This question already has answers here:
Closed 11 years ago.
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How to select the nth row in a SQL database table?
I have a table that I want to select Select 1 record and in other command I want to select second row and in other command 3th row and ... how can I select for example 4th record from top of table without having the row number . just I want to select 4th row from top of table . how can I do this ?
you can select the 4th row by this code in MS sql server.
SELECT * FROM (
SELECT
ROW_NUMBER() OVER (ORDER BY DayRangeId ASC) AS rownumber,
DayRangeId
FROM DayRangeTable
) as temptablename
WHERE rownumber = 4
You want the first row, then the second row, then the third row and then the fourth row?
Can't you select the top n rows, and itterate through them?
I'm not sure what you mean by without having the row number though. Do you mean "Select the 4th row without knowing I need the 4th row"???
If you really need a piece of SQL that can select a specific row...
;WITH
sequenced_data AS (
SELECT ROW_NUMBER() OVER (ORDER BY <whatever fields>) AS row_id, * FROM myTable
)
SELECT
TOP 1 *
FROM
sequenced_data
WHERE
row_id = #n
Related
This question already has answers here:
Get top 1 row of each group
(19 answers)
Closed 3 months ago.
Is there a way to use TOP and WHERE multiple times? similar as a for loop to create a table?
I am using the following query to create a table that contains the top 26 records where the value of the column [code] is 11:
SELECT TOP 26 [date]
,[group]
,[code]
,[pol]
,[relation]
FROM [database].[table1] WHERE group in ('A','B','C',...,'Z') and code = '11'
The problem with this query is that I get 26 records with the value of the column [group] equal to A. This happens because there are thousands of records that meet that criterion.
Ideally, I would like the top 1 of each group (A to Z) with the value of code 11. I could achieve that by running the query above 26 times using TOP 1 and a different value of group, but this is impractical.
Is there any way to run this query multiple times to get the desired table?
Thanks in advance!
You can use a CTE to assign a ROW_NUMBER(), then only return rows WHERE ROW_NUMBER() = 1. You may need to adjust the PARTITION/ORDER BY depending on your data and what you're expecting your result set to be.
;WITH cte AS
(
SELECT *,
ROW_NUMBER() OVER (PARTITION BY [group], code ORDER BY code ASC) AS rn
FROM table1
)
SELECT *
FROM cte
WHERE rn = 1 AND code = 11
Demo here.
Assuming that your database support WITH TIES, then no need for a subquery. You can just order by with row_number:
SELECT TOP 1 WITH TIES
[date]
,[group]
,[code]
,[pol]
,[relation]
FROM [database].[table1]
WHERE group in ('A','B','C',...,'Z') and code = '11'
ORDER BY ROW_NUMBER() OVER(PARTITION BY [group] ORDER BY [date] desc)
You did not tell which column should be used to identify the top row per group, so I used a descending sort on the date. For each group, ROW_NUMBER assigns 1 to the row with the latest date ; TOP 1 WITH TIES then selects all such rows.
This question already has answers here:
How to get the top 10 values in postgresql?
(4 answers)
Closed 7 months ago.
i am using postgresql. i want to be able to select or delete the first n rows after sorting the ascendingly according to the timestamp of insertion.
i found an example but it uses a table named logtable, i do not have that logtable.
in otherwords, i want to sort the records in the table based on the time of data insertion into the table then selected the first N rows
please let me know how to achieve this task
You could use ROW_NUMBER:
WITH cte AS (
SELECT *, ROW_NUMBER() OVER (ORDER BY ts_insertion) rn
FROM yourTable
)
SELECT *
FROM cte
WHERE rn <= N -- replace N with your actual limit value
ORDER BY ts_insertion;
If you only want the first n rows, you can use LIMIT n, a window function is not necessary.
You can also use OFFSET:
SELECT column1, column2
FROM table_name
ORDER BY column1 ASC
LIMIT row_count OFFSET row_to_skip;
This question already has answers here:
How do I limit the number of rows returned by an Oracle query after ordering?
(14 answers)
Closed 5 years ago.
I want to select the number of rows which are greater than 3 by rownum function i_e "(rownum>3)"
for example if there are 25 rows and I want to retrieve the last 22 rows by rownum function.
but when I write the
select * from test_table where rownum>3;
it retrieve no row.
can any one help me to solve this problem.
thanks in advance
In RDBMS there is no first or last rows. What you calls "raws" , actually is set(sets), they can be ordered or not. rownum is a function, which is just enumerates result set, it makes sense only after set is calculated, to order your set of data (rows) you should do it in your query before rownum call, you must tell DB what means for the order in particular select statement.
It is not working because: for the first row assumes the ROWNUM of 1 and since your WHERE clause is ROWNUM>3 then this reduces to 1>3 and the row is discarded. The subsequent row will then be tested against a ROWNUM of 1 (since the previous row is no longer in the output and now does not require a row number), which will again fail the test and be discarded. Repeat, ad nauseum and all rows fail the WHERE clause filter and are discarded.
If you want to assign the rows a ROWNUM then you need to do this is a sub-query:
SELECT * -- Finally, in the outer query, filter on the assigned ROWNUM
FROM (
SELECT t.*, -- First, in the inner sub-query, apply a ROWNUM
ROWNUM AS rn
FROM test_table t
)
WHERE rn > 3;
Or, if you want to order the results before numbering:
SELECT * -- Finally, in the outer query, filter on the assigned ROWNUM
FROM (
SELECT t.*, -- Second, in the next level sub-query, apply a ROWNUM
ROWNUM AS rn
FROM (
SELECT * -- First, in the inner-most sub-query, apply an order
FROM test_table
ORDER BY some_column
) t
)
WHERE rn > 3;
select * from (select rownum as rn, t.* from test_table t) where rn > 3
see this article for more samples
On Top-n and Pagination Queries By Tom Kyte
This question already has answers here:
How to delete duplicate rows in SQL Server?
(26 answers)
Closed 6 years ago.
I have a table with 82,535 rows, where 65,087 rows are unique by ID. When I pull the entire result set of 82,535 and copy to Excel and remove duplicates, it shows that there are 17,448 duplicates. But when I'm using the query below I'm getting different results:
SELECT
BLD_ID, COUNT(BLD_ID) AS [BLD_ID COUNT]
FROM
Project.BreakageAnalysisOutcome_SentToAIM
GROUP BY
BLD_ID
HAVING
COUNT(BLD_ID) > = 2
This query returns a value of 17,364
I know for sure that the number of unique BLD_ID is 65,087
Most likely reason fro that is duplicate record may have more than 2 occurrence.
find duplicate count
Select COUNT(BLD_ID)- COUNT( DISTINCT BLD_ID)
From Project.BreakageAnalysisOutcome_SentToAIM
Use CTE with a Row_Number Function instead of count with group by clause and filer by Row_Number > 1.
;WITH cte
AS
(
SELECT ID,
ROW_NUMBER() OVER(PARTITION BY ID ORDER BY ID) AS Rn
FROM [Table1]
)
DELETE cte WHERE Rn > 1
This question already has answers here:
How to delete duplicate rows in SQL Server?
(26 answers)
Closed 6 years ago.
I have a table and need to delete entire row where ID occurs second and subsequent times, but leave the first occurrence of suCustomerIDBy the way. M table has ID which is a primary key and CustometID which is duplicated. So I need to remove all rows with duplicated CustomerID.
Delete From Table1 where ID IN (select ID From Table1 where count(distinct CutomerID) >=2 group by CustomerID)
The code above will delete all id including the first occurrence of each of the IDs, but I need to keep their first occurrence. Please advise.
This code should give you what you need.
There may be better ways to do it if you can provide the full table schema for Table1
If you obtain the row number and then just ignore the first ones:
;WITH cte
AS
(
SELECT ID,
ROW_NUMBER() OVER(PARTITION BY ID ORDER BY ID) AS Rn
FROM [Table1]
)
DELETE cte WHERE Rn > 1
delete a from(
Select dense_rank() OVER(PARTITION BY id ORDER BY id) AS Rn,*
from Table1)a
where a.Rn>1