I need to perform TREAMMEAN in Access, which does not have this function.
In a table I have many Employees, each has many records.
I need to TRIMMEAN Values for each Employee separately.
Following queries perform TOP 10 percent for all records:
qry_data_TOP10_ASC
qry_data_TOP10_DESC
unionqry_TOP10_ASCandDESC
qry_data_ALL_minus_union_qry
After that, I can use Avg (Average).
But I don't know how to do it for each employee.
Visualization:
Note:
This question is edited to simplify problem.
You don't really give information in your pseudo code about your data fields but using your example that DOES have basic field information I can suggest the following should work as you described
It assumes field1 is your unique record ID - but you make no mention of which fields are keys
SELECT AVG(qry_data.field2) FROM qry_data WHERE qry_data.field1 NOT IN
(SELECT * FROM
(SELECT TOP 10 PERCENT qry_data.field1, qry_data.field2
FROM qry_data
ORDER BY qry_data.field2 ASC)
UNION
(SELECT TOP 10 PERCENT qry_data.field1, qry_data.field2
FROM qry_data
ORDER BY qry_data.field2 DESC)
)
This should give you what you want, the two sub-queries should correlate the TOP 10s (ascending and descending) for every employee. The two NOT INs should then remove those from the Table1 records and then you group the Employees and Average the Scores.
SELECT Table1.Employee, AVG(Table1.Score) AS AvgScore
FROM Table1
WHERE ID NOT IN
(
SELECT TOP 10 ID
FROM Table1 a
WHERE a.Employee = Table1.Employee
ORDER BY Score ASC, Employee, ID
)
AND ID NOT IN
(
SELECT TOP 10 ID
FROM Table1 b
WHERE b.Employee = Table1.Employee
ORDER BY Score DESC, Employee, ID
)
GROUP BY Table1.Employee;
Related
I have this table called item:
| PERSON_id | ITEM_id |
|------------------|----------------|
|------CP2---------|-----A03--------|
|------CP2---------|-----A02--------|
|------HB3---------|-----A02--------|
|------BW4---------|-----A01--------|
I need an SQL statement that would output the person with the most Items. Not really sure where to start either.
I advice you to use inner query for this purpose. the inner query is going to include group by and order by statement. and outer query will select the first statement which has the most items.
SELECT * FROM
(
SELECT PERSON_ID, COUNT(*) FROM TABLE1
GROUP BY PERSON_ID
ORDER BY 2 DESC
)
WHERE ROWNUM = 1
here is the fiddler link : http://sqlfiddle.com/#!4/4c4228/5
Locating the maximum of an aggregated column requires more than a single calculation, so here you can use a "common table expression" (cte) to hold the result and then re-use that result in a where clause:
with cte as (
select
person_id
, count(item_id) count_items
from mytable
group by
person_id
)
select
*
from cte
where count_items = (select max(count_items) from cte)
Note, if more than one person shares the same maximum count; more than one row will be returned bu this query.
i have a table with a bunch of customer IDs. in a customer table is also these IDs but each id can be on multiple records for the same customer. i want to select the most recently used record which i can get by doing order by <my_field> desc
say i have 100 customer IDs in this table and in the customers table there is 120 records with these IDs (some are duplicates). how can i apply my order by condition to only get the most recent matching records?
dbms is sql server 2000.
table is basically like this:
loc_nbr and cust_nbr are primary keys
a customer shops at location 1. they get assigned loc_nbr = 1 and cust_nbr = 1
then a customer_id of 1.
they shop again but this time at location 2. so they get assigned loc_nbr = 2 and cust_Nbr = 1. then the same customer_id of 1 based on their other attributes like name and address.
because they shopped at location 2 AFTER location 1, it will have a more recent rec_alt_ts value, which is the record i would want to retrieve.
You want to use the ROW_NUMBER() function with a Common Table Expression (CTE).
Here's a basic example. You should be able to use a similar query with your data.
;WITH TheLatest AS
(
SELECT *, ROW_NUMBER() OVER (PARTITION BY group-by-fields ORDER BY sorting-fields) AS ItemCount
FROM TheTable
)
SELECT *
FROM TheLatest
WHERE ItemCount = 1
UPDATE: I just noticed that this was tagged with sql-server-2000. This will only work on SQL Server 2005 and later.
Since you didn't give real table and field names, this is just psuedo code for a solution.
select *
from customer_table t2
inner join location_table t1
on t1.some_key = t2.some_key
where t1.LocationKey = (select top 1 (LocationKey) as LatestLocationKey from location_table where cust_id = t1.cust_id order by some_field)
Use an aggregate function in the query to group by customer IDs:
SELECT cust_Nbr, MAX(rec_alt_ts) AS most_recent_transaction, other_fields
FROM tableName
GROUP BY cust_Nbr, other_fields
ORDER BY cust_Nbr DESC;
This assumes that rec_alt_ts increases every time, thus the max entry for that cust_Nbr would be the most recent entry.
By using time and date we can take out the recent detail for the customer.
use the column from where you take out the date and the time for the customer.
eg:
SQL> select ename , to_date(hiredate,'dd-mm-yyyy hh24:mi:ss') from emp order by to_date(hiredate,'dd-mm-yyyy hh24:mi:ss');
Let's say I have an app that determine the winners in a prize drawing. All entries are entered into a table indicating their employeeID. Each employee can enter the drawing multiple times. I select from the table, order by newid to get a random sort. I assume the more entries (database records) an employee has the better chance he will end up in the top 5 of my query each time I run it. So far so good. However, because each employee has multiple records, there is a good chance he will come up multiple times in the top 5. I need the ability to return 5 unique records from the randomly sorted results.
How do I get 5 unique rows while still ensuring those with multiple drawing entries get a heavier weighting in the selection?
My base query:
SELECT TOP 5 employeeID
FROM events
TABLESAMPLE(1000 ROWS)
ORDER BY CHECKSUM(NEWID());
Kinda what I am trying to do:
SELECT TOP 5 *
FROM events
WHERE employeeID IN (SELECT employeeID
FROM events
TABLESAMPLE(1000 ROWS)
ORDER BY CHECKSUM(NEWID())
)
ORDER BY CHECKSUM(NEWID())
But of course I cannot do an order by in the subquery.
Any solution must take into account 2 things:
If an employee enter multiple tickets, his chance of winning increases relative to other.
Everyone can only win once
Here's my approach:
;WITH
tmp1 AS
(
SELECT EmployeeID,
ROW_NUMBER() OVER (ORDER BY NEWID()) AS SortOrder
FROM Events
),
tmp2 AS
(
SELECT EmployeeID,
MIN(SortOrder) AS WinOrder
FROM tmp1
GROUP BY EmployeeID
)
SELECT TOP 5 *
FROM tmp2
ORDER BY WinOrder
The SQL Fiddle gives employees 1 & 5 higher chances to win, but they will only win once each, regardless of how many times they enter.
Here's a fairly simple way to get what you're after:
select top 5 EmployeeID
from
(
select EmployeeID, row_number() over (order by newid()) DrawOrder
from Events
) wins
group by EmployeeID
order by min(DrawOrder)
Let's say I have a bases with a table:
-courses (key: name [ofthecourse], other attributes: year in which the course takes place)
I want to complete a query looking for an answer to the question:
On which year of study there is a maximum number of courses?
Normally, the query would be:
SELECT TOP 1 STUDYEAR
FROM COURSES
GROUP BY STUDYEAR
ORDER BY COUNT(CNO) DESC;
But my question is, which query could complete this without using the TOP 1 phrase?
You can use an inner query to get the maximum count. The only difference is though that it can return more than one record if they have the same count.
SELECT STUDYEAR
FROM COURSES
GROUP BY STUDYEAR
HAVING COUNT(CNO) = (SELECT MAX(CNOCount) FROM
(SELECT COUNT(CNO) CNOCount
FROM COURSES
GROUP BY STUDYEAR) X)
Another version with only one inner query:
SELECT STUDYEAR
FROM
(SELECT STUDYEAR, ROW_NUMBER() OVER (ORDER BY COUNT(CNO) DESC) RowNumber
FROM COURSES
GROUP BY STUDYEAR) X
WHERE RowNumber = 1
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"