I have a table made up of dates and sales totals for the particular date. I would like to be able to query the table and select the following: max sales, the date associated with the max sale figure, sum of all sales, and the minimum date in the table. One additional complication is that there are duplicate max values. I don't care which max value is chosen but I just want one at random. This is for Oracle.
Here is what I tried. It was using a sub query.
Select sales, date, min(date), sum(sales) from table
Where sales = (select distinct(max(sales)) from table)
select
max(sales),
max(date_) keep (dense_rank first order by sales desc),
sum(sales),
min(date_)
from
table_
See also This SQL Fiddle
Related
I'm trying to find the nth greatest value in each group in a table; is there an efficient way to do this in SQL? (specifically Google BigQuery, if that's relevant)
For example, suppose we had a table sales with two fields, customer_id and amount, where each record corresponds to the sale of an item to a customer for a given amount. If I wanted the top sale to each customer, I could do
SELECT customer_id, MAX(amount) top_amount
FROM sales
GROUP BY customer_id;
If I instead wanted the 5th greatest value for each customer, is there an efficient/idiomatic way to do that in SQL?
Consider below approach
SELECT customer_id, array_agg(amount order by amount desc limit 5)[safe_offset(4)] top_5th_amount
FROM sales
GROUP BY customer_id;
Yet another option with use of nth_value() function
SELECT distinct customer_id,
nth_value(amount, 5) over win top_5th_amount
FROM sales
window win as (partition by customer_id order by amount desc rows between unbounded preceding and unbounded following )
You can use qualify:
select s.*
from sales s
where 1=1
qualify row_number() over (partition by customer_id order by amount desc) = 5;
Note: You question is unclear on how to handle tied amounts. This treats them as separate amounts (so the 5th could be the same as the 1st). If you want the 5th largest distinct value, use dense_rank() instead.
I have a table that contains unique transactions along with the year of the transaction and the employee who executed it. I need to find the employee with most transactions in each year.
I need a table with each year, the employee w/ most transactions in that year, and the number of transactions they had in that year.
This is as close as I am able to get without producing an error. I am unable to select the employee without producing an aggregation error.
select year, max(num_trans)
from (select year, employee, count(trans_id) as num_trans
from transactions
group by year, employee) as x
group by year
I am curious about how to work around this.
Use distinct on:
select distinct on (year) year, employee, count(*) as num_trans
from transactions
group by year, employee
order by year, count(*) desc;
distinct on is a handy Postgres extension to standard SQL that keeps the first row in a group of rows. The groups are defined by the distinct on key(s). Which row is first is determined by the order by.
I am trying to pull the most recent sale amount for each salesperson. The salespeople have made a sale on multiple days, I only want the most recent one.
My attempt below:
SELECT salesperson, amount
FROM table
WHERE date = (SELECT MAX(date) FROM table);
Use correlated subquery :
SELECT t.salesperson, t.amount
FROM table t
WHERE t.date = (SELECT MAX(t1.date)
FROM table t1
WHERE t1.salesperson = t.salesperson -- for each salesperson
);
If you are using PostgreSQL, you can take advantage of DISTINCT ON:
SELECT DISTINCT ON (salesperson) salesperson, amount
FROM table t
ORDER BY salesperson, date DESC
This will return only one row for each salesperson. The ORDER BY clause says to return the one with the largest date for that salesperson.
Unfortunately, DISTINCT ON is not supported by other databases.
I have table which looks similar to this:
I want to build query for searching all records from this table with the given date (let's say 5.12.2019) and with earlier dates but group by materialID.
Example: select all materials with date 6.12.2019 should show all materials with this date (or materials with earlier dates) group by material id with the biggest date Result should look like this:
Problem: I want to group my results by MaterialID with the biggest date. So in this example I don't want to show materials with the same id with earlier dates.
For the same example:
Question: How to build query like this using SQL and also JPQL? Because i would like to use this query in Hibernate so i need also JPQL query.
Thanks for your help.
This is a special case of a "top N per category" query. You want to show the maximum date per material id. In SQL (would also work in JPQL):
SELECT SUM(Amount), SUM(Price), MaterialId, MAX(Date)
FROM t
GROUP BY MaterialId
Note that with this technique, you cannot also display the ID, or MAX(ID), as the IDs and dates are not necessarily both monotonously increasing. If you still want the ID displayed as in your example, then write this SQL query (I don't think this can be done in JPQL):
SELECT MAX(ID), SUM(Amount), SUM(Price), MaterialId, MAX(Date)
FROM (
SELECT last_value(ID) OVER (
PARTITION BY MaterialId
ORDER BY Date, ID
RANGE BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING
) AS ID,
Amount,
Price,
MaterialId,
SELECT last_value(Date) OVER (
PARTITION BY MaterialId
ORDER BY Date, ID
RANGE BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING
) AS Date
FROM t
) t
GROUP BY MaterialId
Afternoon
I am trying to return the min value/ max values in SQL Server 2005 when I have multiple dates that are the same but the values in the Owed column are all different. I've already filtered the table down by my select statement into a temp table for a different query, when I've then tried to mirror I have all the duplicated dates that you can see below.
I now have a table that looks like:
ID| Date |Owes
-----------------
1 20110901 89
1 20110901 179
1 20110901 101
1 20110901 197
1 20110901 510
2 20111001 10
2 20111001 211
2 20111001 214
2 20111001 669
My current query:
Drop Table #Temp
Select Distinct Convert(Varchar(8), DateAdd(dd, Datediff(DD,0,DateDue),0),112)as Date
,ID
,Paid
Into #Temp
From Table
Where Paid <> '0'
Select ,Id
,Date
,Max(Owed)
,Min(Owed)
From #Temp
Group by ID, Date, Paid
Order By ID, Date, Paid
This doesn't strip out any of my dates that are the same, I'm new to SQL but I'm presuming its because my owed column has different values. I basically want to be able to pull back the first record as this will always be my minimum paid and my last record will always be my maximum owed to work out my total owed by ID.
I'm new to SQL so would like to understand what I've done wrong for my future knowledge of structuring queries?
Many Thanks
In your "select into"statement, you don't have an Owed column?
GROUP BY is the normal way you "strip out values that are the same". If you group by ID and Date, you will get one row in your result for each distinct pair of values in those two columns. Each row in the results represents ALL the rows in the underlying table, and aggregate functions like MIN, MAX, etc. can pull out values.
SELECT id, date, MAX(owes) as MaxOwes, MIN(owes) as minOwes
FROM myFavoriteTable
GROUP BY id, date
In SQL Server 2005 there are "windowing functions" that allow you to use aggregate functions on groups of records, without grouping. An example below. You will get one row for each row in the table:
SELECT id, date, owes,
MAX(Owes) over (PARTITION BY select, id) AS MaxOwes,
MIN(Owes) over (PARTITION BY select, id) AS MinOwes
FROM myfavoriteTable
If you name a column "MinOwes" it might sound like you're just fishing tho.
If you want to group by date you can't also group by ID, too, because ID is probably unique. Try:
Select ,Date
,Min(Owed) AS min_date
,Max(Owed) AS max_date
From #Temp
Group by Date
Order By Date
To get additional values from the row (your question is a bit vague there), you could utilize window functions:
SELECT DISTINCT
,Date
,first_value(ID) OVER (PARTITION BY Date ORDER BY Owed) AS min_owed_ID
,last_value(ID) OVER (PARTITION BY Date ORDER BY Owed) AS max_owed_ID
,first_value(Owed) OVER (PARTITION BY Date ORDER BY Owed) AS min_owed
,last_value(Owed) OVER (PARTITION BY Date ORDER BY Owed) AS max_owed
FROM #Temp
ORDER BY Date;