Here is the situation where I have a table in bigquery like following.
As in the table we have record 1 and 3 with the same id but different first_name (Say the person with the id one changed his first_name) all other fields are same in both of the records (1 and 3) Now I need to select one records out of those 2 how can I do that. I tried self join but that is discarding both of the records, group_by will not work because the records is not duplicate only the Id is duplicate same with the distinct.
Thanks!!!!
The query I am using right now is
select * from table t group by 1,2,3,4,5;
You Can use ROW_NUMBER function to assign row numbers to each of your records in the table.
select *
from(
select *, ROW_NUMBER() OVER(PARTITION BY t.id) rn
from t)
Where rn = 1
ROW_NUMBER does not require the ORDER BY clause. Returns the sequential row ordinal (1-based) of each row for each ordered partition. If the ORDER BY clause is unspecified then the result is non-deterministic.
If you have record created date or modified dates you can use those in the ORDER BY clause to alway pick up the latest records.
SQL tables represent unordered sets. There is no first row unless you have a column that specifies the ordering. Let me assume you have such a column.
If you want a particular row, you can use aggregation with an order by:
select array_agg(t order by ? asc limit 1)[ordinal(1)].*
from t
group by id;
? is the column that specifies the ordering.
You can also leave out the order by:
select array_agg(t limit 1)[ordinal(1)].*
from t
group by id;
I have a table with data where in Column A I have groups of repeating Data (one after another).
I want to select only first row of each group based on values in column A only (no other criteria). Mind you, I want all corresponding columns selected also for the mentioned new found row (I don't want to exclude them).
Can someone help me with a proper query.
Here is a sample:
SAMPLE
Thanks!
#standardSQL
SELECT row.*
FROM (
SELECT ARRAY_AGG(t LIMIT 1)[OFFSET(0)] row
FROM `project.dataset.table` t
GROUP BY columnA
)
you can try smth like this:
#standardSQL
SELECT
* EXCEPT(rn)
FROM (
SELECT
*,
ROW_NUMBER() OVER(PARTITION BY columnA ORDER BY columnA) AS rn
FROM
your_dataset.your_table)
WHERE rn = 1
that will return:
Row columnA col2 ...
1 AC1001 Z_Creation
2 ACO112BISPIC QN
...
Add LIMIT 1 at the end of the query
something like
SELECT name, year FROM person_table ORDER BY year LIMIT 1
You can now use qualify for a more concise solution:
select
*
from
your_dataset.your_table
where true
qualify ROW_NUMBER() OVER(PARTITION BY columnA ORDER BY columnA) = 1
In BigQuery the physical sequence of rows is not significant. “BigQuery does not guarantee a stable ordering of rows in a table. Only the result of a query with an explicit ORDER BY clause has well-defined ordering.”[1].
First, you need to define which property will determine the first row of your group, then you can run Vasily Bronsky’s query by changing ORDER BY with that property. Which means either you should add another column to the table to store the order of the rows or select one from the columns you have.
I have a table that look like this:
The problem is I need to get the last record with duplicates in the column "NRODENUNCIA".
You can use MAX(DENUNCIAID), along with GROUP BY... HAVING to find the duplicates and select the row with the largest DENUNCIAID:
SELECT MAX(DENUNCIAID), NRODENUNCIA, FECHAEMISION, ADUANA, MES, NOMBREESTADO
FROM YourTable
GROUP BY NRODENUNCIA, FECHAEMISION, ADUANA, MES, NOMBREESTADO
HAVING COUNT(1) > 1
This will only show rows that have at least one duplicate. If you want to see non-duplicate rows too, just remove the HAVING COUNT(1) > 1
There are a number of solutions for your problem. One is to use row_number.
Note that I've ordered by DENUNCIID in the OVER clause. This defines the "Last Record" as the one that has the largest DENUNCIID. If you want to define it differently you'd need to change the field that is being ordered.
with dupes as (
SELECT
ROW_NUMBER() OVER (Partition by NRODENUNCIA ORDER BY DENUNCIID DESC) RN,
*
FROM
YourTable
)
SELECT * FROM dupes where rn = 1
This only get's the last record per dupe.
If you want to only include records that have dupes then you change the where clause to
WHERE rn =1
and NRODENUNCIA in (select NRODENUNCIA from dupes where rn > 1)
I want to update the top 10 values of a column in table. I have three columns; id, account and accountrank. To get the top 10 values I can use the following:
SELECT * FROM accountrecords
ORDER BY account DESC
LIMIT 10;
What I would like to do is to set the value in accountrank to be a series of 1 - 10, based on the magnitude of account. Is this possible to do in PostgreSQL?
WITH cte AS (
SELECT id, row_number() OVER (ORDER BY account DESC NULLS LAST) AS rn
FROM accountrecords
ORDER BY account DESC NULLS LAST
LIMIT 10
)
UPDATE accountrecords a
SET accountrank = cte.rn
FROM cte
WHERE cte.id = a.id;
Joining in a table expression is typically faster than correlated subqueries. It is also shorter.
With the window function row_number() distinct numbers are guaranteed. Use rank() (or possibly dense_rank()) if you want rows with equal values for account to share the same number.
Only if there can be NULL values in account, you need to append NULLS LAST for descending sort order, or NULL values sort on top:
Sort by column ASC, but NULL values first?
If there can be concurrent write access, the above query is subject to a race condition. Consider:
Atomic UPDATE .. SELECT in Postgres
Postgres UPDATE … LIMIT 1
However, if that was the case, the whole concept of hard-coding the top ten would be a dubious approach to begin with.
Use a CTE instead of a plain subquery to enforce the LIMIT reliably. See links above.
Sure, you can use your select statement in a subquery. Generating the rank-order isn't trivial, but here's at least one way to do it. I haven't tested this, but off the top of my head:
update accountrecords
set accountrank =
(select count(*) + 1 from accountrecords r where r.account > account)
where id in (select id from accountrecords order by account desc limit 10);
This has the quirk that if two records have the same value for account, then they will get the same rank. You could consider that a feature... :-)
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)