Of course my data set is more complex, but this is essentially what I have:
+--------+--------+-------+
| SEQ_NO | FILTER | VALUE |
+--------+--------+-------+
| 1 | 'A' | 5 |
| 2 | 'A' | 10 |
| 3 | 'A' | 15 |
+--------+--------+-------+
Here is my query:
SELECT MAX(SEQ_NO)
, FILTER
, VALUE
FROM TABLE
GROUP BY FILTER
, VALUE
This returns my entire data set. How can I alter my query so that it only returns the record with the highest SEQ_NO ?
SELECT t1.*
FROM Table AS t1
INNER JOIN
(
SELECT MAX(SEQ_NO) MAXSeq
, FILTER
, VALUE
FROM TABLE
GROUP BY FILTER
, VALUE
) t2 ON t1.SEQ_NO = t2.MAXSeq
AND t1.FILTER = t2.FILTER
AND t1.VALUE = t2.VALUE
Or using row_number:
SELECT *
FROM
(
SELECT *,
row_number() over(partition by FILTER, VALUE
order by SEQ_NO desc) as rn
FROM table
) t
WHERE rn = 1
From Oracle 12C:
SELECT SEQ_NO
, FILTER
, VALUE
FROM TABLE
ORDER BY SEQ_NO DESC
FETCH FIRST 1 ROWS ONLY;
You can use ROWNUM in oracle:
select *
from
( select *
from yourTable
order by SEQ_NO desc ) as t
where ROWNUM = 1;
This should work
SELECT TOP 1 *
FROM TABLE
ORDER BY SEQ_NO DESC
If I understand correctly, you want the top SEQ_NO per filter?
i've created this in SQL Server and converted to Oracle
SELECT a.SEQ_NO,
a.FILTER,
a.VALUE
FROM (
SELECT SEQ_NO,
FILTER,
VALUE,
MAX(SEQ_NO) OVER (PARTITION BY FILTER) m
FROM TABLE
) a
WHERE SEQ_NO = m
Using mysql
SELECT SEQ_NO
, VALUE
, FILTER
FROM TABLE
Order by SEQ_NO DESC LIMIT 1
Related
I'm trying to select data from my database from the highest sequence number, I have been struggling with this for a while and cant get it to work.
The database has a lot of Columns with data. I only want data from the row with the highest sequence number to search in, because the data from lower sequences is not of any value for me. Unfortunately the rows from the lower sequences can not be deleted.
Database looks like this:
-----------------------------
| ID | SEQ | rest of the data
-----------------------------
| 1 | 1 | ..
| 1 | 2 | ....
| 2 | 1 | ..
| 1 | 3 | ....
| 3 | 1 | ..
| 1 | 2 | ....
| 4 | 1 | ........
My question is, how can i select only the ID's with the highest sequence number and search in those rows with the WHERE clause?
On oracle11g you can use:
SELECT *
FROM (
SELECT YOUR_TABLE.*, RANK() OVER (PARTITION BY ID oRDER BY SEQ DESC) RN
FROM YOUR_TABLE) A
WHERE RN=1;
SELECT *
FROM (
SELECT t.*,
ROW_NUMBER() OVER ( PARTITION BY ID ORDER BY SEQ DESC ) AS rn
FROM your_table t
)
WHERE rn = 1
or
SELECT ID,
MAX( seq ) AS seq,
MAX( other_column_1 ) KEEP ( DENSE_RANK LAST ORDER BY seq ) AS other_column_1,
MAX( other_column_2 ) KEEP ( DENSE_RANK LAST ORDER BY seq ) AS other_column_2
-- ...
FROM your_table
GROUP BY id
or
SELECT *
FROM your_table t
WHERE seq IN ( SELECT MAX( seq )
FROM your_table x
WHERE x.id = t.id )
or
SELECT t.*
FROM your_table t
INNER JOIN ( SELECT id, MAX( seq ) AS seq
FROM your_table
GROUP BY id ) x
ON ( x.id = t.id AND x.seq = t.seq )
From this table:
Number Value
1 a
2 b
3 a
2 c
2 b
3 a
2 b
I need to get count of all duplicate rows by Number and Value, i.e. 5.
Thanks.
I think this query is what you want:
SELECT SUM(t.cnt)
FROM
(
SELECT COUNT(*) cnt
FROM table_name
GROUP BY number, value
HAVING COUNT(*) > 1
)t;
Maybe something like this?
select value,number,max(cnt) as Count_distinct from (
select *,row_number () over (partition by value,number order by number) as cnt
from #sample
)t
group by value,number
Output
+---------------------------------+
| Value | Number | Count_Distinct |
| a | 1 | 1 |
| b | 2 | 3 |
| c | 2 | 1 |
| a | 3 | 2 |
+---------------------------------+
Select
count(distinct Number) as Distinct_Numbers,
count(distinct Value) as Distinct_Values
from
Table
This shows how many distinct values are in each column. Does this help?
Give a row number partition by both the columns and order by both the columns. Then count the number of rows where row number greater than 1.
Query
;with cte as(
select [rn] = row_number() over(
partition by [Number], [Value]
order by [Number], [Value]
), *
from [your_table_name]
)
select count(*) from cte
where [rn] > 1;
I think you mean number of unique number - value pairs, you can use:
SELECT count(*)
FROM
(SELECT ROW_NUMBER() OVER (PARTITION BY number, value ORDER BY (select 1)) from mytable rnk) i
where i.rnk = 1
May be this query may help you
select * from [dbo].[Sample_table1]
;WITH
DupContactRecords(number,value,DupsCount)
AS
(
SELECT number,value, COUNT() AS TotalCount FROM [Sample_table1] GROUP BY number,value HAVING COUNT() > 1
)
--to get the duplicats
/*select * from DupContactRecords*/
SELECT sum(DupsCount) FROM DupContactRecords
I have Table1 with three columns:
Key | Date | Price
----------------------
1 | 26-May | 2
1 | 25-May | 2
1 | 24-May | 2
1 | 23 May | 3
1 | 22 May | 4
2 | 26-May | 2
2 | 25-May | 2
2 | 24-May | 2
2 | 23 May | 3
2 | 22 May | 4
I want to select the row where value 2 was last updated (24-May). The Date was sorted using RANK function.
I am not able to get the desired results. Any help will be appreciated.
SELECT *
FROM (SELECT key, DATE, price,
RANK() over (partition BY key order by DATE DESC) AS r2
FROM Table1 ORDER BY DATE DESC) temp;
Another way of looking at the problem is that you want to find the most recent record with a price different from the last price. Then you want the next record.
with lastprice as (
select t.*
from (select t.*
from table1 t
order by date desc
) t
where rownum = 1
)
select t.*
from (select t.*
from table1 t
where date > (select max(date)
from table1 t2
where t2.price <> (select price from lastprice)
)
order by date asc
) t
where rownum = 1;
This query looks complicated. But, it is structured so it can take advantage of indexes on table1(date). The subqueries are necessary in Oracle pre-12. In the most recent version, you can use fetch first 1 row only.
EDIT:
Another solution is to use lag() and find the most recent time when the value changed:
select t1.*
from (select t1.*
from (select t1.*,
lag(price) over (order by date) as prev_price
from table1 t1
) t1
where prev_price is null or prev_price <> price
order by date desc
) t1
where rownum = 1;
Under many circumstances, I would expect the first version to have better performance, because the only heavy work is done in the innermost subquery to get the max(date). This verson has to calculate the lag() as well as doing the order by. However, if performance is an issue, you should test on your data in your environment.
EDIT II:
My best guess is that you want this per key. Your original question says nothing about key, but:
select t1.*
from (select t1.*,
row_number() over (partition by key order by date desc) as seqnum
from (select t1.*,
lag(price) over (partition by key order by date) as prev_price
from table1 t1
) t1
where prev_price is null or prev_price <> price
order by date desc
) t1
where seqnum = 1;
You can try this:-
SELECT Date FROM Table1
WHERE Price = 2
AND PrimaryKey = (SELECT MAX(PrimaryKey) FROM Table1
WHERE Price = 2)
This is very similar to the second option by Gordon Linoff but introduces a second windowed function row_number() to locate the most recent row that changed the price. This will work for all or a range of keys.
select
*
from (
select
*
, row_number() over(partition by Key order by [date] DESC) rn
from (
select
*
, NVL(lag(Price) over(partition by Key order by [date] DESC),0) prevPrice
from table1
where Key IN (1,2,3,4,5) -- as an example
)
where Price <> prevPrice
)
where rn = 1
apologies but I haven't been able to test this at all.
Say we have a table T
+------+
| NUMM |
+------+
| 1 |
| 5 |
| 3 |
| 8 |
+------+
I want the nearest bigger number from the column numm to be in column numm1.
The result will look like this
+-------------+
| NUMM | NUMM1|
+-------------+
| 1 | 3 |
| 3 | 5 |
| 5 | 8 |
+-------------+
I wrote a query like this and it works. But i would like to know if there is a better way for sollution.
select numm, numm + min(dif) as numm1
from (select distinct a.numm numm, b.numm numm1, b.numm - a.numm dif
from (select *
from T
where numm != (select max(numm) from T )) a
join T b
on 1 = 1)
where dif > 0
group by numm
If you want to get the direct successor, you can use the lead() windowing function:
select * from (
select
numm,
lead(numm) over (order by numm) as numm1
from t
)
where numm1 is not null
order by numm;
If it's oracle, you can use row_number() function to rank, then inner join with [left_table].rank = [right_table].rank - 1:
SELECT a.numm,
b.numm
FROM
(SELECT numm, row_number() over(order by numm) AS rank FROM pn_test) a
INNER JOIN
(SELECT numm, row_number() over(order by numm) AS rank FROM pn_test) b
ON a.rank = b.rank - 1;
Try the below query if you have repeated records in your table:
WITH CTE_ABC
AS (
SELECT DISTINCT NUMM
FROM [Table]
)
,CTE_XYZ
AS (
SELECT *
FROM (
SELECT NUMM
,lead(NUMM) OVER (
ORDER BY NUMM
) AS numm1
FROM CTE_ABC
) A
WHERE numm1 IS NOT NULL
)
SELECT A.NUMM
,B.numm1
FROM [Table] A
LEFT JOIN CTE_XYZ B
ON A.columnId = B.columnId
WHERE B.numm1 IS NOT NULL
Frank's answer is probably the best when there are no duplicate numbers, but if you can end up with duplicates here's one possible solution:
with t1 as (
select numm
, dense_rank() over (order by numm) rnk
from t
)
select t1.numm
, t2.numm numm1
from t1
join (select distinct numm, rnk-1 rnk from t1) t2
on t1.rnk = t2.rnk;
In this solution the DENSE_RANK analytic function is first used in T1 to give every distinct NUMM a sequential number (RNK). In the second stage T1 is join on RNK to the distinct set of numm and rnk-1 values from t1.
Looking for better performance, this might do the job:
with t1 as (
select numm
, dense_rank() over (order by numm) rnk
, row_number() over (partition by numm order by rownum) ord
from t
)
select t1.numm
, t2.numm numm1
from t1
join t1 t2
on t1.rnk = t2.rnk-1
and t2.ord = 1;
Here I added a way to grab one record for each numm to the subfactored query and eliminated the distinct operation in T2. With a limited data set of just 5 records and no indexes it has a cost of 9 vs a cost of 10 for the prior query.
I need to select data base upon three conditions
Find the latest date (StorageDate Column) from the table for each record
See if there is more then one entry for date (StorageDate Column) found in first step for same ID (ID Column)
and then see if DuplicateID is = 2
So if table has following data:
ID |StorageDate | DuplicateTypeID
1 |2014-10-22 | 1
1 |2014-10-22 | 2
1 |2014-10-18 | 1
2 |2014-10-12 | 1
3 |2014-10-11 | 1
4 |2014-09-02 | 1
4 |2014-09-02 | 2
Then I should get following results
ID
1
4
I have written following query but it is really slow, I was wondering if anyone has better way to write it.
SELECT DISTINCT(TD.RecordID)
FROM dbo.MyTable TD
JOIN (
SELECT T1.RecordID, T2.MaxDate,COUNT(*) AS RecordCount
FROM MyTable T1 WITH (nolock)
JOIN (
SELECT RecordID, MAX(StorageDate) AS MaxDate
FROM MyTable WITH (nolock)
GROUP BY RecordID)T2
ON T1.RecordID = T2.RecordID AND T1.StorageDate = T2.MaxDate
GROUP BY T1.RecordID, T2.MaxDate
HAVING COUNT(*) > 1
)PT ON TD.RecordID = PT.RecordID AND TD.StorageDate = PT.MaxDate
WHERE TD.DuplicateTypeID = 2
Try this and see how the performance goes:
;WITH
tmp AS
(
SELECT *,
RANK() OVER (PARTITION BY ID ORDER BY StorageDate DESC) AS StorageDateRank,
COUNT(ID) OVER (PARTITION BY ID, StorageDate) AS StorageDateCount
FROM MyTable
)
SELECT DISTINCT ID
FROM tmp
WHERE StorageDateRank = 1 -- latest date for each ID
AND StorageDateCount > 1 -- more than 1 entry for date
AND DuplicateTypeID = 2 -- DuplicateTypeID = 2
You can use analytic function rank , can you try this query ?
Select recordId from
(
select *, rank() over ( partition by recordId order by [StorageDate] desc) as rn
from mytable
) T
where rn =1
group by recordId
having count(*) >1
and sum( case when duplicatetypeid =2 then 1 else 0 end) >=1