Get First Row of SubQuery - sql

I have a table with clients and I need to show the first country of each client like a default country.
Table Clients
ID_Client | Name_Client
1 | Mike
2 | Jon
3 | Ben
Table Countries
ID_Country | ID_Client | Name_Country
1 | 1 | France
2 | 1 | USA
3 | 1 | England
4 | 2 | Portugal
5 | 2 | Spain
6 | 3 | Germany
I only want the first row of each country.
Example:
Mike - France
Jon - Portugal
Ben - Germany
I tried this but only retry 1 row:
SELECT Name_Client, Name_Country
FROM Clients
WHERE ID_Client =
(SELECT TOP 1 ID_Client
From Countries
order by ID_Country)
Result:
Mike | France

Your sub-query isn't correlated, try to use APPLY :
SELECT c.Name_Client, cn.Name_Country
FROM Clients C OUTER APPLY
(SELECT TOP (1) cn.Name_Country
From Countries CN
WHERE cn.ID_Client = c.ID_Client
ORDER BY cn.ID_Country
) cn;

You can use Outer APPLY or CROSS APPLY as per Below example,
SELECT C.Name_Client + ' - ' + V.Name_Country
FROM Clients C
OUTER APPLY(
SELECT TOP 1 * FROM Countries CO WHERE CO.ID_Client = C.ID_Client
ORDER BY ID_Country
) V

SQL Fiddle
MS SQL Server 2017 Schema Setup:
CREATE TABLE CLIENTS (ID_Client int,Name_Client varchar(255))
CREATE TABLE COUNTRIES(ID_Country int,ID_Client int,Name_Country varchar(255))
INSERT INTO CLIENTS(ID_Client,Name_Client)VALUES(1,'Mike'),(2,'Jon'),(3,'Ben')
INSERT INTO COUNTRIES(ID_Country,ID_Client,Name_Country)VALUES(1,1,'France'),
(2,1,'USA'),(3,1,'England'),
(4,2,'Portugal'),(5,2,'Spain'),
(6,3,'Germany')
Query 1:
with CTE as
( select c.Name_Client, ct.Name_Country,
row_number() over(partition by c.ID_Client order by ct.ID_Country) as rn
FROM CLIENTS c
left join COUNTRIES ct on c.ID_Client=ct.ID_Client
)
select CTE.Name_Client,CTE.Name_Country from CTE where rn=1
Results:
| Name_Client | Name_Country |
|-------------|--------------|
| Mike | France |
| Jon | Portugal |
| Ben | Germany |

Rather than having a nested query like this try this instead.
SELECT clients.Name_Client, countries.Name_Country
FROM Clients clients
LEFT JOIN Countries countries on countries.ID_Client = clients.ID_Client
By performing the join you are matching the record and not having a subquery go off and get a bucket load of bad data.

use row_number()
with cte as
( select c.Name_Client, cn.Name_Country,
row_number() over(partition by c.ID_Client order by ID_Country) rn clients c join country cn on c.client_id=cn.client_id
) select * form cte where rn=1

You can use common table expression and filter on the row number partitioned by the id_client. Like this the row_number is incremented until a new id_client is found.
WITH cte_countries AS(
SELECT ROW_NUMBER() OVER (PARTITION BY ID_Client ORDER BY ID_Client) AS Row#,
Countries.*
FROM Countries)
SELECT * FROM Clients cli
JOIN cte_countries coun ON cli.ID_Client = coun.ID_Client
WHERE coun.Row# = 1

You can try this
SELECT * FROM (
SELECT Name_Client, Name_Country ,
RwNumbr = ROW_NUMBER() OVER(PARTITION by c.ID_Client order by cl.ID_Country)
from CLIENTS c
JOIN COUNTRIES cl
ON cl.ID_Client = c.ID_Client
) as f
WHERE f.RwNumbr = 1

Related

Select first rows where condition [duplicate]

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

SQL Query : how to Select Maximum value of each group of joined tables

I have this problem of returning maximum AGE of players in these 2 tables I have, Table tblplayers (with 34 records) when this table is joined to another table called tblClubs (with 9 records).
tblPlayers fields are:
ID(Autonumber) | CLubID(Number) | Player Name(Text) | PlayerAge(Number)
tblClubs fields are:
ID(Autonumber) | ClubName (Text)
Now I need to show Names of players with maximum ages among other players in their own clubs and the club name beside that like this :
Club Name | Player Name | Maximum Age (older player of each club)
please tell me how can i make it?
You can solve this with window functions, if your database supports them:
select c.club_name, p.player_name, p.player_age
from clubs c
inner join (
select
p.*,
rank() over(partition by p.club_id order by p.player_age desc) rn
players p
) p on p.club_id = c.id and p.rn = 1
A common and quite portable alternative is to filter with a subquery:
select
c.club_name,
t.player_name,
t.player_age
from players p
inner join clubs c on c.id = p.club_id
where p.age = (select max(p1.age) from players p1 where p1.club_id = p.club_id)
SQL Fiddle
MS SQL Server 2017 Schema Setup:
CREATE TABLE tblPlayers (ID INT, ClubID INT,PlayerName VARCHAR(255)
,PlayerAge INT)
CREATE TABLE tblClubs (ID int,ClubName VARCHAR(255))
INSERT INTO tblPlayers(ID,ClubID,PlayerName
,PlayerAge) VALUES (1,1,'John',30)
,(2,1,'Mark',25)
,(3,1,'Albert',36)
,(4,2,'David',33)
,(5,2,'John',31)
INSERT INTO tblClubs(ID, ClubName) VALUES(1,'TEAM 1')
,(2,'TEAM 2')
Query 1:
SELECT
*
FROM
(SELECT A.*,RANK() OVER
(PARTITION BY A.ClubID ORDER BY A.PlayerAge desc) as rn
FROM tblPlayers A
INNER JOIN tblClubs B ON A.ClubID=B.ID) t
WHERE
t.rn = 1
Results:
| ID | ClubID | PlayerName | PlayerAge | rn |
|----|--------|------------|-----------|----|
| 3 | 1 | Albert | 36 | 1 |
| 4 | 2 | David | 33 | 1 |

Oracle SQL Get unique symbols from table

I have table with descriptions of smth. For example:
My_Table
id description
================
1 ABC
2 ABB
3 OPAC
4 APEЧ
I need to get all unique symbols from all "description" columns.
Result should look like that:
symbol
================
A
B
C
O
P
E
Ч
And it shoud work for all languages, so, as I see, regular expressions cant help.
Please help me. Thanks.
with cte (c,description_suffix) as
(
select substr(description,1,1)
,substr(description,2)
from mytable
where description is not null
union all
select substr(description_suffix,1,1)
,substr(description_suffix,2)
from cte
where description_suffix is not null
)
select c
,count(*) as cnt
from cte
group by c
order by c
or
with cte(n) as
(
select level
from dual
connect by level <= (select max(length(description)) from mytable)
)
select substr(t.description,c.n,1) as c
,count(*) as cnt
from mytable t
join cte c
on c.n <= length(description)
group by substr(t.description,c.n,1)
order by c
+---+-----+
| C | CNT |
+---+-----+
| A | 4 |
| B | 3 |
| C | 2 |
| E | 1 |
| O | 1 |
| P | 2 |
| Ч | 1 |
+---+-----+
Create a numbers table and populate it with all the relevant ids you'd need (in this case 1..maxlength of string)
SELECT DISTINCT
locate(your_table.description, numbers.id) AS symbol
FROM
your_table
INNER JOIN
numbers
ON numbers.id >= 1
AND numbers.id <= CHAR_LENGTH(your_table.description)
SELECT DISTINCT(SUBSTR(ll,LEVEL,1)) OP --Here DISTINCT(SUBSTR(ll,LEVEL,1)) is used to get all distinct character/numeric in vertical as per requirment
FROM
(
SELECT LISTAGG(DES,'')
WITHIN GROUP (ORDER BY ID) ll
FROM My_Table --Here listagg is used to convert all values under description(des) column into a single value and there is no space in between
)
CONNECT BY LEVEL <= LENGTH(ll);

Count how many times a value appears in tables SQL

Here's the situation:
So, in my database, a person is "responsible" for job X and "linked" to job Y. What I want is a query that returns: name of person, his ID and he number of jobs it's linked/responsible. So far I got this:
select id_job, count(id_job) number_jobs
from
(
select responsible.id
from responsible
union all
select linked.id
from linked
GROUP BY id
) id_job
GROUP BY id_job
And it returns a table with id in the first column and number of occurrences in the second. Now, what I can't do is associate the name of person to the table. When i put that in the "select" from beginning it gives me all the possible combinations... How can I solve this? Thanks in advance!
Example data and desirable output:
| Person |
id | name
1 | John
2 | Francis
3 | Chuck
4 | Anthony
| Responsible |
process_no | id
100 | 2
200 | 2
300 | 1
400 | 4
| Linked |
process_no | id
101 | 4
201 | 1
301 | 1
401 | 2
OUTPUT:
| OUTPUT |
id | name | number_jobs
1 | John | 3
2 | Francis | 3
3 | Chuck | 0
4 | Anthony | 2
Try this way
select prs.id, prs.name, count(*) from Person prs
join(select process_no, id
from Responsible res
Union all
select process_no, id
from Linked lin ) a on a.id=prs.id
group by prs.id, prs.name
I would recommend aggregating each of the tables by the person and then joining the results back to the person table:
select p.*, coalesce(r.cnt, 0) + coalesce(l.cnt, 0) as numjobs
from person p left join
(select id, count(*) as cnt
from responsible
group by id
) r
on r.id = p.id left join
(select id, count(*) as cnt
from linked
group by id
) l
on l.id = p.id;
select id, name, count(process_no) FROM (
select pr.id, pr.name, res.process_no from Person pr
LEFT JOIN Responsible res on pr.id = res.id
UNION
select pr.id, pr.name, lin.process_no from Person pr
LEFT JOIN Linked lin on pr.id = lin.id) src
group by id, name
order by id
Query ain't tested, give it a shot, but this is the way you want to go

Group by minimum value in one field while selecting distinct rows

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