I want to reverse the order in SQLÂ Server of results after using desc. For example:
SELECT TOP 3 * FROM table ORDER BY id DESC
returns results:
505
504
503
But then I want to flip the results to look like this:
503
504
505
I tried
SELECT * FROM (SELECT TOP 3 * FROM table ORDER BY id DESC) ORDER BY id ASC
But that did not work. How can I fix it?
That should work as long as you alias the subquery.
SELECT q.*
FROM (SELECT TOP 3 *
FROM table
ORDER BY id DESC) q
ORDER BY q.id ASC
I think you forgot the subselect alias
SELECT *
FROM (
SELECT TOP 3 *
FROM table
ORDER BY id DESC
) s
ORDER BY id ASC
SELECT * FROM (SELECT TOP 3 * FROM table ORDER BY id DESC) As AliasName ORDER BY id ASC
;WITH cte
AS
(
SELECT *, ROW_NUMBER() OVER(ORDER BY id DESC) rank
FROM table
)
SELECT *
FROM cte
WHERE rank <= 3
ORDER BY id ASC
SELECT * FROM (SELECT TOP 3 * FROM table ORDER BY id DESC) AS r ORDER BY r.id ASC
I figured it out. I needed to make a temporary table and have a name using AS.
SELECT *
FROM (
SELECT *
FROM table
ORDER BY ID DESC
) TMP
ORDER BY TMP.ID ASC
Related
I need to retrieve the last few entries from a table. I can retrieve them using:
SELECT TOP n *
FROM table
ORDER BY id DESC
That I looked everywhere and that's the only answer I could find, But that way I get them in reverse order. I need them in the same order as they are in the table because it's for a messaging interface.
Use a derived table:
select id, ...
from
(
select top n id, ...
from t
order by id desc
) dt
order by id
I suggest you to use a ROW_NUMBER() like this:
SELECT *
FROM (
SELECT
*, ROW_NUMBER() OVER (ORDER BY id DESC) AS RowNo
FROM
yourTable
) AS t
WHERE
(RowNO < #n)
ORDER BY
id
My requirement is to get each client's latest order, and then get top 100 records.
I wrote one query as below to get latest orders for each client. Internal query works fine. But I don't know how to get first 100 based on the results.
SELECT * FROM (
SELECT id, client_id, ROW_NUMBER() OVER(PARTITION BY client_id ORDER BY create_time DESC) rn
FROM order
) WHERE rn=1
Any ideas? Thanks.
Assuming that create_time contains the time the order was created, and you want the 100 clients with the latest orders, you can:
add the create_time in your innermost query
order the results of your outer query by the create_time desc
add an outermost query that filters the first 100 rows using ROWNUM
Query:
SELECT * FROM (
SELECT * FROM (
SELECT
id,
client_id,
create_time,
ROW_NUMBER() OVER(PARTITION BY client_id ORDER BY create_time DESC) rn
FROM order
)
WHERE rn=1
ORDER BY create_time desc
) WHERE rownum <= 100
UPDATE for Oracle 12c
With release 12.1, Oracle introduced "real" Top-N queries. Using the new FETCH FIRST... syntax, you can also use:
SELECT * FROM (
SELECT
id,
client_id,
create_time,
ROW_NUMBER() OVER(PARTITION BY client_id ORDER BY create_time DESC) rn
FROM order
)
WHERE rn = 1
ORDER BY create_time desc
FETCH FIRST 100 ROWS ONLY)
you should use rownum in oracle to do what you seek
where rownum <= 100
see also those answers to help you
limit in oracle
select top in oracle
select top in oracle 2
As Moneer Kamal said, you can do that simply:
SELECT id, client_id FROM order
WHERE rownum <= 100
ORDER BY create_time DESC;
Notice that the ordering is done after getting the 100 row. This might be useful for who does not want ordering.
Update:
To use order by with rownum you have to write something like this:
SELECT * from (SELECT id, client_id FROM order ORDER BY create_time DESC) WHERE rownum <= 100;
First 10 customers inserted into db (table customers):
select * from customers where customer_id <=
(select min(customer_id)+10 from customers)
Last 10 customers inserted into db (table customers):
select * from customers where customer_id >=
(select max(customer_id)-10 from customers)
Hope this helps....
To select top n rows updated recently
SELECT *
FROM (
SELECT *
FROM table
ORDER BY UpdateDateTime DESC
)
WHERE ROWNUM < 101;
Try this:
SELECT *
FROM (SELECT * FROM (
SELECT
id,
client_id,
create_time,
ROW_NUMBER() OVER(PARTITION BY client_id ORDER BY create_time DESC) rn
FROM order
)
WHERE rn=1
ORDER BY create_time desc) alias_name
WHERE rownum <= 100
ORDER BY rownum;
Or TOP:
SELECT TOP 2 * FROM Customers; //But not supported in Oracle
NOTE: I suppose that your internal query is fine. Please share your output of this.
I used this code to select last 5 rows from table:
SELECT TOP(5) * FROM tbl_reg ORDER BY Id DESC
But I want to reorder this result ( the last five rows returned ) By Id ASC
How can I do this?
Use a subquery:
select t.*
from (SELECT TOP(5) * FROM tbl_reg ORDER BY Id DESC) t
order by id ASC;
Here is my query:
SELECT TOP 8 id, rssi1, date
FROM history
WHERE (siteName = 'CCL03412')
ORDER BY id DESC
This the result:
How can I reverse this table based on date (Column2) by using SQL?
You can use the first query to get the matching ids, and use them as part of an IN clause:
SELECT id, rssi1, date
FROM history
WHERE id IN
(
SELECT TOP 8 id
FROM history
WHERE (siteName = 'CCL03412')
ORDER BY id DESC
)
ORDER BY date ASC
You could simply use a sub-query. If you apply a TOP clause the nested ORDER BY is allowed:
SELECT X.* FROM(
SELECT TOP 8 id, Column1, Column2
FROM dbo.History
WHERE (siteName = 'CCL03412')
ORDER BY id DESC) X
ORDER BY Column2
Demo
The SELECT query of a subquery is always enclosed in parentheses. It
cannot include a COMPUTE or FOR BROWSE clause, and may only include an
ORDER BY clause when a TOP clause is also specified.
Subquery Fundamentals
try the below :
select * from (SELECT TOP 8 id, rssi1, date
FROM history
WHERE (siteName = 'CCL03412')
ORDER BY id DESC ) aa order by aa.date DESC
didn't run it, but i think it should go well
WITH cte AS
(
SELECT id, rssi1, date, RANK() OVER (ORDER BY ID DESC) AS Rank
FROM history
WHERE (siteName = 'CCL03412')
)
SELECT id, rssi1, date
FROM cte
WHERE Rank <= 8
ORDER BY Date DESC
I have not run this but i think it will work. Execute and let me know if you face error
select id, rssi1, date from (SELECT TOP 8 id, rssi1, date
FROM history
WHERE (siteName = 'CCL03412')
ORDER BY id DESC) order by date ;
I need a SQL query that returns the top 2 Plans by PlanDate per ClientID. This is all on one table where PlanID is the PrimaryID, ClientID is a foreignID.
This is what I have so far -->
SELECT *
FROM [dbo].[tblPlan]
WHERE [PlanID] IN (SELECT TOP (2) PlanID FROM [dbo].[tblPlan] ORDER BY [PlanDate] DESC)
This, obviously, only returns 2 records where I actually need up to 2 records per ClientID.
This can be done using ROW_NUMBER:
SELECT PlanId, ClientId, PlanDate FROM (
SELECT ROW_NUMBER() OVER (PARTITION BY ClientId ORDER BY PlanDate DESC) rn, *
FROM [dbo].[tblPlan]
) AS T1
WHERE rn <=2
Add any other columns you need to the select to get those too.
Edit, Dec 2011. Corrected CROSS APPLY solution
Try both to see what is best
SELECT *
FROM
( -- distinct ClientID values
SELECT DISTINCT ClientID
FROM [dbo].[tblPlan]
) P1
CROSS APPLY
( -- top 2 per ClientID
SELECT TOP (2) P2.PlanID
FROM [dbo].[tblPlan] P2
WHERE P1.ClientID = P2.ClientID
ORDER BY P2.[PlanDate] DESC
) foo
Or
;WITH cTE AS (
SELECT
*,
ROW_NUMBER () OVER (PARTITION BY clientid ORDER BY [PlanDate] DESC) AS Ranking
FROM
[dbo].[tblPlan]
)
SELECT * FROM cTE WHERE Ranking <= 2