I have a table like this below and i need the result to be like this when i run the query
Results
title | count
----------------
foo | 3
bar | 2
Table
customer_id | title
-------------------
55 | foo
22 | foo
55 | bar <-- duplicate
23 | bar
55 | bar <-- duplicate
23 | foo
UPDATE Thank you all for the quick response!
The trick is to count the distinct customer ids, so you won't count the double Foo for customer 55.
If you need to, you can order the results by that count too, or you can just leave out the order by clause.
select
title,
count(DISTINCT customerid) as `count`
from
yourTable
group by
title
order by
`count` desc
It's as easy as this:
select A.title, count(*) as count -- an usual count(*)
from
(select distinct customer_id, title -- this query removes the duplicates
from table) A
group by A.title -- an usual group by
For SQL Server you can do it like this.
select
t.title ,
count(*)
from your_table as t
group by
t.title
order by count(*) DESC
Related
In Presto, there's an arbitrary() aggregate function to select any arbitrary row in a given group. If there's no group by clause, then I can use distinct on. With group by, every selected column must be in the group by or be an aggregated column. E.g.:
| id | foo |
| 1 | 123 |
| 1 | 321 |
select id, arbitrary(foo), count(*)
from mytable
group by id
Fiddle
It doesn't matter if it returns 1, 123, 2 or 1, 321, 2. Something like min() or max() works, but it's a lot slower.
Does something like arbitrary() exist in Postgres?
select m.foo,b.id,b.cnt from mytable m
join (select id, count(*) cnt
from mytable
group by id) b using (id) limit 1;
If not explicit mention asc, desc all the order is not guaranteed. Therefore in the above query the foo's appearance is arbitrary.
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 want count how many rows use the same catid and order the query by total.
id | catid | name
0 | 1 | foo
1 | 1 | bar
2 | 2 | paint
I've tried COUNT(catid) but this requires a GROUP BY, and I do not want to compress rows.
How may I do this?
Do you want window functions?
select t.*, count(*) over (partition by catid) as cat_cnt
from t
order by cat_cnt, catid;
I should note that if you don't want to see the total, you can put the window function in the order by:
select *
from t
order by count(*) over (partition by catid), catid
Maybe you could run the GROUP BY as a separate SELECT, then JOIN?
E.g.
select orig.*, summ.totals
from t
join (select count(cat_id) totals
from t
group by cat_id) summ
on t.cat_id = summ.cat_id;
i have table storing product price information, the table looks similar to, (no is the primary key)
no name price date
1 paper 1.99 3-23
2 paper 2.99 5-25
3 paper 1.99 5-29
4 orange 4.56 4-23
5 apple 3.43 3-11
right now I want to select all the rows where the "name" field appeared more than once in the table. Basically, i want my query to return the first three rows.
I tried:
SELECT * FROM product_price_info GROUP BY name HAVING COUNT(*) > 1
but i get an error saying:
column "product_price_info.no" must appear in the GROUP BY clause or be used in an aggregate function
SELECT *
FROM product_price_info
WHERE name IN (SELECT name
FROM product_price_info
GROUP BY name HAVING COUNT(*) > 1)
Try this:
SELECT no, name, price, "date"
FROM (
SELECT no, name, price, "date",
COUNT(*) OVER (PARTITION BY name) AS cnt
FROM product_price_info ) AS t
WHERE t.cnt > 1
You can use the window version of COUNT to get the population of each name partition. Then, in an outer query, filter out name partitions having a population that is less than 2.
Window Functions are really nice for this.
SELECT p.*, count(*) OVER (PARTITION BY name) FROM product p;
For a full example:
CREATE TABLE product (no SERIAL, name text, price NUMERIC(8,2), date DATE);
INSERT INTO product(name, price, date) values
('paper', 1.99, '2017-03-23'),
('paper', 2.99, '2017-05-25'),
('paper', 1.99, '2017-05-29'),
('orange', 4.56, '2017-04-23'),
('apple', 3.43, '2017-03-11')
;
WITH report AS (
SELECT p.*, count(*) OVER (PARTITION BY name) as count FROM product p
)
SELECT * FROM report WHERE count > 1;
Gives:
no | name | price | date | count
----+--------+-------+------------+-------
1 | paper | 1.99 | 2017-03-23 | 3
2 | paper | 2.99 | 2017-05-25 | 3
3 | paper | 1.99 | 2017-05-29 | 3
(3 rows)
Self join version, use a sub-query that returns the name's that appears more than once.
select t1.*
from tablename t1
join (select name from tablename group by name having count(*) > 1) t2
on t1.name = t2.name
Basically the same as IN/EXISTS versions, but probably a bit faster.
SELECT name, count(name)
FROM product_price_info
GROUP BY name
HAVING COUNT(name) > 1
LIMIT 3
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