I have a picture as above, I would like to keep the item 2 and item 4 that have max count and name duplicated, but I need to keep the ID column. how to write the SQL script and get the result as below?
ID Name Count
2 A 5
4 B 5
Pretty sure this is what you want:
SELECT id, name, count
FROM my_table mt1
WHERE count = (SELECT MAX(count) FROM my_table mt2 WHERE mt1.name = mt2.name);
EDIT: If, however, as Chris suggests it does have duplicates:
SELECT MAX(id), name, count
FROM my_table mt1
WHERE count = (SELECT MAX(count) FROM my_table mt2 WHERE mt1.name = mt2.name)
GROUP BY name, count;
Output is:
+------+------+-------+
| id | name | count |
+------+------+-------+
| 2 | A | 5 |
| 4 | B | 5 |
+------+------+-------+
SELECT ID, Name, Count
FROM myTable
WHERE (SELECT COUNT(*) FROM myTable t2
WHERE t2.Name = myTable.Name
AND (t2.Count > myTable.Count
OR (t2.Count = myTable.Count AND t2.ID > myTable.ID))) = 0;
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 am not sure how to write the title of this question so the title might sound really confusing so please look at here. let suppose we have these two tables.
The first table is id, name, gender. This is the id of their game like login where name and gender is their name and gender in real life.
Where in the second table, name, and gender refer to in-game gender and name refer to in-game name as well.
id | name | gender id | name | Gender
---+------+---- ----+-------------+--------
1 | A | F 1 | a | F
2 | B | M and 1 | b | F
3 | C | M 2 | c | F
4 | D | M 3 | d | M
3 | e | M
3 | f | F
4 | g | M
4 | h | M
We want to select the id, name, and gender(in the first table), and the number of characters that an id had but that doesn't have the same gender as real-life. This might sound really confusing so here is what should be the output is
id | name | gender| #Character
---+-------+-------+-----------
2 B M 1
3 C M 3
id1: The reason it doesn't output id 1 is that id = 1 is F and she created 2 characters in the game, but both of the characters are F so she didn't 'switch' her gender so we do not print the row.
id2: We select id 2 because id = 2 is M in real-life, but he in-game character is F so we select this row.
id3: He is M in real life and one of his characters is F so we select this row
id4: He is M and none of his character is F so we don't need to print it on the screen.
Hopefully, you get what I'm trying to do here.
select id, name, gender,
(select count(*) from table_2 as t2 where t2.id = t1.id group by id) as #Character
from table_1 as t1
order by login;
Above query will print every id, name and gender and number of characters an id had, but this is not what I wanted. What should I change my code so that it works as what I Intended to do?
Actually it is simple.
We want to select the id, name, and gender(in the first table), and ... characters that an id had but that doesn't have the same gender as real-life.
select
*
from
t1
where
exists (select 1 from t2 where t1.id = t2.id and t1.gender <> t2.gender)
We want to select the id, name, and gender(in the first table), and the number of characters ...
select
*,
(select count(*) from t2 where t1.id = t2.id) as "#Character"
from
t1
where
exists (select 1 from t2 where t1.id = t2.id and t1.gender <> t2.gender)
Now that I understand the problem:
select t1.*, count(*)
from table_1 t1 join
table_2 t2
on t2.login = t1.login
group by t1.id
having count(*) filter (where t2.gender <> t1.gender) > 0;
In other words, you can filter in the having clause.
I think id and login could be confused -- your sample query does not match the sample data. But you should get the idea.
Try this:
select id,
(select count(*) from table_2 as t2 where t2.id = t1.id group by id having count(*) >1) as #Character
from table_1 as t1
order by login;
I am trying to make a sql query. I got some results from 2 tables below. Below results are good for me. Now I want those values which is present in each group. for example, A and B is present in each group(in each ID). so i want only A and B in result. and also i want make my query dynamic. Could anyone help?
| ID | Value |
|----|-------|
| 1 | A |
| 1 | B |
| 1 | C |
| 1 | D |
| 2 | A |
| 2 | B |
| 2 | C |
| 3 | A |
| 3 | B |
In the following query, I have placed your current query into a CTE for further use. We can try selecting those values for which every ID in your current result appears. This would imply that such values are associated with every ID.
WITH cte AS (
-- your current query
)
SELECT Value
FROM cte
GROUP BY Value
HAVING COUNT(DISTINCT ID) = (SELECT COUNT(DISTINCT ID) FROM cte);
Demo
The solution is simple - you can do this in two ways at least. Group by letters (Value), aggregate IDs with SUM or COUNT (distinct values in ID). Having that, choose those letters that have the value for SUM(ID) or COUNT(ID).
select Value from MyTable group by Value
having SUM(ID) = (SELECT SUM(DISTINCT ID) from MyTable)
select Value from MyTable group by Value
having COUNT(ID) = (SELECT COUNT(DISTINCT ID) from MyTable)
Use This
WITH CTE
AS
(
SELECT
Value,
Cnt = COUNT(DISTINCT ID)
FROM T1
GROUP BY Value
)
SELECT
Value
FROM CTE
WHERE Cnt = (SELECT COUNT(DISTINCT ID) FROM T1)
I have the following data structure
ID | REFID | NAME
1 | 100 | A
2 | 101 | B
3 | 101 | C
With
SELECT DISTINCT REFID, ID, NAME
FROM my_table
ORDER BY ID
I would like to have the following result:
1 | 100 | A
2 | 101 | B
Colum NAME and ID should contain the MIN or FIRST value.
But actually I get stuck at using MIN/FIRST here.
I welcome every tipps :-)
select id,
refid,
name
from (select id,
refid,
name,
row_number() over(partition by refid order by name) as rn
from my_table)
where rn = 1
order by id
You can use a subquery to do this.
WITH Q AS
( SELECT MIN(NAME) AS NAME, REFID FROM T GROUP BY REFID )
SELECT T.ID, T.REFID, T.NAME
FROM T
JOIN Q
ON (T.NAME = Q.NAME)
Also, note that SQL tables have no order. So there's no "First" value.
I'll try to explain the type of the query that I want:
Assume I have a table like this:
| ID | someID | Number |
|----|--------|--------|
| 1 | 1 | 10 |
| 2 | 1 | 11 |
| 3 | 1 | 14 |
| 4 | 2 | 10 |
| 5 | 2 | 13 |
Now, I want to find the someID that have a specific numbers (For example query for numbers 10, 11, 14 will return someID 1 and query for numbers 10, 13 will return 2). But, if someID contains all the query numbers but also more numbers, it will not return by the query. (For example query for 10, 11 will return nothing).
Is it possible?
SELECT t1.someId
FROM yourTable t1
WHERE t1.number IN (10,14,11)
GROUP BY t1.someID
HAVING COUNT(DISTINCT t1.ID) = (SELECT COUNT(DISTINCT t2.ID) FROM yourTable t2 WHERE t1.someID=t2.someID)
Example Fiddle
select someID
from yourtable
where number in (10,11,14)
and not exists (select * from yourtable t2 where number not in(10,11,14)
and t2.someid=yourtable.someid)
group by someID
having count(distinct ID) = 3
Where 3 is the number of items you are querying for
Yes, once you get the query numbers into a table variable (say it's called #QNums, with one column named QNum)) try
Select distinct someId
From table t
Where exists (Select * from #QNums
where QNum = t.Number)
And not Exists (Select * From table t2
Where someId = t.someId
And not exists(Select * From #QNums
where QNum = t3.Number))