I have a table
User | Phone | Value
Peter | 0 | 1
Peter | 456 | 2
Peter | 456 | 3
Paul | 456 | 7
Paul | 789 | 10
I want to select MAX value for every user, than it also lower than a tresshold
For tresshold 8, I want result to be
Peter | 456 | 3
Paul | 456 | 7
I have tried the GROUP BY with HAVING, but I am getting
column "phone" must appear in the GROUP BY clause or be used in an aggregate function
Similar query logic works in MySQL, but I am not quite sure how to operate with GROUP BY in PostgreSQL. I dont want to GROUP BY phone.
After I have results from "juergen d" solution, I came up with this which gives me the same results faster
SELECT DISTINCT ON(user) user, phone, value
FROM table
WHERE value < 8
ORDER BY user, value DESC;
select t1.*
from your_table t1
join
(
select user, max(value) as max_value
from your_table
where value < 8
group by user
) t2 on t1.user = t2.user and t1.value = t2.max_value
Alternatively, you could use a ranking function:
select * from
(
select *, RANK() OVER (partition by [user] ORDER BY t.value desc ) as value_rank from test_table as t
where t.value < 8
) as t1
where value_rank = 1
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 table like this :
id name value
1 roger 43
2 phil 12
3 zac 14
4 phil 42
5 maurice 450
...
and i'm trying to retrieve the max value for each name in order to do a join later.
I'm expecting the intermediate result to be something like this :
name value
roger 43
zac 14
phil 42
maurice 450
And this is easily achieved using and select name,max(value) from table group by name
My issue is that i NEED the id in order to later be able to do my join. but if i add my id to the aggregate/ group by it will mess up the result and will show all values since the id will be different.
So the true expect result is more like this :
id name value
1 roger 43
3 zac 14
4 phil 42
5 maurice 450
I have seen many question regarding similar issues but none where the id need to be retrieved but not included in the group by since i want uniqness for the name and only need the id for my join.
From what you describe, you just want the max of id:
select max(id) as id, name, max(value)
from table
group by name;
Or, what I think you want is the row with the max value:
select distinct on (name) t.*
from t
order by name, value desc;
With not exists:
select min(t.id) id, t.name, t.value
from tablename t
where not exists (
select 1 from tablename
where name = t.name and value > t.value
)
group by t.name, t.value
order by id
See the demo.
Results:
| id | name | value |
| --- | ------- | ----- |
| 1 | roger | 43 |
| 3 | zac | 14 |
| 4 | phil | 42 |
| 5 | maurice | 450 |
Try this:
WITH X (n, v) AS (
SELECT name, MAX(value) FROM tbl GROUP BY name
)
SELECT T.*
FROM tbl AS T INNER JOIN X ON T.name = X.n AND T.value = X.v
I heed your help with the following:
I have a table like this:
Table_Values
ID | Value | Date
1 | ASD | 01-Jan-2019
2 | ZXC | 10-Jan-2019
3 | ASD | 01-Jan-2019
4 | QWE | 05-Jan-2019
5 | RTY | 15-Jan-2019
6 | QWE | 29-Jan-2019
That I need is to get the values that are duplicated and have a different Date, for example the value "QWE" is duplicated and has different date:
ID | Value | Date
4 | QWE | 05-Jan-2019
6 | QWE | 29-Jan-2019
With EXISTS:
select * from Table_Values t
where exists (
select 1 from Table_Values
where value = t.value and date <> t.date
)
Using Join:
select
t1.*
from
Table_Values t1
join
Table_Values t2
on t1.Value = t2.Value
and t1.Date <> t2.Date
However, Exists approach is better.
You want all rows where there is more than one date per value. You can use COUNT OVER for this.
One method (featured as of Oracle 12c):
select id, value, date
from mytable
order by case when count(distinct date) over (partition by value) > 1 then 1 else 2 end
fetch first row with ties
But you'll have to put this into a subquery (derived table / cte), if you want the result sorted.
And another method without FETCH FIRST clause (valid as of Oracle 8i):
select id, value, date
from
(
select id, value, date, count(distinct date) over (partition by value) as cnt
from mytable
)
where cnt > 1
order by id, value, date;
forpas' solution with EXISTS may be faster, though. Well, pick whichever method you like better :-)
With EXISTS, "correlated subquery" is used. So I don't think it's better than JOIN.
However, Oracle optimizer could re-write "EXISTS" to JOIN.
I like to use JOIN in classic way :)
SELECT t1.*
FROM table_values t1, table_values t2
WHERE t1.f_value = t2.f_value
AND t1.f_date <> t2.f_date
ORDER BY 1;
I have inherited a SQL Server table in the (abbreviated) form of (includes sample data set):
| SID | name | Invite_Date |
|-----|-------|-------------|
| 101 | foo | 2013-01-06 |
| 102 | bar | 2013-04-04 |
| 101 | fubar | 2013-03-06 |
I need to select all SID's and the Invite_date, but if there is a duplicate SID, then just get the latest entry (by date).
So the results from the above would look like:
101 | fubar | 2013-03-06
102 | bar | 2013-04-04
Any ideas please.
N.B the Invite_date column has been declared as a nvarchar, so to get it in a date format I am using CONVERT(DATE, Invite_date)
You can use a ranking function like ROW_NUMBER or DENSE_RANK in a CTE:
WITH CTE AS
(
SELECT SID, name, Invite_Date,
rn = Row_Number() OVER (PARTITION By SID
Order By Invite_Date DESC)
FROM dbo.TableName
)
SELECT SID, name, Invite_Date
FROM CTE
WHERE RN = 1
Demo
Use Row_Number if you want exactly one row per group and Dense_Rank if you want all last Invite_Date rows for each group in case of repeating max-Invite_Dates.
select t1.*
from your_table t1
inner join
(
select sid, max(CONVERT(DATE, Invite_date)) mdate
from your_table
group by sid
) t2 on t1.sid = t2.sid and CONVERT(DATE, t1.Invite_date) = t2.mdate
select
SID,name,MAX(Invite_date)
FROM
Table1
GROUP BY
SID
http://sqlfiddle.com/#!2/6b6f66/1
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