, abs(value), seqnum) as cnt
from t
)
delete from tt
where cnt = 2;
I have a project through which I'm creating a game powered by a database.
The database has data entered like this:
(ID, Name) || (1, PhotoID),(1,PhotoID),(1,PhotoID),(2,PhotoID),(2,PhotoID) and so on. There are thousands of entries.
This is my current SQL statement:
$sql = "SELECT TOP 8 * FROM Image WHERE Hidden = '0' ORDER BY NEWID()";
But this can also produce results with matching IDs, where I need to have each result have a unique ID (that is I need one result from each group).
How can I change my query to grab one result from each group?
Thanks!
Since ORDER BY NEWID() will result in tablescan anyway, you might use row_number() to isolate first in group:
; with randomizer as (
select id,
name,
row_number() over (partition by id
order by newid()) rn
from Image
where hidden = 0
)
select top 8
id,
name
from randomizer
where rn = 1
-- Added by mellamokb's suggestion to allow groups to be randomized
order by newid()
Sql Fiddle playground thanks to mellamokb.
Looks like this may work, but I can't vouch for performance:
SELECT TOP 8 ID,
(select top 1 name from image i2
where i2.id = i1.id order by newid())
FROM Image i1
WHERE hidden = '0'
group by ID
ORDER BY NEWID();
Demo: http://www.sqlfiddle.com/#!3/657ad/6
If you have an index on the ID column and want to take advantage of the index and avoid a full table scan, do your randomization on the key values first:
WITH IDs AS
(
SELECT DISTINCT ID
FROM Image
WHERE Hidden = '0'
),
SequencedIDs AS
(
SELECT ID, ROW_NUMBER() OVER (ORDER BY NEWID()) AS Seq
FROM IDs
),
ImageGroups AS
(
SELECT i.*, ROW_NUMBER() OVER (PARTITION BY i.ID ORDER BY NEWID()) Seq
FROM SequencedIDs s
INNER JOIN Image i
ON i.ID = s.ID
WHERE s.Seq < 8
AND i.Hidden = '0'
)
SELECT *
FROM ImageGroups
WHERE Seq = 1
This should drastically reduce the cost over the table scan approach, although I don't have a schema big enough that I can test with - so try running some statistics in SSMS and make sure ID is actually indexed for this to be effective.
select * from (select * from photos order by rand()) as _SUB group by _SUB.id;
select ID, Name from (select ID, Name, row_number() over
(partition by ID, Name order by ID) as ranker from Image where Hidden = 0 ) Z where ranker = 1
order by newID()