I need to get the latest price of an item (as part of a larger select statement) and I can't quite figure it out.
Table:
ITEMID DATE SALEPRICE
1 1/1/2014 10
1 2/2/2014 20
2 3/3/2014 15
2 4/4/2014 13
I need the output of the select to be '20' when looking for item 1 and '13' when looking for item 2 as per the above example.
I am using Oracle SQL
The most readable/understandable SQL (in my opinion) would be this:
select salesprice from `table` t
where t.date =
(
select max(date) from `table` t2 where t2.itemid = t.itemid
)
and t.itemid = 1 -- change item id here;
assuming your table's name is table and you only have one price per day and item (else the where condition would match more than one row per item). Alternatively, the subselect could be written as a self-join (should not make a difference in performance).
I'm not sure about the OVER/PARTITION used by the other answers. Maybe they could be optimized to better performance depending on the DBMS.
Maybe something like this:
Test data
DECLARE #tbl TABLE(ITEMID int,DATE DATETIME,SALEPRICE INT)
INSERT INTO #tbl
VALUES
(1,'1/1/2014',10),
(1,'2/2/2014',20),
(2,'3/3/2014',15),
(2,'4/4/2014',13)
Query
;WITH CTE
AS
(
SELECT
ROW_NUMBER() OVER(PARTITION BY ITEMID ORDER BY [DATE] DESC) AS rowNbr,
tbl.*
FROM
#tbl AS tbl
)
SELECT
*
FROM
CTE
WHERE CTE.rowNbr=1
Try this!
In sql-server may also work in Oracle sql
select * from
(
select *,rn=row_number()over(partition by ITEMID order by DATE desc) from table
)x
where x.rn=1
You need Row_number() to allocate a number to all records which is partition by ITEMID so each group will get a RN,then as you are ordering by date desc to get Latest record
SEE DEMO
Related
i´m working with a table that looks like this:
Start
https://i.stack.imgur.com/uibc3.png
My desired result would look like this:
Result
https://i.stack.imgur.com/v0sic.png
So i´m triyng to select the max value from two "combined" colums. If the values are the same amount (Part C), the outcome doesn't matter.
I tried to order the table by max value and then using distinct but the result didn't turn out as expected
Could you please offer a solution or some insight to this? Thanks in advance!
Use row_number():
select *
from (
select t.*, row_number() over(partition by part order by amount desc, zone) rn
from mytable t
) t
where rn = 1
For each part, this gives you the row with the highest amount; if there are top ties, column zone is used to break them.
If you want to allow ties, then use rank() instead, like:
rank() over(partition by part order by amount desc) rn
You can achieve this by using SUB Query
DECLARE #T TABLE(
PART VARCHAR(50),
ZONE VARCHAR(10),
Amt INT)
Insert Into #T Values ('PartA','71H',1),('PartA','75H',2),('PartB','98D',1),('PartB','98A',3),('PartC','75H',1),('PartC','52H',1)
SELECT M.PART,MIN(M.Zone) AS ZONE,S.AMOUNT
FROM #T M
INNER JOIN (
SELECT Part,MAX(Amt) as AMOUNT From #T
GROUP BY PART) S ON S.AMOUNT=M.Amt AND S.PART=M.PART
GROUP BY M.PART,S.AMOUNT
ORDER BY M.PART
i've got a table that i need to return about 14 column values but only return 1 row for the duplicates on some of the columns.
The second problem is that between the duplicates i need to keep the one that has the biggest int in one of the columns that is not required to be unique.
Since the Table is somewhat big, I am seeking advice into doing this in the most efficient way.
should i be doing a group by?
my table is somewhat like this, i will simplify the number of columns.
ID(UniqueIdentifier) | ACCID(UniqueIdentifier) | DateTime(DateTime) | distance(int)|type(int)
28761188-0886-E911-822F-DD1FA635D450 1238FD8A-BD00-411A-A81C-0F6F5C026BCC 2019-06-03 14:04:41.000 2 3
41761188-0886-E911-822F-DD1FA635D450 1238FD8A-BD00-411A-A81C-0F6F5C026BCC 2019-06-03 14:04:41.000 1 3
I should be only selecting when ACCID and DATETIME is unique, the column ID in primary so will never be duplicate, and i need to keep the row with the biggest distance.
You can use the ROW_NUMBER() window function, as in:
select *
from (
select
id,
accid,
datetime,
distance,
type,
row_number() over(partition by accid, datetime order by type desc) as rn
from t
) x
where rn = 1
If you want to show multiple "ties", then replace ROW_NUMBER() by RANK().
I would suggest a correlated subquery with the right index as the fastest method:
select t.*
from t
where t.id = (select top (1) t2.id
from t t2
where t2.ACCID = t.ACCID
order by t2.distance desc
) ;
The best index is on (ACCID, distance desc, id).
I have the following table:
ItemID Price
1 10
2 20
3 12
4 10
5 11
I need to find the second lowest price. So far, I have a query that works, but i am not sure it is the most efficient query:
select min(price)
from table
where itemid not in
(select itemid
from table
where price=
(select min(price)
from table));
What if I have to find third OR fourth minimum price? I am not even mentioning other attributes and conditions... Is there any more efficient way to do this?
PS: note that minimum is not a unique value. For example, items 1 and 4 are both minimums. Simple ordering won't do.
SELECT MIN( price )
FROM table
WHERE price > ( SELECT MIN( price )
FROM table )
select price from table where price in (
select
distinct price
from
(select t.price,rownumber() over () as rownum from table t) as x
where x.rownum = 2 --or 3, 4, 5, etc
)
Not sure if this would be the fastest, but it would make it easier to select the second, third, etc... Just change the TOP value.
UPDATED
SELECT MIN(price)
FROM table
WHERE price NOT IN (SELECT DISTINCT TOP 1 price FROM table ORDER BY price)
To find out second minimum salary of an employee, you can use following:
select min(salary)
from table
where salary > (select min(salary) from table);
This is a good answer:
SELECT MIN( price )
FROM table
WHERE price > ( SELECT MIN( price )
FROM table )
Make sure when you do this that there is only 1 row in the subquery! (the part in brackets at the end).
For example if you want to use GROUP BY you will have to define even further using:
SELECT MIN( price )
FROM table te1
WHERE price > ( SELECT MIN( price )
FROM table te2 WHERE te1.brand = te2.brand)
GROUP BY brand
Because GROUP BY will give you multiple rows, otherwise you will get the error:
SQL Error [21000]: ERROR: more than one row returned by a subquery used as an expression
I guess a simplest way to do is using offset-fetch filter from standard sql, distinct is not necessary if you don't have repeat values in your column.
select distinct(price) from table
order by price
offset 1 row fetch first 1 row only;
no need to write complex subqueries....
In amazon redshift use limit-fetch instead for ex...
Select distinct(price) from table
order by price
limit 1
offset 1;
You can either use one of the following:-
select min(your_field) from your_table where your_field NOT IN (select distinct TOP 1 your_field from your_table ORDER BY your_field DESC)
OR
select top 1 ColumnName from TableName where ColumnName not in (select top 1 ColumnName from TableName order by ColumnName asc)
I think you can find the second minimum using LIMIT and ORDER BY
select max(price) as minimum from (select distinct(price) from tableName order by price asc limit 2 ) --or 3, 4, 5, etc
if you want to find third or fourth minimum and so on... you can find out by changing minimum number in limit. you can find using this statement.
You can use RANK functions,
it may seem complex query but similar results like other answers can be achieved with the same,
WITH Temp_table AS (SELECT ITEM_ID,PRICE,RANK() OVER (ORDER BY PRICE) AS
Rnk
FROM YOUR_TABLE_NAME)
SELECT ITEM_ID FROM Temp_table
WHERE Rnk=2;
Maybe u can check the min value first and then place a not or greater than the operator. This will eliminate the usage of a subquery but will require a two-step process
select min(price)
from table
where min(price) <> -- "the min price you previously got"
You have three fields ID, Date and Total. Your table contains multiple rows for the same day which is valid data however for reporting purpose you need to show only one row per day. The row with the highest ID per day should be returned the rest should be hidden from users (not returned).
To better picture the question below is sample data and sample output:
ID, Date, Total
1, 2011-12-22, 50
2, 2011-12-22, 150
The correct result is:
2, 2012-12-22, 150
The correct output is single row for 2011-12-22 date and this row was chosen because it has the highest ID (2>1)
Assuming that you have a database that supports window functions, and that the date column is indeed just date (and not datetime), then something like:
SELECT
* --TODO - Pick columns
FROM
(
SELECT ID,[Date],Total,ROW_NUMBER() OVER (PARTITION BY [Date] ORDER BY ID desc) rn
FROM [Table]
) t
WHERE
rn = 1
Should produce one row per day - and the selected row for any given day is that with the highest ID value.
SELECT *
FROM table
WHERE ID IN ( SELECT MAX(ID)
FROM table
GROUP BY Date )
This will work.
SELECT *
FROM tableName a
INNER JOIN
(
SELECT `DATE`, MAX(ID) maxID
FROM tableName
GROUP BY `DATE`
) b ON a.id = b.MaxID AND
a.`date` = b.`date`
SQLFiddle Demo
Probably
SELECT * FROM your_table ORDER BY ID DESC LIMIT 1
Select MAX(ID),Data,Total from foo
for MySQL
Another simple way is
SELECT TOP 1 * FROM YourTable ORDER BY ID DESC
And, I think this is the most simple way!
SELECT * FROM TABLE_SUM S WHERE S.ID =
(
SELECT MAX(ID) FROM TABLE_SUM
WHERE CDATE = GG.CDATE
GROUP BY CDATE
)
I need, if possible, a t-sql query that, returning the values from an arbitrary table, also returns a incremental integer column with value = 1 for the first row, 2 for the second, and so on.
This column does not actually resides in any table, and must be strictly incremental, because the ORDER BY clause could sort the rows of the table and I want the incremental row in perfect shape always.
The solution must run on SQL Server 2000
For SQL 2005 and up
SELECT ROW_NUMBER() OVER( ORDER BY SomeColumn ) AS 'rownumber',*
FROM YourTable
for 2000 you need to do something like this
SELECT IDENTITY(INT, 1,1) AS Rank ,VALUE
INTO #Ranks FROM YourTable WHERE 1=0
INSERT INTO #Ranks
SELECT SomeColumn FROM YourTable
ORDER BY SomeColumn
SELECT * FROM #Ranks
Order By Ranks
see also here Row Number
You can start with a custom number and increment from there, for example you want to add a cheque number for each payment you can do:
select #StartChequeNumber = 3446;
SELECT
((ROW_NUMBER() OVER(ORDER BY AnyColumn)) + #StartChequeNumber ) AS 'ChequeNumber'
,* FROM YourTable
will give the correct cheque number for each row.
Try ROW_NUMBER()
http://msdn.microsoft.com/en-us/library/ms186734.aspx
Example:
SELECT
col1,
col2,
ROW_NUMBER() OVER (ORDER BY col1) AS rownum
FROM tbl
It is ugly and performs badly, but technically this works on any table with at least one unique field AND works in SQL 2000.
SELECT (SELECT COUNT(*) FROM myTable T1 WHERE T1.UniqueField<=T2.UniqueField) as RowNum, T2.OtherField
FROM myTable T2
ORDER By T2.UniqueField
Note: If you use this approach and add a WHERE clause to the outer SELECT, you have to added it to the inner SELECT also if you want the numbers to be continuous.