I have the following data set:
| EMAIL | SIGNUP_DATE |
| A#ABC.COM | 1/1/2021 |
| B#ABC.COM | 1/2/2021 |
| C#ABC.COM | 1/3/2021 |
In order to find the running total of email signups as of a certain day, I ran the following sql query:
select
signup_date,
count(email) OVER (order by signup_date ASC) as running_total_signups
I got the following results:
| SIGNUP_DATE | RUNNING_TOTAL_SIGNUPS |
| 1/1/21 | 1 |
| 1/2/21 | 2 |
| 1/3/21 | 3 |
However for my next step, I want to be able to see not just the running total signups, but the actual signup names themselves. Therefore I want to run the same window function (count(email) OVER (order by signup_date ASC)) but instead of a count(email) just a select distinct email. This would hopefully result in the following output:
| SIGNUP_DATE | RUNNING_TOTAL_SIGNUPS |
| 1/1/21 | a#abc.com |
| 1/2/21 | a#abc.com |
| 1/2/21 | b#abc.com |
| 1/3/21 | a#abc.com |
| 1/3/21 | b#abc.com |
| 1/3/21 | c#abc.com |
How would I do this? I'm getting an error on this code:
select
signup_date,
distinct email OVER (order by signup_date ASC) as running_total_signups
One way would be to cross-join the results and filter the joined table having a total <= to the running total:
with counts as (
select *,
Count(*) over (order by SIGNUP_DATE asc) as tot
from t
)
select c1.EMAIL, c1.SIGNUP_DATE
from counts c1
cross join counts c2
where c2.tot <= c1.tot
I want to run the same window function (count(email) OVER (order by
signup_date ASC)) but instead of a count(email) just a select distinct
email
Why do you want COUNT() window function?
It has nothing to do with with your reqirement.
All you need is a simple self join:
SELECT t1.SIGNUP_DATE, t2.EMAIL
FROM tablename t1 INNER JOIN tablename t2
ON t2.SIGNUP_DATE <= t1.SIGNUP_DATE
ORDER BY t1.SIGNUP_DATE, t2.EMAIL;
which will work for your sample data, but just in case there are more than 1 rows for each day in your table you should use:
SELECT t1.SIGNUP_DATE, t2.EMAIL
FROM (SELECT DISTINCT SIGNUP_DATE FROM tablename) t1 INNER JOIN tablename t2
ON t2.SIGNUP_DATE <= t1.SIGNUP_DATE
ORDER BY t1.SIGNUP_DATE, t2.EMAIL;
See the demo.
It's actually slightly simpler than Stu proposed:
select
x2.signup_date,
x1.email
from
signups x1
INNER JOIN signups x2 ON x1.signup_date <= x2.signup_date
order by signup_date
If you join the table to itself but for any date that is less than or equal to, it causes a half cartesian explosion. The lowest dated row matches with only itself. The next one matches with itself and the earlier one, so one of the table aliases has its data repeated.. This continues adding more rows to the explosion as the dates increase:
In this resultset we can see we want the emails from x1, and the dates from x2
Related
Here's what I'm trying to do. Let's say I have this table t:
key_id | id | record_date | other_cols
1 | 18 | 2011-04-03 | x
2 | 18 | 2012-05-19 | y
3 | 18 | 2012-08-09 | z
4 | 19 | 2009-06-01 | a
5 | 19 | 2011-04-03 | b
6 | 19 | 2011-10-25 | c
7 | 19 | 2012-08-09 | d
For each id, I want to select the row containing the minimum record_date. So I'd get:
key_id | id | record_date | other_cols
1 | 18 | 2011-04-03 | x
4 | 19 | 2009-06-01 | a
The only solutions I've seen to this problem assume that all record_date entries are distinct, but that is not this case in my data. Using a subquery and an inner join with two conditions would give me duplicate rows for some ids, which I don't want:
key_id | id | record_date | other_cols
1 | 18 | 2011-04-03 | x
5 | 19 | 2011-04-03 | b
4 | 19 | 2009-06-01 | a
How about something like:
SELECT mt.*
FROM MyTable mt INNER JOIN
(
SELECT id, MIN(record_date) AS MinDate
FROM MyTable
GROUP BY id
) t ON mt.id = t.id AND mt.record_date = t.MinDate
This gets the minimum date per ID, and then gets the values based on those values. The only time you would have duplicates is if there are duplicate minimum record_dates for the same ID.
I could get to your expected result just by doing this in mysql:
SELECT id, min(record_date), other_cols
FROM mytable
GROUP BY id
Does this work for you?
To get the cheapest product in each category, you use the MIN() function in a correlated subquery as follows:
SELECT categoryid,
productid,
productName,
unitprice
FROM products a WHERE unitprice = (
SELECT MIN(unitprice)
FROM products b
WHERE b.categoryid = a.categoryid)
The outer query scans all rows in the products table and returns the products that have unit prices match with the lowest price in each category returned by the correlated subquery.
I would like to add to some of the other answers here, if you don't need the first item but say the second number for example you can use rownumber in a subquery and base your result set off of that.
SELECT * FROM
(
SELECT
ROW_NUM() OVER (PARTITION BY Id ORDER BY record_date, other_cols) as rownum,
*
FROM products P
) INNER
WHERE rownum = 2
This also allows you to order off multiple columns in the subquery which may help if two record_dates have identical values. You can also partition off of multiple columns if needed by delimiting them with a comma
This does it simply:
select t2.id,t2.record_date,t2.other_cols
from (select ROW_NUMBER() over(partition by id order by record_date)as rownum,id,record_date,other_cols from MyTable)t2
where t2.rownum = 1
If record_date has no duplicates within a group:
think of it as of filtering. Simpliy get (WHERE) one (MIN(record_date)) row from the current group:
SELECT * FROM t t1 WHERE record_date = (
select MIN(record_date)
from t t2 where t2.group_id = t1.group_id)
If there could be 2+ min record_date within a group:
filter out non-min rows (see above)
then (AND) pick only one from the 2+ min record_date rows, within the given group_id. E.g. pick the one with the min unique key:
AND key_id = (select MIN(key_id)
from t t3 where t3.record_date = t1.record_date
and t3.group_id = t1.group_id)
so
key_id | group_id | record_date | other_cols
1 | 18 | 2011-04-03 | x
4 | 19 | 2009-06-01 | a
8 | 19 | 2009-06-01 | e
will select key_ids: #1 and #4
SELECT p.* FROM tbl p
INNER JOIN(
SELECT t.id, MIN(record_date) AS MinDate
FROM tbl t
GROUP BY t.id
) t ON p.id = t.id AND p.record_date = t.MinDate
GROUP BY p.id
This code eliminates duplicate record_date in case there are same ids with same record_date.
If you want duplicates, remove the last line GROUP BY p.id.
This a old question, but this can useful for someone
In my case i can't using a sub query because i have a big query and i need using min() on my result, if i use sub query the db need reexecute my big query. i'm using Mysql
select t.*
from (select m.*, #g := 0
from MyTable m --here i have a big query
order by id, record_date) t
where (1 = case when #g = 0 or #g <> id then 1 else 0 end )
and (#g := id) IS NOT NULL
Basically I ordered the result and then put a variable in order to get only the first record in each group.
The below query takes the first date for each work order (in a table of showing all status changes):
SELECT
WORKORDERNUM,
MIN(DATE)
FROM
WORKORDERS
WHERE
DATE >= to_date('2015-01-01','YYYY-MM-DD')
GROUP BY
WORKORDERNUM
select
department,
min_salary,
(select s1.last_name from staff s1 where s1.salary=s3.min_salary ) lastname
from
(select department, min (salary) min_salary from staff s2 group by s2.department) s3
I have the following table
| id | date | team |
|----|------------|------|
| 1 | 2019-01-05 | A |
| 2 | 2019-01-05 | A |
| 3 | 2019-01-01 | A |
| 4 | 2019-01-04 | B |
| 5 | 2019-01-01 | B |
How can I query the table to receive the most recent values for the teams?
For example, the result for the above table would be ids 1,2,4.
In this case, you can use window functions:
select t.*
from (select t.*, rank() over (partition by team order by date desc) as seqnum
from t
) t
where seqnum = 1;
In some databases a correlated subquery is faster with the right indexes (I haven't tested this with Postgres):
select t.*
from t
where t.date = (select max(t2.date) from t t2 where t2.team = t.team);
And if you wanted only one row per team, then the canonical answer is:
select distinct on (t.team) t.*
from t
order by t.team, t.date desc;
However, that doesn't work in this case because you want all rows from the most recent date.
If your dataset is large, consider the max analytic function in a subquery:
with cte as (
select
id, date, team,
max (date) over (partition by team) as max_date
from t
)
select id
from cte
where date = max_date
Notionally, max is O(n), so it should be pretty efficient. I don't pretend to know the actual implementation on PostgreSQL, but my guess is it's O(n).
One more possibility, generic:
select * from t join (select max(date) date,team from t
group by team) tt
using(date,team)
Window function is the best solution for you.
select id
from (
select team, id, rank() over (partition by team order by date desc) as row_num
from table
) t
where row_num = 1
That query will return this table:
| id |
|----|
| 1 |
| 2 |
| 4 |
If you to get it one row per team, you need to use array_agg function.
select team, array_agg(id) ids
from (
select team, id, rank() over (partition by team order by date desc) as row_num
from table
) t
where row_num = 1
group by team
That query will return this table:
| team | ids |
|------|--------|
| A | [1, 2] |
| B | [4] |
I've got two tables wchich I need to join with Select and I've got a problem.
The tables look like that:
table_price
Product_ID | Buy_date | Buy_price |
1 | 16.10.01 | 2.50 |
1 | 16.11.02 | 3.20 |
2 | 16.10.31 | 3.80 |
table expire_date
Product_ID | Count | Exp_date |
1 | 1000 | 17.10.01|
1 | 500 | 17.11.31|
2 | 500 | 17.11.01|
I need to write a select in Oracle PL/SQL wchich gives me following results:
Product_ID| Count | Exp_date| last_buy_price|
1 | 1000 | 17.10.01| 3.20 |
1 | 500 | 17.31.31| 3.20 |
2 | 500 | 17.11.01| 3.80 |
It means that it will give me every expire date with count of product from table expire_date and match it with last buy price from table_price with product_id (always with last buy price, ordered by column buy_date)
Please guys help me, I've tried so many codes and I still can't get satysfying results
A correlated subquery using keep is possibly the most performant method:
select ed.*,
(select max(p.buy_price) keep (dense_rank first order by p.buy_date desc)
from table_price p
where p.product_id = ed.product_id
) as last_buy_price
from expire_date ed;
You could, of course, also express this in the from clause:
select ed.*, p.last_buy_price
from expire_date ed left join
(select p.product_id,
max(p.buy_price) keep (dense_rank first order by p.buy_date desc) as last_buy_price
from table_price p
) p
on p.product_id = ed.product_id;
You can use ROW_NUMBER() :
SELECT ed.*,
tp.buy_price as last_buy_price
FROM expire_date ed
JOIN(SELECT s.*,
ROW_NUMBER() OVER(PARTITION BY s.product_id ORDER BY s.buy_date DESC) as rnk
FROM table_price s) tp
ON(ed.product_id = tp.product_id and tp.rnk = 1 )
I have a table that stores the lines on a contract. Each contract line his it's own unique ID, it also has the ID of its parent contract. Example:
+-------------+---------+
| contract_id | line_id |
+-------------+---------+
| 1111 | 100 |
| 1111 | 101 |
| 1111 | 102 |
+-------------+---------+
I have another table that stores the historical changes to contract lines. For example, every time the number of units on a contract line is changed a new row is added to the table. Example:
+-------------+---------+--------------+-------+
| contract_id | line_id | date_changed | units |
+-------------+---------+--------------+-------+
| 1111 | 100 | 2016-01-01 | 1 |
| 1111 | 100 | 2016-02-01 | 2 |
| 1111 | 100 | 2016-03-01 | 3 |
+-------------+---------+--------------+-------+
As you can see the contract line with ID 100 belonging to the contract with ID 1111 has been edited 3 times over 3 months. The current value is 3 units.
I'm running a query against the contract lines table to select all data. I want to join to the historical data table and select the most recent row for each contract line and show the units in my results. How do I do this?
Expected results (there would single results for 101 and 102 as well):
+-------------+---------+-------+
| contract_id | line_id | units |
+-------------+---------+-------+
| 1111 | 100 | 3 |
+-------------+---------+-------+
I've tried the query below with a left join but it returns 3 rows instead of 1.
Query:
SELECT *, T1.units
FROM contract_lines
LEFT JOIN (
SELECT contract_id, line_id, units, MAX(date_changed) AS maxdate
FROM contract_history
GROUP BY contract_id, line_id, units) AS T1
ON contract_lines.contract_id = T1.contract_id
AND contract_lines.line_id = T1.line_id
Actual results:
+-------------+---------+-------+
| contract_id | line_id | units |
+-------------+---------+-------+
| 1111 | 100 | 1 |
| 1111 | 100 | 2 |
| 1111 | 100 | 3 |
+-------------+---------+-------+
An extra join to contract_history along with maxdate will work
SELECT contract_lines.*,T2.units
FROM contract_lines
LEFT JOIN (
SELECT contract_id, line_id, MAX(date_changed) AS maxdate
FROM contract_history
GROUP BY contract_id, line_id) AS T1
JOIN contract_history T2 ON
T1.contract_id=T2.contract_id and
T1.line_id= T2.line_id and
T1.maxdate=T2.date_changed
ON contract_lines.contract_id = T1.contract_id
AND contract_lines.line_id = T1.line_id
Output
This is my preferred style because it doesn't require self joining and cleanly expresses your intent. Also, it competes very well with the ROW_NUMBER() method in terms of performance.
select a.*
, b.units
from contract_lines as a
join (
select a.contract_id
, a.line_id
, a.units
, Max(a.date_changed) over(partition by a.contract_id, a.line_id) as max_date_changed
from contract_history as a
) as b
on a.contract_id = b.contract_id
and a.line_id = b.line_id
and b.date_changed = b.max_date_changed;
Another possible solution to this. This uses RANK to sort/filter this. Similar to what you did, just a different tact.
SELECT contract_lines.*, T1.units
FROM contract_lines
LEFT JOIN (
SELECT contract_id, line_id, units,
RANK() OVER (PARTITION BY contract_id, line_id ORDER BY date_changed DESC) AS [rank]
FROM contract_history) AS T1
ON contract_lines.contract_id = T1.contract_id
AND contract_lines.line_id = T1.line_id
AND T1.rank = 1
WHERE T1.units IS NOT NULL
You could change this to a INNER JOIN and remove the IS NOT NULL in the WHERE clause if you expect data to be present all the time.
Glad you figured it out!
Try this simple query:
SELECT TOP 1 T1.*
FROM contract_lines T0
INNER JOIN contract_history T1
ON T0.contract_id = T1.contract_id and
T0.line_id = T1.line_id
ORDER BY date_changed DESC
As always seems to be the way after spending an hour looking at it and shouting at StackOverflow for having a rare period of maintenance I solve my own problem not long after posting a question.
In an effort to help anyone else who's stuck I'll show what I found. It might not be an efficient way to achieve this so if someone has a better suggestion I'm all ears.
I adapted the answer from here: T-SQL Subquery Max(Date) and Joins
SELECT *,
Units = (SELECT TOP 1 units
FROM contract_history
WHERE contract_lines.contract_id = contract_history.contract_id
AND contract_lines.line_id = contract_history.line_id
ORDER BY date_changed DESC
)
FROM ....
Here's what I'm trying to do. Let's say I have this table t:
key_id | id | record_date | other_cols
1 | 18 | 2011-04-03 | x
2 | 18 | 2012-05-19 | y
3 | 18 | 2012-08-09 | z
4 | 19 | 2009-06-01 | a
5 | 19 | 2011-04-03 | b
6 | 19 | 2011-10-25 | c
7 | 19 | 2012-08-09 | d
For each id, I want to select the row containing the minimum record_date. So I'd get:
key_id | id | record_date | other_cols
1 | 18 | 2011-04-03 | x
4 | 19 | 2009-06-01 | a
The only solutions I've seen to this problem assume that all record_date entries are distinct, but that is not this case in my data. Using a subquery and an inner join with two conditions would give me duplicate rows for some ids, which I don't want:
key_id | id | record_date | other_cols
1 | 18 | 2011-04-03 | x
5 | 19 | 2011-04-03 | b
4 | 19 | 2009-06-01 | a
How about something like:
SELECT mt.*
FROM MyTable mt INNER JOIN
(
SELECT id, MIN(record_date) AS MinDate
FROM MyTable
GROUP BY id
) t ON mt.id = t.id AND mt.record_date = t.MinDate
This gets the minimum date per ID, and then gets the values based on those values. The only time you would have duplicates is if there are duplicate minimum record_dates for the same ID.
I could get to your expected result just by doing this in mysql:
SELECT id, min(record_date), other_cols
FROM mytable
GROUP BY id
Does this work for you?
To get the cheapest product in each category, you use the MIN() function in a correlated subquery as follows:
SELECT categoryid,
productid,
productName,
unitprice
FROM products a WHERE unitprice = (
SELECT MIN(unitprice)
FROM products b
WHERE b.categoryid = a.categoryid)
The outer query scans all rows in the products table and returns the products that have unit prices match with the lowest price in each category returned by the correlated subquery.
I would like to add to some of the other answers here, if you don't need the first item but say the second number for example you can use rownumber in a subquery and base your result set off of that.
SELECT * FROM
(
SELECT
ROW_NUM() OVER (PARTITION BY Id ORDER BY record_date, other_cols) as rownum,
*
FROM products P
) INNER
WHERE rownum = 2
This also allows you to order off multiple columns in the subquery which may help if two record_dates have identical values. You can also partition off of multiple columns if needed by delimiting them with a comma
This does it simply:
select t2.id,t2.record_date,t2.other_cols
from (select ROW_NUMBER() over(partition by id order by record_date)as rownum,id,record_date,other_cols from MyTable)t2
where t2.rownum = 1
If record_date has no duplicates within a group:
think of it as of filtering. Simpliy get (WHERE) one (MIN(record_date)) row from the current group:
SELECT * FROM t t1 WHERE record_date = (
select MIN(record_date)
from t t2 where t2.group_id = t1.group_id)
If there could be 2+ min record_date within a group:
filter out non-min rows (see above)
then (AND) pick only one from the 2+ min record_date rows, within the given group_id. E.g. pick the one with the min unique key:
AND key_id = (select MIN(key_id)
from t t3 where t3.record_date = t1.record_date
and t3.group_id = t1.group_id)
so
key_id | group_id | record_date | other_cols
1 | 18 | 2011-04-03 | x
4 | 19 | 2009-06-01 | a
8 | 19 | 2009-06-01 | e
will select key_ids: #1 and #4
SELECT p.* FROM tbl p
INNER JOIN(
SELECT t.id, MIN(record_date) AS MinDate
FROM tbl t
GROUP BY t.id
) t ON p.id = t.id AND p.record_date = t.MinDate
GROUP BY p.id
This code eliminates duplicate record_date in case there are same ids with same record_date.
If you want duplicates, remove the last line GROUP BY p.id.
This a old question, but this can useful for someone
In my case i can't using a sub query because i have a big query and i need using min() on my result, if i use sub query the db need reexecute my big query. i'm using Mysql
select t.*
from (select m.*, #g := 0
from MyTable m --here i have a big query
order by id, record_date) t
where (1 = case when #g = 0 or #g <> id then 1 else 0 end )
and (#g := id) IS NOT NULL
Basically I ordered the result and then put a variable in order to get only the first record in each group.
The below query takes the first date for each work order (in a table of showing all status changes):
SELECT
WORKORDERNUM,
MIN(DATE)
FROM
WORKORDERS
WHERE
DATE >= to_date('2015-01-01','YYYY-MM-DD')
GROUP BY
WORKORDERNUM
select
department,
min_salary,
(select s1.last_name from staff s1 where s1.salary=s3.min_salary ) lastname
from
(select department, min (salary) min_salary from staff s2 group by s2.department) s3