There is a few posts about it but i can't make it work...
I just want to select just one row per ID, something like row_number() over Partition in oracle but in access.
ty
SELECT a.*
FROM DATA as a
WHERE a.a_sku = (SELECT top 1 b.a_sku
FROM DATA as b
WHERE a.a_sku = b.a_sku)
but i get the same table Data out of it
Sample of table DATA
https://ibb.co/X4492fY
You should try below query -
SELECT a.*
FROM DATA as a
WHERE a.Active = (SELECT b.Active
FROM DATA as b
WHERE a.a_sku = b.a_sku
AND a.Active < b.Active)
If you don't care which record within each group of records with a matching a_sku values is returned, you can use the First or Last functions, e.g.:
select t.a_sku, first(t.field2), first(t.field3), ..., first(t.fieldN)
from data t
group by t.a_sku
Related
I've created a part of the query that returns me the data like in the picture below:
Now, I am trying to select First 2 records (1 and 2) of each group (sap_id, wr_nbr) where "rn" has more than 1.
So, my final table should look like:
I've tried with TOP 2 WITH TIES but it returns me only two records of the whole table.
Any idea how to achieve this?
Thank you in advance.
SELECT b.*
FROM
(SELECT a.[sap_id]
,a.[wr_nbr]
,a.[start_date]
,a.[end_date]
,a.[vs_ind]
,a.[rn]
,COUNT(*) OVER (PARTITION BY a.sap_id, a.wr_nbr) as count_rows
FROM
(SELECT [sap_id]
,[ts_nbr]
,[wr_nbr]
,[check_line]
,[check_nbr]
,[start_date]
,[end_date]
,[vs_ind]
,[rn]
,[rank_ind]
FROM [dbo].[first_two]) a) b
WHERE b.count_rows > 1
AND b.rn <= 2
Final table looks like:
I have the following query:
SELECT
A.POSTCARD_ID, A.STAMP_ID, B.END_DT
FROM
PST_VS_STAMP A
JOIN
STAMP B ON A.POSTCARD_ID = B.POSTCARD_ID
WHERE
B.ACCOUNT LIKE 'AA%'
AND B.END_DT = '9999-12-31'
GROUP BY
A.POSTCARD_ID, A.STAMP_ID, B.END_DT
HAVING
COUNT(A.POSTCARD_ID) < 2
But I get the wrong results.
I want only the postcards ID's where there is 1 record (HAVING < 2) in the PST_VS_STAMP table. How can I query this?
Do the aggregation in the subquery, only on the table where you want one row. Because there is one row, you can use an aggregation function to pull out the value of any column (for one row min(col) is the column's value):
select s.postcard_id, vs.stamp_id, s.end_dt
from stamp s join
(select vs.postcard_id, min(stamp_id) as stamp_id
from pst_vs_stamp vs
group by vs.postcard_id
having count(*) = 1
) s
on vs.POSTCARD_ID = s.POSTCARD_ID
where s.ACCOUNT like 'AA%' and s.END_DT = '9999-12-31';
Query:
SELECT *
FROM [MemberBackup].[dbo].[OriginalBackup]
where ration_card_id in
(
1247881,174772,
808454,2326154
)
Right now the data is ordered by the auto id or whatever clause I'm passing in order by.
But I want the data to come in sequential format as per id's I have passed
Expected Output:
All Data for 1247881
All Data for 174772
All Data for 808454
All Data for 2326154
Note:
Number of Id's to be passed will 300 000
One option would be to create a CTE containing the ration_card_id values and the orders which you are imposing, and the join to this table:
WITH cte AS (
SELECT 1247881 AS ration_card_id, 1 AS position
UNION ALL
SELECT 174772, 2
UNION ALL
SELECT 808454, 3
UNION ALL
SELECT 2326154, 4
)
SELECT t1.*
FROM [MemberBackup].[dbo].[OriginalBackup] t1
INNER JOIN cte t2
ON t1.ration_card_id = t2.ration_card_id
ORDER BY t2.position DESC
Edit:
If you have many IDs, then neither the answer above nor the answer given using a CASE expression will suffice. In this case, your best bet would be to load the list of IDs into a table, containing an auto increment ID column. Then, each number would be labelled with a position as its record is being loaded into your database. After this, you can join as I have done above.
If the desired order does not reflect a sequential ordering of some preexisting data, you will have to specify the ordering yourself. One way to do this is with a case statement:
SELECT *
FROM [MemberBackup].[dbo].[OriginalBackup]
where ration_card_id in
(
1247881,174772,
808454,2326154
)
ORDER BY CASE ration_card_id
WHEN 1247881 THEN 0
WHEN 174772 THEN 1
WHEN 808454 THEN 2
WHEN 2326154 THEN 3
END
Stating the obvious but note that this ordering most likely is not represented by any indexes, and will therefore not be indexed.
Insert your ration_card_id's in #temp table with one identity column.
Re-write your sql query as:
SELECT a.*
FROM [MemberBackup].[dbo].[OriginalBackup] a
JOIN #temps b
on a.ration_card_id = b.ration_card_id
order by b.id
Please find below image for make understanding my issues. I have a table as shown below picture. I need to get only highlighted (yellow) records. What is the best method to find these records?
In SQL Server 2012+, you can use the lead() and lag() functions. However, this is not available in SQL Server 2008. Here is a method using outer apply:
select t.*
from t outer apply
(select top 1 tprev.*
from t tprev
on tprev.time < t.time
order by tprev.time desc
) tprev outer apply
(select top 1 tnext.*
from t tnext
on tnext.time > t.time
order by tnext.time asc
)
where (t.cardtype = 1 and tnext.cardtype = 2) or
(t.cardtype = 2 and tprev.cardtype = 1);
With your sample data, it would also be possible to use self joins on the id column. This seems unsafe, though, because there could be gaps in that columns values.
Havent tried this, but I think it should work. First, make a view of the table in your question, with the rownumber included as one column:
CREATE VIEW v AS
SELECT
ROW_NUMBER() OVER(ORDER BY id) AS rownum,
id,
time,
card,
card_type
FROM table
Then, you can get all the rows of type 1 followed by a row of type 2 like this:
SELECT
a.id,
-- And so on...
FROM v AS a
JOIN v AS b ON b.rownum = a.rownum + 1
WHERE a.card_type = 1 AND b.card_type = 2
And all the rows of type 2 preceded by a row of type 1 like this:
SELECT
b.id,
-- And so on...
FROM v AS b
JOIN v AS a ON b.rownum = a.rownum + 1
WHERE a.card_type = 1 AND b.card_type = 2
To get them both in the same set of results, you can just use UNION ALL. Technically, you don't need the view. You could use nested selects instead, but since you will need to query the table four times it might be nice to have it as a view.
Also, if the ID is continous (it goes 1, 2, 3 without any gaps), you don't need the rownum and can just use the ID instead.
here is a code you can run in sql server
select * from Table_name where id in (1,2,6,7,195,160,164,165)
i have table as shown here in this picture --> http://www.directupload.net/file/d/3710/lj7etq5j_png.htm
I need the correct Query to get only data_id 10.
The query should be like this: Compare the latest date rows (2014-08-08) with the earliest date rows (2014-08-06). If there is a row on 2014-08-08 which is NOT at 2014-08-06, this row should returned.
I already tried it with self-joins and Sub-Selects, but i did't get it work.
Thx for any help!
Maybe something like this is what you're looking for?
select * from Table1
where
data not in (
select data from Table1
where dataOfDate = (select min(dataofdate) from Table1)
)
and dataOfDate = (select max(dataofdate) from Table1)
The first where clause compares the data field of the returned rows to the data field in the set of oldest rows and the second where clause limits the set of rows the the newest.
Note that I'm only comparing rows based on thedatafield, so you might have to change the query if you want to includenextTableIdin the comparison.
Here is a sample SQL Fiddle.
How about something like this:
SELECT d1.* FROM Dates d1 LEFT JOIN Dates d2 ON d1.nextTableId = d2.nextTableId WHERE d1.dataofDate = '2014-08-08' AND d2.dataofDate = '2014-08-06' AND d2.data_id IS NULL;
SELECT `data_id`
FROM `my_table`
WHERE `dataOfDate` = (SELECT MAX(`dataOfDate`) FROM `my_table`)
AND `nextTableId` NOT IN (
SELECT `nextTableId` FROM `my_table` WHERE `dataOfDate` = (SELECT MIN(`dataOfDate`) FROM `my_table`)
)
select all rows with max date
that don't have values amongst rows with min date
edit ops, too late