I have a table, myTable that has two fields in it ID and patientID. The same patientID can be in the table more than once with a different ID. How can I make sure that I get only ONE instance of every patientID.?
EDIT: I know this isn't perfect design, but I need to get some info out of the database and today and then fix it later.
You could use a CTE with ROW_NUMBER function:
WITH CTE AS(
SELECT myTable.*
, RN = ROW_NUMBER()OVER(PARTITION BY patientID ORDER BY ID)
FROM myTable
)
SELECT * FROM CTE
WHERE RN = 1
It sounds like you're looking for DISTINCT:
SELECT DISTINCT patientID FROM myTable
you can get the same "effect" with GROUP BY:
SELECT patientID FROM myTable GROUP BY patientID
The simple way would be to add LIMIT 1 to the end of your query. This will ensure only a single row is returned in the result set.
WITH CTE AS
(
SELECT tableName.*,ROW_NUMBER() OVER(PARTITION BY patientID ORDER BY patientID) As 'Position' FROM tableName
)
SELECT * FROM CTE
WHERE
Position = 1
Related
How can I get the Latest amount, I already had some queries but instead it shows two records ,Im expecting to show only the the '7370' current amount
you can use correlated subquery
select * from tablename a where lasttime in (select max(lasttime) from tablename b where a.id=b.id)
OR you can use row_number()
select * from
(
select *,row_number() over(partition by id order by lasttime desc) as rn from tablename
)A where rn=1
Just add Top 1 before your fields.
Select TOP 1 fields from table
SELECT TOP 1 currentBalance FROM DBO.tbl_billing ORDER BY [date]
I've seen many topics about this and none of them is what I'm looking for.
Say we have this simple table:
CREATE TABLE A (
id INT,
date DATETIME
);
I want to retrieve the MAX value after grouping.
So I do it as follow:
DECLARE #tmpTable TABLE(id INT, count INT);
INSERT INTO #tmpTable SELECT id, COUNT(*) FROM A GROUP BY id;
SELECT MAX(count) FROM #tmpTable;
Is there a better way of doing that?
I've seen a solution in a book that I'm reading that they do it as follows:
SELECT MAX(count) FROM (SELECT COUNT(*) AS count FROM A GROUP BY id);
But this won't work :/ Could be that it works in newer T-SQL servers? Currently I'm using 2008 R2.
You can make use of TOP
SELECT TOP 1 Id,COUNT(*) AS MAXCOUNT
FROM A
GROUP BY Id
ORDER BY MAXCOUNT DESC
If you wants the result with same max count use TOP WITH TIES
SELECT TOP 1 WITH TIES Id,COUNT(*) AS MAXCOUNT
FROM A
GROUP BY Id
ORDER BY MAXCOUNT DESC
Is there a better way of doing that?
We could try using analytic functions:
WITH cte AS (
SELECT id, COUNT(*) cnt, ROW_NUMBER() OVER (ORDER BY COUNT(*) DESC) rn
FROM A
GROUP BY id
)
SELECT cnt
FROM cte
WHERE rn = 1;
This approach is to turn out a row number, ordered descending by the count, during your original aggregation query by id. The id with the highest count then should be the first record (and this result should hold valid even if more than one id be tied for the highest count).
Regarding your original max query, see the answer by #apomene, and you are just missing an alias.
You also need to add an alias name for your sub query. Try like:
SELECT MAX(sub.count1) FROM (SELECT COUNT(*) AS count1 FROM A GROUP BY id) sub;
Is there any optimised way in sql sever to optimse this code, I am trying to find 2nd duplicate
WITH CTE AS (
SELECT *,
ROW_NUMBER() OVER(PARTITION BY id,AN_KEY ORDER BY [ENTITYID]) AS [rn]
FROM [data].[dbo].[TRANSFER]
)
select *
INTO dbo.#UpSingle
from CTE
where RN=2
UPDATE:
As GurV pointed out - this query doesn't solve the problem. It will only give you the items that have exactly two duplicates, but not the row where the second duplicate lies.
I am just going to leave this here from reference purposes.
Original Answer
Why not try something like this from another SO post: Finding duplicate values in a SQL table
SELECT
id, AN_KEY, COUNT(*)
FROM
[data].[dbo].[TRANSFER]
GROUP BY
id, AN_KEY
HAVING
COUNT(*) = 2
I gather from your original SQL that the cols you would want to group by on are :
Id
AN_KEY
Here is another way to get the the second duplicate row (in the order of increasing ENTITYID of course):
select *
from [data].[dbo].[TRANSFER] a
where [ENTITYID] = (
select min([ENTITYID])
from [data].[dbo].[TRANSFER] b
where [ENTITYID] > (
select min([ENTITYID])
from [data].[dbo].[TRANSFER] c
where b.id = c.id
and b.an_key = c.an_key
)
and a.id = b.id
and a.an_key = b.an_key
)
Provided there is an index on id, an_key and ENTITYID columns, performance of both your query and this should be acceptable.
Let me assume that this query does what you want:
WITH CTE AS (
SELECT t.*,
ROW_NUMBER() OVER (PARTITION BY id, AN_KEY
ORDER BY [ENTITYID]) AS [rn]
FROM [data].[dbo].[TRANSFER] t
)
SELECT *
INTO dbo.#UpSingle
FROM CTE
WHERE RN = 2;
For performance, you want a composite index on [data].[dbo].[TRANSFER](id, AN_KEY, ENTITYID).
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
I'm working with Sql server 2008.i have a table contains following columns,
Id,
Name,
Date
this table contains more than one record for same id.i want to get distinct id having maximum date.how can i write sql query for this?
Use the ROW_NUMBER() function and PARTITION BY clause. Something like this:
SELECT Id, Name, Date FROM (
SELECT *, ROW_NUMBER() OVER (PARTITION BY Id ORDER BY Date desc) AS ROWNUM
FROM [MyTable]
) x WHERE ROWNUM = 1
If you need only ID column and other columns are NOT required, then you don't need to go with ROW_NUMBER or MAX or anything else. You just do a Group By over ID column, because whatever the maximum date is you will get same ID.
SELECT ID FROM table GROUP BY ID
--OR
SELECT DISTINCT ID FROM table
If you need ID and Date columns with maximum date, then simply do a Group By on ID column and select the Max Date.
SELECT ID, Max(Date) AS Date
FROM table
GROUP BY ID
If you need all the columns but 1 line having Max. date then you can go with ROW_NUMBER or MAX as mentioned in other answers.
SELECT *
FROM table AS M
WHERE Exists(
SELECT 1
FROM table
WHERE ID = M.ID
HAVING M.Date = Max(Date)
)
One way, using ROW_NUMBER:
With CTE As
(
SELECT Id, Name, Date, Rn = Row_Number() Over (Partition By Id
Order By Date DESC)
FROM dbo.TableName
)
SELECT Id --, Name, Date
FROM CTE
WHERE Rn = 1
If multiple max-dates are possible and you want all you could use DENSE_RANK instead.
Here's an overview of sql-server's ranking function: http://technet.microsoft.com/en-us/library/ms189798.aspx
By the way, CTE is a common-table-expression which is similar to a named sub-query. I'm using it to be able to filter by the row_number. This approach allows to select all columns if you want.
select Max(Date) as "Max Date"
from table
group by Id
order by Id
Try with Max(Date) and GROUP BY the other two columns (the ones with repeating data)..
SELECT ID, Max(Date) as date, Name
FROM YourTable
GROUP BY ID, Name
You may try with this
DECLARE #T TABLE(ID INT, NAME VARCHAR(50),DATE DATETIME)
INSERT INTO #T VALUES(1,'A','2014-04-20'),(1,'A','2014-04-28')
,(2,'A2','2014-04-22'),(2,'A2','2014-04-24')
,(3,'A3','2014-04-20'),(3,'A3','2014-04-28')
,(4,'A4','2014-04-28'),(4,'A4','2014-04-28')
,(5,'A5','2014-04-28'),(5,'A5','2014-04-28')
SELECT T.ID FROM #T T
WHERE T.DATE=(SELECT MAX(A.DATE)
FROM #T A
WHERE A.ID=T.ID
GROUP BY A.ID )
GROUP BY T.ID
select id, max(date) from NameOfYourTable group by id;