I have a oracle database with over 2 million rows and 200 columns. I'm trying to query data in five columns where one of the columns is equal to the most recent date. This query below works but seems to be taking long (over 2 min) to process. Is there a different logic I can use to speed up the query?
SELECT a,b,c,date,e FROM table a WHERE a.date = (SELECT MAX(date) FROM table)
try rownum
SELECT * FROM (
SELECT A,B,C,DATE,E FROM TABLE A WHERE A.DATE DESC
)
WHERE ROWNUM=1
you can also use another similar solution
with tbl1 as
(SELECT A,B,C,DATE,E,
first_value(date) over( order by date desc) maxdate
FROM TABLE A)
select A,B,C,DATE,E from tbl1 where date = maxdate
YES !!!
Set Index "date" to your table as I_TABLE_DATE, after that change your query to this
SELECT --+ index_desc(a I_TABLE_DATE)
a,b,c,date,e
FROM table a
WHERE a.date = (SELECT --+ index_desc(b I_TABLE_DATE)
b.Date
FROM table b
where Rownum = 1)
it will be more faster because, during getting maximum date there will be only scanning index by descening, and your main query will work by index ascening and you dont need scan all table
Related
I have a table where limits were sanctioned to the customer
I am trying to get the output as below picture i.e. total amount sanctioned till particular date
I am trying below code but this sums the total sanction amount
select gam.id, sum(SANCTION_AMOUNT) from gam
join (select ID,ACCOUNT_OPEN_DATE from gam where ACCOUNT_OPEN_DATE between'01-04-2019' and '30-04-2019' AND SCHEME_CODE IN ('SB','CCKLY')) ) action
on( gam.ACCOUNT_OPEN_DATE <=action.ACCOUNT_OPEN_DATE and gam.id=action.cust_id) group by gam.id;
In Oracle, this can be a way:
select id, sanction_amount, scheme_code, account_open_date,
sum(sanction_amount) over (partition BY ID order by account_open_date) as total_sanction_amount
from gam
order by account_open_date
Not sure your database is MySQL or Oracle, But this below script is workable in most of the database. Just adjust the table and column names accordingly.
You can check MySQL DEMO HERE
SELECT *,
(
SELECT SUM(sanction_Amount)
FROM Your_Table B
WHERE B.ID = A.ID
AND B.acc_open_date <= A.acc_open_date
) Total_sanction_Amount
FROM Your_Table A
I have a SQL database table, "Helium_Test_Data", that has multiple entries based on the KeyID column (the KeyID represents a single tested part ). I need to query the entries and only show one entry per KeyID (part) based on the earliest creation date-time (format example is 2018-12-29 08:22:11.123). This is because the same part was tested several times but the first reading is the one I need to use. Here is the query currently tried:
SELECT mt.*
FROM Helium_Test_Data mt
INNER JOIN
(
SELECT
KeyID,
MIN(DateTime) AS DateTime
FROM Helium_Test_Data
WHERE PSNo='11166565'
GROUP BY KeyID
) t ON mt.KeyID = t.KeyID AND mt.DateTime = t.DateTime
WHERE PSNo='11167197'
AND (mt.DateTime > '2018-12-29 07:00')
AND (mt.DateTime < '2018-12-29 18:00') AND OK=1
ORDER BY KeyId,DateTime
It returns only the rows that have no duplicate KeyID present in the table whereas I need one row per every single KeyID (duplicate or not). And for the duplicate ones, I need the earliest date.
Thanks in advance for the help.
use row_number() window function which support most dbms
select * from
(
select *,row_number() over(partition by KeyID order by DateTime) rn
from Helium_Test_Data
) t where t.rn=1
or you could use corelated subquery
select t1.* from Helium_Test_Data t1
where t1.DateTime= (select min(DateTime)
from Helium_Test_Data t2
where t2.KeyID=t1.KeyID
)
I am using a sql to select data from a table for a particular range of records.
I am using rownum to implement greater and less than logic to arrive at the required set of data.
With the use of below sql I can fetch the records from 21 to 40 from my table. In total for this condition table contains 100 rows.
Through this sql I want to fetch an indicator(value) which tells that there are more records for this condition.
I could not find any solution in google.
Sql -
select * from ( select rownum rnum, a.* from(SELECT TO_CHAR(D.DATE,'YYYYMMDD'),D.TYPE,
TO_CHAR(D.VDATE,'YYYYMMDD'),D.AMT,D.PARTICULAR,D.NUM,D.ID,
D.CODE,D.INFO FROM MySCHEMA
.MYTABLE D WHERE D.DATE >= TO_CHAR(TO_DATE('20160701','YYYYMMDD'),'DD-MON-RRRR')
AND D.DATE <= TO_CHAR(TO_DATE('20161105','YYYYMMDD'),'DD-MON-RRRR') AND D.XDATE >= TO_CHAR(TO_DATE('20160701','YYYYMMDD'),'DD-MON-RRRR')
AND D.XDATE <= TO_CHAR(TO_DATE('20161105','YYYYMMDD'),'DD-MON-RRRR')
AND D.FLG='Y' AND D.TYPE IN('D','C')
AND
D.ACI = 'CO6'
ORDER BY D.DATE DESC
)
a where rownum <= 40 ) where rnum >= 21;
You can add a count(*) over () total_rows to your inner select. That will tell you how many rows the inner query would return without the rownum predicates. That is going to mean, however, that every time you ask for a page of results, Oracle has to execute the inner query completely and then discard all the rows you aren't fetching. That's going to be more expensive than what you are currently doing
select *
from ( select rownum rnum, a.*
from(SELECT count(1) over () total_rows,
<<the rest of your query>>
This is a known question but the best solution I've found is something like:
SELECT TOP N *
FROM MyTable
ORDER BY Id DESC
I've a table with lots of rows. It is not a posibility to use that query because it takes lot of time. So how can I do to select last N rows without using ORDER BY?
EDIT
Sorry duplicated question of this one
You can get SQL server to select the last N rows with the following query:
select * from tbl_name order by id desc limit N;
I tested JonVD's code, but found it was very slow, 6s.
This code took 0s.
SELECT TOP(5) ORDERID, CUSTOMERID, OrderDate
FROM Orders where EmployeeID=5
Order By OrderDate DESC
You can do it by using the ROW NUMBER BY PARTITION Feature also. A great example can be found here:
I am using the Orders table of the Northwind database... Now let us retrieve the Last 5 orders placed by Employee 5:
SELECT ORDERID, CUSTOMERID, OrderDate
FROM
(
SELECT ROW_NUMBER() OVER (PARTITION BY EmployeeID ORDER BY OrderDate DESC) AS OrderedDate,*
FROM Orders
) as ordlist
WHERE ordlist.EmployeeID = 5
AND ordlist.OrderedDate <= 5
If you want to select last numbers of rows from a table.
Syntax will be like
select * from table_name except select top
(numbers of rows - how many rows you want)* from table_name
These statements work but differrent ways. thank you guys.
select * from Products except select top (77-10) * from Products
in this way you can get last 10 rows but order will show descnding way
select top 10 * from products
order by productId desc
select * from products
where productid in (select top 10 productID from products)
order by productID desc
select * from products where productID not in
(select top((select COUNT(*) from products ) -10 )productID from products)
First you most get record count from
Declare #TableRowsCount Int
select #TableRowsCount= COUNT(*) from <Your_Table>
And then :
In SQL Server 2012
SELECT *
FROM <Your_Table> As L
ORDER BY L.<your Field>
OFFSET <#TableRowsCount-#N> ROWS
FETCH NEXT #N ROWS ONLY;
In SQL Server 2008
SELECT *
FROM
(
SELECT ROW_NUMBER() OVER(ORDER BY ID) AS sequencenumber, *
FROM <Your_Table>
Order By <your Field>
) AS TempTable
WHERE sequencenumber > #TableRowsCount-#N
In a very general way and to support SQL server here is
SELECT TOP(N) *
FROM tbl_name
ORDER BY tbl_id DESC
and for the performance, it is not bad (less than one second for more than 10,000 records On Server machine)
Is "Id" indexed? If not, that's an important thing to do (I suspect it is already indexed).
Also, do you need to return ALL columns? You may be able to get a substantial improvement in speed if you only actually need a smaller subset of columns which can be FULLY catered for by the index on the ID column - e.g. if you have a NONCLUSTERED index on the Id column, with no other fields included in the index, then it would have to do a lookup on the clustered index to actually get the rest of the columns to return and that could be making up a lot of the cost of the query. If it's a CLUSTERED index, or a NONCLUSTERED index that includes all the other fields you want to return in the query, then you should be fine.
select * from (select top 6 * from vwTable order by Hours desc) T order by Hours
Here's something you can try without an order by but I think it requires that each row is unique. N is the number of rows you want, L is the number of rows in the table.
select * from tbl_name except select top L-N * from tbl_name
As noted before, which rows are returned is undefined.
EDIT: this is actually dog slow. Of no value really.
A technique I use to query the MOST RECENT rows in very large tables (100+ million or 1+ billion rows) is limiting the query to "reading" only the most recent "N" percentage of RECENT ROWS. This is real world applications, for example I do this for non-historic Recent Weather Data, or recent News feed searches or Recent GPS location data point data.
This is a huge performance improvement if you know for certain that your rows are in the most recent TOP 5% of the table for example. Such that even if there are indexes on the Tables, it further limits the possibilites to only 5% of rows in tables which have 100+ million or 1+ billion rows. This is especially the case when Older Data will require Physical Disk reads and not only Logical In Memory reads.
This is well more efficient than SELECT TOP | PERCENT | LIMIT as it does not select the rows, but merely limit the portion of the data to be searched.
DECLARE #RowIdTableA BIGINT
DECLARE #RowIdTableB BIGINT
DECLARE #TopPercent FLOAT
-- Given that there is an Sequential Identity Column
-- Limit query to only rows in the most recent TOP 5% of rows
SET #TopPercent = .05
SELECT #RowIdTableA = (MAX(TableAId) - (MAX(TableAId) * #TopPercent)) FROM TableA
SELECT #RowIdTableB = (MAX(TableBId) - (MAX(TableBId) * #TopPercent)) FROM TableB
SELECT *
FROM TableA a
INNER JOIN TableB b ON a.KeyId = b.KeyId
WHERE a.Id > #RowIdTableA AND b.Id > #RowIdTableB AND
a.SomeOtherCriteria = 'Whatever'
MS doesn't support LIMIT in t-sql. Most of the times i just get MAX(ID) and then subtract.
select * from ORDERS where ID >(select MAX(ID)-10 from ORDERS)
This will return less than 10 records when ID is not sequential.
This query returns last N rows in correct order, but it's performance is poor
select *
from (
select top N *
from TableName t
order by t.[Id] desc
) as temp
order by temp.[Id]
use desc with orderby at the end of the query to get the last values.
This may not be quite the right fit to the question, but…
OFFSET clause
The OFFSET number clause enables you to skip over a number of rows and then return rows after that.
That doc link is to Postgres; I don't know if this applies to Sybase/MS SQL Server.
DECLARE #MYVAR NVARCHAR(100)
DECLARE #step int
SET #step = 0;
DECLARE MYTESTCURSOR CURSOR
DYNAMIC
FOR
SELECT col FROM [dbo].[table]
OPEN MYTESTCURSOR
FETCH LAST FROM MYTESTCURSOR INTO #MYVAR
print #MYVAR;
WHILE #step < 10
BEGIN
FETCH PRIOR FROM MYTESTCURSOR INTO #MYVAR
print #MYVAR;
SET #step = #step + 1;
END
CLOSE MYTESTCURSOR
DEALLOCATE MYTESTCURSOR
In order to get the result in ascending order
SELECT n.*
FROM
(
SELECT *
FROM MyTable
ORDER BY id DESC
LIMIT N
) n
ORDER BY n.id ASC
I stumpled acros this issue while using SQL server
What i did to resolve it is order the results descending and giving row number to the results of that, After i filtered the results and turned them around again.
SELECT *
FROM (
SELECT *
,[rn] = ROW_NUMBER() OVER (ORDER BY [column] DESC)
FROM [table]
) A
WHERE A.[rn] < 3
ORDER BY [column] ASC
Easy copy paste answer
To display last 3 rows without using order by:
select * from Lms_Books_Details where Book_Code not in
(select top((select COUNT(*) from Lms_Books_Details ) -3 ) book_code from Lms_Books_Details)
Try using the EXCEPT syntax.
Something like this:
SELECT *
FROM clientDetails
EXCEPT
(SELECT TOP (numbers of rows - how many rows you want) *
FROM clientDetails)
If I have a table with columns id, name, score, date
and I wanted to run a sql query to get the record where id = 2 with the earliest date in the data set.
Can you do this within the query or do you need to loop after the fact?
I want to get all of the fields of that record..
If you just want the date:
SELECT MIN(date) as EarliestDate
FROM YourTable
WHERE id = 2
If you want all of the information:
SELECT TOP 1 id, name, score, date
FROM YourTable
WHERE id = 2
ORDER BY Date
Prevent loops when you can. Loops often lead to cursors, and cursors are almost never necessary and very often really inefficient.
SELECT TOP 1 ID, Name, Score, [Date]
FROM myTable
WHERE ID = 2
Order BY [Date]
While using TOP or a sub-query both work, I would break the problem into steps:
Find target record
SELECT MIN( date ) AS date, id
FROM myTable
WHERE id = 2
GROUP BY id
Join to get other fields
SELECT mt.id, mt.name, mt.score, mt.date
FROM myTable mt
INNER JOIN
(
SELECT MIN( date ) AS date, id
FROM myTable
WHERE id = 2
GROUP BY id
) x ON x.date = mt.date AND x.id = mt.id
While this solution, using derived tables, is longer, it is:
Easier to test
Self documenting
Extendable
It is easier to test as parts of the query can be run standalone.
It is self documenting as the query directly reflects the requirement
ie the derived table lists the row where id = 2 with the earliest date.
It is extendable as if another condition is required, this can be easily added to the derived table.
Try
select * from dataset
where id = 2
order by date limit 1
Been a while since I did sql, so this might need some tweaking.
Using "limit" and "top" will not work with all SQL servers (for example with Oracle).
You can try a more complex query in pure sql:
select mt1.id, mt1."name", mt1.score, mt1."date" from mytable mt1
where mt1.id=2
and mt1."date"= (select min(mt2."date") from mytable mt2 where mt2.id=2)