I need a query which returns the number of sportsman in each team.
It should only display teams with a name longer than the average length of names and surnames of sportsman.
I have two tables teams and sportmans.
Here's what I tried:
SELECT TIMOVI,SPORTISTI COUNT(IME_PREZIME)
FROM TIMOVI
JOIN SPORTISTI
ON SIFRA=TIM,
GROUP BY TIMOVI
WHERE LENGTH(naziv)> ((AVG(LENGHT(ime_preizime)))/2)
How about a query like this:
SELECT t.name, COUNT(s.id)
FROM team t
INNER JOIN sportsman s ON s.team = t.id
WHERE LENGTH(t.name) > (
SELECT AVG(LENGTH(s2.name)) FROM sportsman s2
)
GROUP BY t.id
For schema like this:
CREATE TABLE sportsman (
id INTEGER,
team INTEGER,
name VARCHAR(50)
);
CREATE TABLE team (
id INTEGER,
name VARCHAR(50)
);
SQL Fiddle
Related
I have two tables:
CREATE TABLE stores (
stores_id varchar PRIMARY KEY,
owner_id varchar
);
CREATE TABLE sets (
sets_id varchar PRIMARY KEY,
stores_id varchar not null,
owner_id varchar not null,
item_id varchar not null,
);
How do I make a request that shows the number of items on the sets in stores?
With selection by owner.
For example:
select
stores.*,
count(sets.item_id)
from stores
LEFT OUTER JOIN sets on stores.owner_id = sets.owner_id
where
stores.owner_id = 'e185775fc4f5'
GROUP BY stores.owner_id;
Thank you.
I think you'd need to join on both the store and the owner, then COUNT(DISTINCT item_id)
select
st.owner_id,
st.stores_id,
count(distinct se.item_id)
from stores st left join
sets se
on st.owner_id = se.owner_id
and st.stores_id = se.stores_id
group by st.owner_id, st.stores_id;
That will give a table that shows the owner, the store, then the number of items
Is this what you want?
select st.stores_id, count(se.item_id)
from stores st left join
sets se
on st.owner_id = se.owner_id
where st.owner_id = 'e185775fc4f5'
group by st.stores_id;
I have three tables: groups and people and groups_people which forms a many-to-many relationship between groups and people.
Schema:
CREATE TABLE groups (
id SERIAL PRIMARY KEY,
name TEXT
);
CREATE TABLE people (
id SERIAL PRIMARY KEY,
name TEXT,
join_date TIMESTAMP
);
CREATE TABLE groups_people (
group_id INT REFERENCES groups(id),
person_id INT REFERENCES people(id)
);
When I want to query for the latest 10 people who recenlty joined the group which has id = 1:
WITH person_ids AS (SELECT person_id FROM groups_people WHERE group_id = 1)
SELECT * FROM people WHERE id = ANY(SELECT person_id FROM person_ids)
ORDER BY join_date DESC LIMIT 10;
The query needs to scan all of the joined people then ordering them before selecting. That would be slow if the group containing too many people.
Is there anyway to work around it?
Schema (re-)design to allow same person joining multiple group
Since you mentioned that the relationship between groups and people
is many-to-many, I think you may want to move join_date to groups_people
(from people) because the same person can join different groups and each
such event has its own join_date
So I would change the schema to
CREATE TABLE people (
id SERIAL PRIMARY KEY,
name TEXT --, -- change
-- join_date TIMESTAMP -- delete
);
CREATE TABLE groups_people (
group_id INT REFERENCES groups(id),
person_id INT REFERENCES people(id), -- change
join_date TIMESTAMP -- add
);
Query
select
p.id
, p.name
, gp.join_date
from
people as p
, groups_people as gp
where
p.id = gp.person_id
and gp.group_id=1
order by gp.join_date desc
limit 10
Disclaimer: The above query is in MySQL syntax (the question was originally tagged with MySQL)
This seems much easier to write as a simple join with order by and limit:
select p.*
from people p join
groups_people gp
on p.id = gp.person_id
where gp.group_id = 1
order by gp.join_date desc
limit 10; -- or fetch first 10 rows only
Try rewriting using EXISTS
SELECT *
FROM people p
WHERE EXISTS (SELECT 1
FROM groups_people ps
WHERE p.id = ps.person_id and group_id = 1)
ORDER BY join_date DESC
LIMIT 10;
This is probably pretty simple and dumb to ask but Im just not getting there right now. I have a db for cocktails and want to check which cocktails I can make with the available ingredients:
Get the names of all cocktails where every ingredient is in stock
These are my tables:
create table cocktails
(
name TEXT PRIMARY KEY
)
create table ingredients
(
name TEXT PRIMARY KEY
)
create table cocktail_ingredients
(
cocktail_name TEXT ,
ingredient_name TEXT ,
amount INTEGER ,
FOREIGN KEY ( cocktail_name ) REFERENCES cocktails( name ) ,
FOREIGN KEY ( ingredient_name ) REFERENCES ingredients( name )
)
create table ingredients_in_stock
(
ingredient_name TEXT ,
FOREIGN KEY ( ingredient_name ) REFERENCES ingredients ( name )
)
And this is my code so far:
SELECT ci.cocktail_name
FROM cocktail_ingredients ci
WHERE ci.ingredient_name IN ( SELECT iis.ingredient_name
FROM ingredients_in_stock iis
)
GROUP BY ci.cocktail_name
HAVING COUNT(*) = ( SELECT COUNT(*)
FROM ingredients_in_stock
)
;
You can use a LEFT JOIN and a IN clause for this. Something like this:
SELECT name FROM cocktails WHERE Name NOT IN(
SELECT DISTINCT ci.cocktail_name FROM cocktail_ingredients ci LEFT JOIN ingredients_in_stock istk
ON ci.ingredient_name=istk.ingredient_name WHERE istk.ingredient_name IS NULL)
This query inverts the logic: List the cocktails where none of it's ingredients are missing on the ingredients_in_stock table. Hope the idea helps you
A correlated subquery should work:
select cocktail_name as all_ingredients_in_stock
from cocktail_ingredients ci
inner join ingredients_in_stock iis
on ci.ingredient_name = iis.ingredient_name
group by cocktail_name
having count(*) =
(select count(*)
from cocktail_ingredients
where cocktail_name = ci.cocktail_name
)
Sample SQL Fiddle
You could just say something like this:
select ci.name
from cocktail_ingredients ci
left join ingredients_in_stock iis on iis.ingredient_name = ci.ingredient_name
group by ci.name
having count(ci.ingredient_name) = sum( case
when iis.ingredient_name is not null
then 1
else 0
end
)
In the having clause,
The count(ci.ingredient_name) gives you the total number of ingredients required for the cocktail
The sum() expression gives you the count of in-stock ingredients used by the cocktail.
I'm trying to determine a correct way to isolate rows within a table that have the same values in 2 columns.
There are two tables, one (Name) with the person's names and IDs, and the other one (Nation) with people's IDs and their nations. I join the two tables with inner join, and now the new table columns consist of an ID, first name, last name, and nation. If I want to find pairs of people who have the same last name and are from the same nation, why isn't
select ID, FName, LName, Nation
from (Name inner join Nation on Name.ID = Nation.ID)
group by Name, Nation
having count(Name) > 1 and count(Nation) > 1
working?
I'm aiming for the result to be a table with columns:
ID -------First--------------- Last ---------Nation
where the last names and nations will be identical pairs while first names will be different.
I feel like the group by part isnt appropriate, but is there even an alternate way? Thanks for any help.
If you are using MS SQL Server:
select
*
from
(
select
Name.*,
Nation.Nation,
cnt = count(*) over(partition by LName, Nation)
from Name
join Nation on Nation.ID = Name.ID
) t
where cnt > 1
Try this:
SELECT * FROM (
SELECT Name.ID, Name.FName, Name.LName, Nation.Nation
FROM Name
INNER JOIN Nation ON (Name.ID = Nation.ID)
) a
INNER JOIN (
SELECT Name.ID, Name.FName, Name.LName, Nation.Nation
FROM Name
INNER JOIN Nation ON (Name.ID = Nation.ID)
) b ON (a.LName = b.LName AND a.Nation = b.Nation)
WHERE a.ID < b.ID
As Simon Righarts hinted, something's not right with the design.
Scenario 1)
If a name can have multiple nations, you would have 3 tables implementing an n:m relationship.
CREATE TABLE name (name_id int, name text, ...);
CREATE TABLE nation (nation_id int, nation text, ...);
CREATE TABLE nationality (name_id int references name(name_id)
,nation_id int references nation(nation_id)
... );
Query for the scenario:
SELECT a.name_id, a.fname, a.lname, n.nation
FROM name a
JOIN nationality na USING (name_id)
JOIN nation n USING (nation_id)
JOIN (
SELECT a.lname, na.nation_id
FROM name a
JOIN nationality na USING (name_id)
GROUP BY 1,2
HAVING count(*) > 1) x USING (lname, nation_id)
Scenario 2)
If a name can only have one nation, there would be a column nation_id in the table name:
CREATE TABLE name (name_id int
,name text
,nation_id int references nation(nation_id), ...);
CREATE TABLE nation (nation_id int, nation text, ...);
Query for this scenario:
SELECT a.name_id, a.fname, a.lname, n.nation
FROM name a
JOIN nation n USING (nation_id)
JOIN (
SELECT a.lname, a.nation_id
FROM name a
GROUP BY 1,2
HAVING count(*) > 1) x USING (lname, nation_id);
All multiple occurrences are included here, not just "pairs" - assuming you meant that.
Your actual description doesn't fit either scenario.
Here are two table in parent/child relationship.
What i need to do is to select students with there average mark:
CREATE TABLE dbo.Students(
Id int NOT NULL,
Name varchar(15) NOT NULL,
CONSTRAINT PK_Students PRIMARY KEY CLUSTERED
(
CREATE TABLE [dbo].[Results](
Id int NOT NULL,
Subject varchar(15) NOT NULL,
Mark int NOT NULL
)
ALTER TABLE [dbo].[Results] WITH CHECK ADD CONSTRAINT [FK_Results_Students] FOREIGN KEY([Id])
REFERENCES [dbo].[Students] ([Id])
I wrote a query like this :
SELECT name , coalesce(avg(r.[mark]),0) as Avmark
FROM students s
LEFT JOIN results r ON s.[id]=r.[id]
GROUP BY s.[name]
ORDER BY ISNULL(AVG(r.[mark]),0) DESC;
But the result is that all of students with there avg mark in desc order.What i need is to restrict result set with students that have the highest average mark agaist other,i.e.if the are two students with avg mark 50 and 1 with 25 i need to display only those students with 50.If there are only one student with highest avg mark- only he must appear in result set.How can i do this in best way?
SQL Server 2005+, using CTEs:
WITH grade_average AS (
SELECT r.id,
AVG(r.mark) 'avg_mark'
FROM RESULTS r
GROUP BY r.id),
highest_average AS (
SELECT MAX(ga.avg_mark) 'highest_avg_mark'
FROM grade_average ga)
SELECT DISTINCT
s.name,
ga.avg_mark
FROM STUDENTS s
JOIN grade_average ga ON ha.id = s.id
JOIN highest_average ha ON ha.highest_avg_mark = ga.avg_mark
Non-CTE equivalent:
SELECT DISTINCT
s.name,
ga.avg_mark
FROM STUDENTS s
JOIN (SELECT r.id,
AVG(r.mark) 'avg_mark'
FROM RESULTS r
GROUP BY r.id) ga ON ha.id = s.id
JOIN SELECT MAX(ga.avg_mark) 'highest_avg_mark'
FROM (SELECT r.id,
AVG(r.mark) 'avg_mark'
FROM RESULTS r
GROUP BY r.id) ga) ha ON ha.highest_avg_mark = ga.avg_mark
If you're using a relatively new version of MS SQL server, you can use WITH to make this simple to write:
WITH T AS (
SELECT
name,
coalesce(avg(r.[mark]),0) as mark
FROM students s
LEFT JOIN results r ON s.[id]=r.[id]
GROUP BY s.[name])
SELECT name as 'ФИО', mark as 'Средний бал'
FROM T
WHERE T.mark = (SELECT MAX(mark) from T)
Is it as simple as this? For all versions of SQL Server 2000+
SELECT TOP 1 WITH TIES
name, ISNULL(avg(r.[mark]),0) as AvMark
FROM
students s
LEFT JOIN
results r ON s.[id]=r.[id]
GROUP BY
s.[name]
ORDER BY
ISNULL(avg(r.[mark]),0) DESC;
SELECT name as 'ФИО',
coalesce(avg(r.[mark]),0) as 'Средний бал'
FROM students s
LEFT JOIN results r
ON s.[id]=r.[id]
GROUP BY s.[name]
HAVING AVG(r.[mark]) >= 50
ORDER BY ISNULL(AVG(r.[mark]),0) DESC
about HAVING clause