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
Related
I have two sets of tables (i.e. a.1, a.2, a.3, b.1, b.2, b.3, etc) created using slightly different logic. The analogous table in the two schemas have the exact same columns (i.e. a.1 has the same columns as b.1). My belief is that the tables in the two schemas should contain the exact same information, but I want to test that belief. Therefore I want to write a query that compares two analogous tables and returns lines that are not in both tables. Is there an easy way to write a query to do that without manually writing the join? In other words, can I have a query that can produce the results that I want where I only have to change the table names I want to compare while leaving the rest of the query unchanged?
To be a bit more explicit, I'm looking to do something like the following:
select *
from a.1
where (all columns in a.1) not in (select * from b.1);
If I could write something like this then all I would have to do to compare a.2 to b.2 would be to change the table names. However, it's not clear to me how to come up with the (all columns in a.1) piece in a general way.
Based on a recommendation in the comments, I've created the following showing the kind of thing I'd like to see:
https://dbfiddle.uk/?rdbms=db2_11.1&fiddle=ad0141b0daf8f8f92e6e3fa8d57e67ad
I was looking for the except clause.
So
select *
from a.1
where (all columns in a.1) not in (select * from b.1);
can be written as
select * from a.1
except
select * from b.1
In db-fiddle I give an explicit exmaple of what I wanted.
If you have a primary key to match rows between the tables, then you can try a full anti-join. For example:
select a.id as aid, b.id as bid
from a
full join b on b.id = a.id
where a.id is null or b.id is null
If the tables are:
A: 1, 2, 3
B: 1, 2, 4
The result is:
AID BID
---- ----
null 4 -- means it's present in B, but not in A
3 null -- means it's present in A, but not in B
See running example at db<>fiddle.
Of course, if your tables do not have a primary key, or if the rows are inconsistent (same PK, different data), then you'll need to adjust the query.
As an alternative you can try this:
select 'a1' t,* from (
select a1.*,row_number() over (partition by c1 order by 1) as rn from a1
minus
select b1.*,row_number() over (partition by c1 order by 1) as rn from b1
)
union all
select 'b1' t,* from (
select b1.*,row_number() over (partition by c1 order by 1) as rn from b1
minus
select a1.*,row_number() over (partition by c1 order by 1) as rn from a1
)
fiddle
edit: you can shorten the query by precalculating the rn part, instead of doing the same calculation again.
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
scenario 1:
I have two tables INFUSION_APP_APPOINTMENT,INFUSION_APP_NURSE_NOTES where
INFUSION_APP_NURSE_NOTES.APPOINTMENT_ID=INFUSION_APP_APPOINTMENT.ID and i want to find out the INFUSION_APP_NURSE_NOTES.ID's where INFUSION_APP_NURSE_NOTES.APPOINTMENT_ID is same.
for eg. if the INFUSION_APP_NURSE_NOTES.APPOINTMENT_ID = 1 and INFUSION_APP_NURSE_NOTES.ID is 12,15,78, then i want to display all the
INFUSION_APP_NURSE_NOTES.ID's where INFUSION_APP_NURSE_NOTES.APPOINTMENT_ID =1.
i use below script
SELECT INFUSION_APP_NURSE_NOTES.APPOINTMENT_ID,INFUSION_APP_NURSE_NOTES.ID
FROM INFUSION_APP_NURSE_NOTES
GROUP BY INFUSION_APP_NURSE_NOTES.APPOINTMENT_ID,INFUSION_APP_NURSE_NOTES.ID
HAVING COUNT(INFUSION_APP_NURSE_NOTES.APPOINTMENT_ID)>1
but it does not gives me any records.
scenario 2:
I am running below script with the intention to get the duplicate records with different INFUSION_APP_NURSE_NOTES.ID's but same INFUSION_APP_NURSE_NOTES.APPOINTMENT_ID.
SELECT INFUSION_APP_NURSE_NOTES.ID,INFUSION_APP_NURSE_NOTES.APPOINTMENT_ID,INFUSION_APP_NURSE_NOTES.TYPE
FROM INFUSION_APP_NURSE_NOTES
WHERE
EXISTS (
SELECT 1 FROM INFUSION_APP_APPOINTMENT
WHERE
INFUSION_APP_NURSE_NOTES.ENABLE=1
AND INFUSION_APP_NURSE_NOTES.APPOINTMENT_ID=INFUSION_APP_APPOINTMENT.ID
GROUP BY INFUSION_APP_NURSE_NOTES.ID
HAVING COUNT(INFUSION_APP_NURSE_NOTES.APPOINTMENT_ID)>1
)
ORDER BY INFUSION_APP_NURSE_NOTES.APPOINTMENT_ID;
but getting below error
SQL Error(164): Each GROUP BY expression must contain at least one
column that is not an outer reference
how to solve it?
i want the only row which has common APPOINTMENT_ID but different n
The question is unclear. Finding duplicates is typically performed using ranking functions like ROW_NUMBER(). This query :
SELECT *,ROW_NUMBER(PARTITION BY APPOINTMENT_ID ORDER BYID) as RN
FROM INFUSION_APP_NURSE_NOTES
WHERE
ENABLE=1
Will rank notes for the same appointment by ID and return 1, 2, 3 etc starting from the earliest note. ORDER BY ID DESC would return 1 for the latest note.
This can be used in a subquery or CTE to find the first, last or or duplicate records, eg :
with notes as (
SELECT *,ROW_NUMBER(PARTITION BY APPOINTMENT_ID ORDER BYID) as RN
FROM INFUSION_APP_NURSE_NOTES
WHERE
ENABLE=1
)
select *
from notes
where RN=1
Will return the first note per appointment while :
where RN>1
Will return only duplicates.
The question doesn't say what should be done with the duplicates though.
If the question is how to return all notes from appointments with multiple notes, a subquery can be used to return the APPOINTMENT_IDs that have more than one note. There's no need to include the INFUSION_APP_APPOINTMENT table though :
SELECT *
FROM INFUSION_APP_NURSE_NOTES
where
ENABLE=1 AND
APPOINTMENT_ID IN ( SELECT APPOINTMENT_ID
FROM INFUSION_APP_NURSE_NOTES
WHERE
ENABLE=1
group by APPOINTMENT_ID
having count(*)>1)
Try this
SELECT INFUSION_APP_NURSE_NOTES.ID,INFUSION_APP_NURSE_NOTES.APPOINTMENT_ID,INFUSION_APP_NURSE_NOTES.TYPE
FROM INFUSION_APP_NURSE_NOTES
WHERE
EXISTS (
SELECT COUNT(B.APPOINTMENT_ID), B.ID
FROM INFUSION_APP_APPOINTMENT A
INNER JOIN INFUSION_APP_NURSE_NOTES B ON B.APPOINTMENT_ID = A.ID
WHERE
B.ENABLE=1
GROUP BY B.ID
HAVING COUNT(B.APPOINTMENT_ID)>1
)
ORDER BY INFUSION_APP_NURSE_NOTES.APPOINTMENT_ID;
I need to update the following query so that it only returns one child record (remittance) per parent (claim).
Table Remit_To_Activate contains exactly one date/timestamp per claim, which is what I wanted.
But when I join the full Remittance table to it, since some claims have multiple remittances with the same date/timestamps, the outermost query returns more than 1 row per claim for those claim IDs.
SELECT * FROM REMITTANCE
WHERE BILLED_AMOUNT>0 AND ACTIVE=0
AND REMITTANCE_UUID IN (
SELECT REMITTANCE_UUID FROM Claims_Group2 G2
INNER JOIN Remit_To_Activate t ON (
(t.ClaimID = G2.CLAIM_ID) AND
(t.DATE_OF_LATEST_REGULAR_REMIT = G2.CREATE_DATETIME)
)
where ACTIVE=0 and BILLED_AMOUNT>0
)
I believe the problem would be resolved if I included REMITTANCE_UUID as a column in Remit_To_Activate. That's the REAL issue. This is how I created the Remit_To_Activate table (trying to get the most recent remittance for a claim):
SELECT MAX(create_datetime) as DATE_OF_LATEST_REMIT,
MAX(claim_id) AS ClaimID,
INTO Latest_Remit_To_Activate
FROM Claims_Group2
WHERE BILLED_AMOUNT>0
GROUP BY Claim_ID
ORDER BY Claim_ID
Claims_Group2 contains these fields:
REMITTANCE_UUID,
CLAIM_ID,
BILLED_AMOUNT,
CREATE_DATETIME
Here are the 2 rows that are currently giving me the problem--they're both remitts for the SAME CLAIM, with the SAME TIMESTAMP. I only want one of them in the Remits_To_Activate table, so only ONE remittance will be "activated" per Claim:
enter image description here
You can change your query like this:
SELECT
p.*, latest_remit.DATE_OF_LATEST_REMIT
FROM
Remittance AS p inner join
(SELECT MAX(create_datetime) as DATE_OF_LATEST_REMIT,
claim_id,
FROM Claims_Group2
WHERE BILLED_AMOUNT>0
GROUP BY Claim_ID
ORDER BY Claim_ID) as latest_remit
on latest_remit.claim_id = p.claim_id;
This will give you only one row. Untested (so please run and make changes).
Without having more information on the structure of your database -- especially the structure of Claims_Group2 and REMITTANCE, and the relationship between them, it's not really possible to advise you on how to introduce a remittance UUID into DATE_OF_LATEST_REMIT.
Since you are using SQL Server, however, it is possible to use a window function to introduce a synthetic means to choose among remittances having the same timestamp. For example, it looks like you could approach the problem something like this:
select *
from (
select
r.*,
row_number() over (partition by cg2.claim_id order by cg2.create_datetime desc) as rn
from
remittance r
join claims_group2 cg2
on r.remittance_uuid = cg2.remittance_uuid
where
r.active = 0
and r.billed_amount > 0
and cg2.active = 0
and cg2.billed_amount > 0
) t
where t.rn = 1
Note that that that does not depend on your DATE_OF_LATEST_REMIT table at all, it having been subsumed into the inline view. Note also that this will introduce one extra column into your results, though you could avoid that by enumerating the columns of table remittance in the outer select clause.
It also seems odd to be filtering on two sets of active and billed_amount columns, but that appears to follow from what you were doing in your original queries. In that vein, I urge you to check the results carefully, as lifting the filter conditions on cg2 columns up to the level of the join to remittance yields a result that may return rows that the original query did not (but never more than one per claim_id).
A co-worker offered me this elegant demonstration of a solution. I'd never used "over" or "partition" before. Works great! Thank you John and Gaurasvsa for your input.
if OBJECT_ID('tempdb..#t') is not null
drop table #t
select *, ROW_NUMBER() over (partition by CLAIM_ID order by CLAIM_ID) as ROW_NUM
into #t
from
(
select '2018-08-15 13:07:50.933' as CREATE_DATE, 1 as CLAIM_ID, NEWID() as
REMIT_UUID
union select '2018-08-15 13:07:50.933', 1, NEWID()
union select '2017-12-31 10:00:00.000', 2, NEWID()
) x
select *
from #t
order by CLAIM_ID, ROW_NUM
select CREATE_DATE, MAX(CLAIM_ID), MAX(REMIT_UUID)
from #t
where ROW_NUM = 1
group by CREATE_DATE
I'm having a bit of a weird question, given to me by a client.
He has a list of data, with a date between parentheses like so:
Foo (14/08/2012)
Bar (15/08/2012)
Bar (16/09/2012)
Xyz (20/10/2012)
However, he wants the list to be displayed as follows:
Foo (14/08/2012)
Bar (16/09/2012)
Bar (15/08/2012)
Foot (20/10/2012)
(notice that the second Bar has moved up one position)
So, the logic behind it is, that the list has to be sorted by date ascending, EXCEPT when two rows have the same name ('Bar'). If they have the same name, it must be sorted with the LATEST date at the top, while staying in the other sorting order.
Is this even remotely possible? I've experimented with a lot of ORDER BY clauses, but couldn't find the right one. Does anyone have an idea?
I should have specified that this data comes from a table in a sql server database (the Name and the date are in two different columns). So I'm looking for a SQL-query that can do the sorting I want.
(I've dumbed this example down quite a bit, so if you need more context, don't hesitate to ask)
This works, I think
declare #t table (data varchar(50), date datetime)
insert #t
values
('Foo','2012-08-14'),
('Bar','2012-08-15'),
('Bar','2012-09-16'),
('Xyz','2012-10-20')
select t.*
from #t t
inner join (select data, COUNT(*) cg, MAX(date) as mg from #t group by data) tc
on t.data = tc.data
order by case when cg>1 then mg else date end, date desc
produces
data date
---------- -----------------------
Foo 2012-08-14 00:00:00.000
Bar 2012-09-16 00:00:00.000
Bar 2012-08-15 00:00:00.000
Xyz 2012-10-20 00:00:00.000
A way with better performance than any of the other posted answers is to just do it entirely with an ORDER BY and not a JOIN or using CTE:
DECLARE #t TABLE (myData varchar(50), myDate datetime)
INSERT INTO #t VALUES
('Foo','2012-08-14'),
('Bar','2012-08-15'),
('Bar','2012-09-16'),
('Xyz','2012-10-20')
SELECT *
FROM #t t1
ORDER BY (SELECT MIN(t2.myDate) FROM #t t2 WHERE t2.myData = t1.myData), T1.myDate DESC
This does exactly what you request and will work with any indexes and much better with larger amounts of data than any of the other answers.
Additionally it's much more clear what you're actually trying to do here, rather than masking the real logic with the complexity of a join and checking the count of joined items.
This one uses analytic functions to perform the sort, it only requires one SELECT from your table.
The inner query finds gaps, where the name changes. These gaps are used to identify groups in the next query, and the outer query does the final sorting by these groups.
I have tried it here (SQL Fiddle) with extended test-data.
SELECT name, dat
FROM (
SELECT name, dat, SUM(gap) over(ORDER BY dat, name) AS grp
FROM (
SELECT name, dat,
CASE WHEN LAG(name) OVER (ORDER BY dat, name) = name THEN 0 ELSE 1 END AS gap
FROM t
) x
) y
ORDER BY grp, dat DESC
Extended test-data
('Bar','2012-08-12'),
('Bar','2012-08-11'),
('Foo','2012-08-14'),
('Bar','2012-08-15'),
('Bar','2012-08-16'),
('Bar','2012-09-17'),
('Xyz','2012-10-20')
Result
Bar 2012-08-12
Bar 2012-08-11
Foo 2012-08-14
Bar 2012-09-17
Bar 2012-08-16
Bar 2012-08-15
Xyz 2012-10-20
I think that this works, including the case I asked about in the comments:
declare #t table (data varchar(50), [date] datetime)
insert #t
values
('Foo','20120814'),
('Bar','20120815'),
('Bar','20120916'),
('Xyz','20121020')
; With OuterSort as (
select *,ROW_NUMBER() OVER (ORDER BY [date] asc) as rn from #t
)
--Now we need to find contiguous ranges of the same data value, and the min and max row number for such a range
, Islands as (
select data,rn as rnMin,rn as rnMax from OuterSort os where not exists (select * from OuterSort os2 where os2.data = os.data and os2.rn = os.rn - 1)
union all
select i.data,rnMin,os.rn
from
Islands i
inner join
OuterSort os
on
i.data = os.data and
i.rnMax = os.rn-1
), FullIslands as (
select
data,rnMin,MAX(rnMax) as rnMax
from Islands
group by data,rnMin
)
select
*
from
OuterSort os
inner join
FullIslands fi
on
os.rn between fi.rnMin and fi.rnMax
order by
fi.rnMin asc,os.rn desc
It works by first computing the initial ordering in the OuterSort CTE. Then, using two CTEs (Islands and FullIslands), we compute the parts of that ordering in which the same data value appears in adjacent rows. Having done that, we can compute the final ordering by any value that all adjacent values will have (such as the lowest row number of the "island" that they belong to), and then within an "island", we use the reverse of the originally computed sort order.
Note that this may, though, not be too efficient for large data sets. On the sample data it shows up as requiring 4 table scans of the base table, as well as a spool.
Try something like...
ORDER BY CASE date
WHEN '14/08/2012' THEN 1
WHEN '16/09/2012' THEN 2
WHEN '15/08/2012' THEN 3
WHEN '20/10/2012' THEN 4
END
In MySQL, you can do:
ORDER BY FIELD(date, '14/08/2012', '16/09/2012', '15/08/2012', '20/10/2012')
In Postgres, you can create a function FIELD and do:
CREATE OR REPLACE FUNCTION field(anyelement, anyarray) RETURNS numeric AS $$
SELECT
COALESCE((SELECT i
FROM generate_series(1, array_upper($2, 1)) gs(i)
WHERE $2[i] = $1),
0);
$$ LANGUAGE SQL STABLE
If you do not want to use the CASE, you can try to find an implementation of the FIELD function to SQL Server.