I have a table of data and want to retrieve the penultimate record.
How is this done?
TABLE: results
-------
30
31
35
I need to get 31.
I've been trying with rownum but it doesn't seem to work.
Assuming you want the second highest number and there are no ties
SELECT results
FROM (SELECT results,
rank() over (order by results desc) rnk
FROM your_table_name)
WHERE rnk = 2
Depending on how you want to handle ties, you may want either the rank, dense_rank, or row_number analytic function. If there are two 35's for example, would you want 35 returned? Or 31? If there are two 31's, would you want a single row returned? Or would you want both 31's returned.
This can use for n th rank ##
select Total_amount from (select Total_amount, rank() over (order by Total_amount desc) Rank from tbl_booking)tbl_booking where Rank=3
Related
I am creating ranks for partitions of my table. Partitions are performed by name column with ordered by its transaction value. While I am generating these partitions and checking count for each of the ranks, I get different number in each rank for every query run I do.
select count(*) FROM (
--
-- Sort and ranks the element of RFM
--
SELECT
*,
RANK() OVER (PARTITION BY name ORDER BY date_since_last_trans desc) AS rfmrank_r,
FROM (
SELECT
name,
id_customer,
cust_age,
gender,
DATE_DIFF(entity_max_date, customer_max_date, DAY ) AS date_since_last_trans,
txncnt,
txnval,
txnval / txncnt AS avg_txnval
FROM
(
SELECT
name,
id_customer,
MAX(cust_age) AS cust_age,
COALESCE(APPROX_TOP_COUNT(cust_gender,1)[OFFSET(0)].VALUE, MAX(cust_gender)) AS gender,
MAX(date_date) AS customer_max_date,
(SELECT MAX(date_date) FROM xxxxx) AS entity_max_date,
COUNT(purchase_amount) AS txncnt,
SUM(purchase_amount) AS txnval
FROM
xxxxx
WHERE
date_date > (
SELECT
DATE_SUB(MAX(date_date), INTERVAL 24 MONTH) AS max_date
FROM
xxxxx)
AND cust_age >= 15
AND cust_gender IN ('M','F')
GROUP BY
name,
id_customer
)
)
)
group by rfmrank_r
For 1st run I am getting
Row f0
1 3970
2 3017
3 2116
4 2118
For 2nd run I am getting
Row f0
1 4060
2 3233
3 2260
4 2145
What can be done, If I need to get same number of partitions getting ranked same for each run
Edit:
Sorry for the blurring of fields
This is the output of field ```query to get this column````
The RANK window function determines the rank of a value in a group of values.
Each value is ranked within its partition. Rows with equal values for the ranking criteria receive the same rank. Drill adds the number of tied rows to the tied rank to calculate the next rank and thus the ranks might not be consecutive numbers.
For example, if two rows are ranked 1, the next rank is 3.
i am having table having 23 records , I am trying to get total count of record and last record also in single query. something like that
select count(*) ,(m order by createdDate) from music m ;
is there any way to pull this out only last record as well as total count in PostgreSQL.
This can be done using window functions
select *
from (
select m.*,
row_number() over (order by createddate desc) as rn,
count(*) over () as total_count
from music
) t
where rn = 1;
Another option would be to use a scalar sub-query and combine it with a limit clause:
select *,
(select count(*) from order_test.orders) as total_count
from music
order by createddate desc
limit 1;
Depending on the indexes, your memory configuration and the table definition might be faster then the two window functions.
No, it's not not possible to do what is being asked, sql does not function that way, the second you ask for a count () sql changes the level of your data to an aggregation. The only way to do what you are asking is to do a count() and order by in a separate query.
Another solution using windowing functions and no subquery:
SELECT DISTINCT count(*) OVER w, last_value(m) OVER w
FROM music m
WINDOW w AS (ORDER BY date DESC RANGE BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING);
The point here is that last_value applies on partitions defined by windows and not on groups defined by GROUP BY.
I did not perform any test but I suspect my solution to be the less effective amongst the three already posted. But it is also the closest to your example query so far.
I need to get the Top 10 results for each Region, Market and Name along with those with highest counts (Gaps). There are 4 Regions with 1 to N Markets. I can get the Top 10 but cannot figure out how to do this without using a Union for every Market. Any ideas on how do this?
SELECT DISTINCT TOP 10
Region, Market, Name, Gaps
FROM
TableName
ORDER BY
Region, Market, Gaps DESC
One approach would be to use a CTE (Common Table Expression) if you're on SQL Server 2005 and newer (you aren't specific enough in that regard).
With this CTE, you can partition your data by some criteria - i.e. your Region, Market, Name - and have SQL Server number all your rows starting at 1 for each of those "partitions", ordered by some criteria.
So try something like this:
;WITH RegionsMarkets AS
(
SELECT
Region, Market, Name, Gaps,
RN = ROW_NUMBER() OVER(PARTITION BY Region, Market, Name ORDER BY Gaps DESC)
FROM
dbo.TableName
)
SELECT
Region, Market, Name, Gaps
FROM
RegionsMarkets
WHERE
RN <= 10
Here, I am selecting only the "first" entry for each "partition" (i.e. for each Region, Market, Name tuple) - ordered by Gaps in a descending fashion.
With this, you get the top 10 rows for each (Region, Market, Name) tuple - does that approach what you're looking for??
I think you want row_number():
select t.*
from (select t.*,
row_number() over (partition by region, market order by gaps desc) as seqnum
from tablename t
) t
where seqnum <= 10;
I am not sure if you want name in the partition by clause. If you have more than one name within a market, that may be what you are looking for. (Hint: Sample data and desired results can really help clarify a question.)
ID Sum Name
a 10 Joe
a 8 Mary
b 21 Kate
b 110 Casey
b 67 Pierce
What would you recommend as the best way to
obtain for each ID the name that corresponds to the largest sum (grouping by ID).
What I tried so far:
select ID, SUM(Sum) s, Name
from Table1
group by ID, Name
Order by SUM(Sum) DESC;
this will arrange the records into groups that have the highest sum first. Then I have to somehow flag those records and keep only those. Any tips or pointers? Thanks a lot
In the end I'd like to obtain:
a 10 Joe
b 110 Casey
You want the row_number() function:
select id, [sum], name
from (select t.*]
row_number() over (partition by id order by [sum] desc) as seqnum
from table1
) t
where seqnum = 1;
Your question is more confusing than it needs to be because you have a column called sum. You should avoid using SQL reserved words for identifiers.
The row_number() function assigns a sequential number to a group of rows, starting with 1. The group is defined by the partition by clause. In this case, all rows with the same id are in the same group. The ordering of the numbers is determined by the order by clause, so the one with the largest value of sum gets the value of 1.
If you might have duplicate maximum values and you want all of them, use the related function rank() or dense_rank().
select *
from
(
select *
,rn = row_number() over (partition by Id order by sum desc)
from table
)x
where x.rn=1
demo
I'm trying to select the estimated hours of a row with the lowest date from a table.
SELECT prev_est_hrs
FROM ( SELECT MIN(change_date), prev_est_hrs
FROM task_history
WHERE task_id = 5
GROUP BY prev_est_hrs
);
However this is returning two rows, why? I thought MIN was supposed to return the lowest only?
Help much appreciated.
You have a GROUP BY clause. The MIN will return the minimum value in each group.
Also, you are only returning the group by value from the outer SELECT.
Mitch is right. One way to get the prev_est_hrs for the record with the earliest change_date, which seems to be what you're trying to find, is with an analytic function:
SELECT prev_est_hrs
FROM (
SELECT prev_est_hrs, ROW_NUMBER() OVER (ORDER BY change_date) AS rn
FROM task_history
WHERE task_id = 5
)
WHERE rn = 1;
You need to consider what should happen if you have two rows with the same date. This would pick one of them at random. If there is some other criteria you could use to break the tie you could add that to the order by clause. If you wanted all matching rows in that case you could use rank() instead; look at dense_rank() as well, they all have their place.
Use this query to do that:
SELECT max(prev_est_hrs) keep (dense_rank first order by change_date)
FROM task_history
WHERE task_id = 5;