I have this query
the table flights also contains price column. I'd like to sum it all up and display. How can I solve this?
Can I do this by taking the values from SELECT * from get_cities; somehow or it should be done in the query?
Table img
I am trying to solve this
Write a query finding all the names of the cities City name can be reached by plane with 3 stops. Display all the cities where the stop took place and the total cost of the trip. Also sum up the journey cost.
WITH RECURSIVE get_cities AS (
SELECT 0 as count, city, cid from cities where CITY = 'Agat'
UNION ALL
SELECT c.count + 1, b.city, b.cid from get_cities c
JOIN flights t on t.departure = c.cid
JOIN cities b on t.arrival = b.cid
WHERE COUNT < 3
)
SELECT cid, sum(price) from get_cities
JOIN flights f on f.fid = cid
GROUP BY cid
;
You can sum the prices directly in the recursive cte :
WITH RECURSIVE get_cities AS (
SELECT 0 as count, array[city] as city_path, array[cid] as cid_path, 0 as total_price
FROM cities
WHERE CITY = 'Agat'
UNION ALL
SELECT c.count + 1, c.city_path || b.city, c.cid_path || b.cid, c.total_price + t.price
FROM get_cities c
JOIN flights t on t.departure = c.cid
JOIN cities b on t.arrival = b.cid
WHERE COUNT < 3
)
SELECT *
FROM get_cities
WHERE count = 2 -- select only the journey with 3 stops ;
Related
I have table with self-related foreign keys and can not get how I can receive firs child or descendant which meet condition. My_table structure is:
id
parent_id
type
1
null
union
2
1
group
3
2
group
4
3
depart
5
1
depart
6
5
unit
7
1
unit
I should for id 1 (union) receive all direct child or first descendant, excluding all groups between first descendant and union. So in this example as result I should receive:
id
type
4
depart
5
depart
7
unit
id 4 because it's connected to union through group with id 3 and group with id 2 and id 5 because it's connected directly to union.
I've tried to write recursive query with condition for recursive part: when parent_id = 1 or parent_type = 'depart' but it doesn't lead to expected result
with recursive cte AS (
select b.id, p.type_id
from my_table b
join my_table p on p.id = b.parent_id
where b.id = 1
union
select c.id, cte.type_id
from my_table c
join cte on cte.id = c.parent_id
where c.parent_id = 1 or cte.type_id = 'group'
)
Here's my interpretation:
if type='group', then id and parent_id are considered in the same group
id#1 and id#2 are in the same group, they're equals
id#2 and id#3 are in the same group, they're equals
id#1, id#2 and id#3 are in the same group
If the above is correct, you want to get all the first descendent of id#1's group. The way to do that:
Get all the ids in the same group with id#1
Get all the first descendants of the above group (type not in ('union', 'group'))
with recursive cte_group as (
select 1 as id
union all
select m.id
from my_table m
join cte_group g
on m.parent_id = g.id
and m.type = 'group')
select mt.id,
mt.type
from my_table mt
join cte_group cg
on mt.parent_id = cg.id
and mt.type not in ('union','group');
Result:
id|type |
--+------+
4|depart|
5|depart|
7|unit |
Sounds like you want to start with the row of id 1, then get its children, and continue recursively on rows of type group. To do that, use
WITH RECURSIVE tree AS (
SELECT b.id, b.type, TRUE AS skip
FROM my_table b
WHERE id = 1
UNION ALL
SELECT c.id, c.type, (c.type = 'group') AS skip
FROM my_table c
JOIN tree p ON c.parent_id = p.id AND p.skip
)
SELECT id, type
FROM tree
WHERE NOT skip
I have the following schema:
expenses
id
name, varchar
cost, double
date, DATE
category_id, int f_key
user_id, int f_key
1
Pizza
22.9
22/08/2022
1
1
2
Pool
34.9
23/08/2022
2
1
categories
id
name, varchar
1
Food
2
Leisure
3
Medicine
4
Fancy food
users_categories(user_id int foreign key, category_id foreign key)
user_id int f_key
category_id int f_key
1
1
1
2
1
3
2
4
And two users with id 1 and 2.
Relation between user and category is many to many.
Problem:
I want to get statistics (total cost amount and count) for all categories. For categories where there are no expenses I want to return 0. Here is my query:
SELECT categories.name as name, count(expenses.name) as count, round(SUM(price)::numeric,2) as sum
FROM expenses
Right JOIN categories ON expenses.category_id = categories.id
and expenses.category_id in (
select users_categories.category_id from users_categories where users_categories.user_id = 1
)
and expenses.id in(
Select expenses.id from expenses
join users_categories on expenses.category_id = users_categories.category_id
and expenses.user_id = 1
AND (extract(year from date) = 2022 OR CAST(2022 AS int) is null)
AND (extract(month from date) = 8 OR CAST(8 AS int) is null)
)
GROUP BY categories.id ORDER BY categories.id
The response is:
name
count
sum
Food
1
22.9
Leisure
1
33.9
Medicine
0
null
Fancy food
0
null
How I should edit my query to eliminate the last row, because this category doesn't belong to the user 1.
In your query you used user_categories as subquery so it will not filter category ids,
Try this Query
SELECT categories.name as name,count(expenses.name) as count, coalesce(round(SUM(price)::numeric,2),0) as sum from
categories
left join users_categories on users_categories.category_id= categories.id
left join expenses ON expenses.category_id = categories.id
AND (extract(year from date) = 2022 OR CAST(2022 AS int) is null)
AND (extract(month from date) = 8 OR CAST(8 AS int) is null)
where users_categories.user_id='1'
GROUP BY categories.name,categories.id ORDER BY categories.id
OUTPUT :
name count sum
Food 1 22.90
Leisure 1 34.90
Medicine 0 0
You want to move expenses.category_id in ... out of the ON condition and into a WHERE clause.
When it is in the ON clause, that means rows which were removed by the in-test just get NULL-fabricated anyway. You want to remove those rows after the NULL-fabrication is done, so that they remain removed. But why do you use that in-test anyway? Seems like it would be much simpler written as another join.
What I understand is that you are trying to get the count and sum of expenses for all the categories related to the user_id 1 within the month of august 2022.
Please try out the following query.
WITH statistics
AS (SELECT e.category_id,
Count(e.*) AS count,
Round(Sum(e.cost), 2) AS sum
FROM expenses e
WHERE e.user_id = 1
AND ( e.date BETWEEN '01/08/2022' AND '31/08/2022' )
GROUP BY e.category_id),
user_category
AS (SELECT uc.category_id,
COALESCE(s.count, 0) AS count,
COALESCE(s.sum, 0) AS sum
FROM users_categories uc
LEFT JOIN statistics s
ON uc.category_id = s.id
WHERE uc.user_id = 1)
SELECT c.NAME,
u.count,
u.sum
FROM categories c
INNER JOIN user_category u
ON u.category_id = c.id;
I have Car table. Car has is_sold and is_shipped. A Car belongs to a dealership, dealership_id (FK).
I want to run a query that tells me the count of sold cars and the count of shipped cars for a given dealership all in one result.
sold_count | shipped_count
10 | 4
The single queries I have look like this:
select count(*) as sold_count
from car
where dealership_id=25 and is_sold=true;
and
select count(*) as shipped_count
from car
where dealership_id=25 and is_shipped=true;
How do I combine the two to get both counts in one result?
This will do:
select dealership_id,
sum(case when is_sold is true then 1 else 0 end),
sum(case when is_shipped is true then 1 else 0 end)
from cars group by dealership_id;
You can use the filter clause of the Aggregate function. (see demo)
select dealership_id
, count(*) filter (where is_sold) cars_sold
, count(*) filter (where is_shipped) cars_shipped
from cars
where dealership_id = 25
group by dealership_id;
You can also using cross join.
select 'hello' as col1, 'world' as col2;
return:
col1 | col2
-------+-------
hello | world
(1 row)
similarly,
with a as
(
select count(*) as a1 from emp where empid> 5),
b as (
select count(*) as a2 from emp where salary > 6000)
select * from a, b;
or you can even apply to different table. like:
with a as
(select count(*) as a1 from emp where empid> 5),
b as
(select count(*) as a2 from ab )
select * from a, b;
with a as
(
select count(*) as sold_count
from car
where dealership_id=25 and is_sold=true
),
b as
(
select count(*) as shipped_count
from car
where dealership_id=25 and is_shipped=true
)
select a,b;
further reading: https://www.postgresql.org/docs/current/queries-table-expressions.html.
https://stackoverflow.com/a/26369295/15603477
I'm trying to perform recursive cte with postgres but I can't wrap my head around it. In terms of performance issue there are only 50 items in TABLE 1 so this shouldn't be an issue.
TABLE 1 (expense):
id | parent_id | name
------------------------------
1 | null | A
2 | null | B
3 | 1 | C
4 | 1 | D
TABLE 2 (expense_amount):
ref_id | amount
-------------------------------
3 | 500
4 | 200
Expected Result:
id, name, amount
-------------------------------
1 | A | 700
2 | B | 0
3 | C | 500
4 | D | 200
Query
WITH RECURSIVE cte AS (
SELECT
expenses.id,
name,
parent_id,
expense_amount.total
FROM expenses
WHERE expenses.parent_id IS NULL
LEFT JOIN expense_amount ON expense_amount.expense_id = expenses.id
UNION ALL
SELECT
expenses.id,
expenses.name,
expenses.parent_id,
expense_amount.total
FROM cte
JOIN expenses ON expenses.parent_id = cte.id
LEFT JOIN expense_amount ON expense_amount.expense_id = expenses.id
)
SELECT
id,
SUM(amount)
FROM cte
GROUP BY 1
ORDER BY 1
Results
id | sum
--------------------
1 | null
2 | null
3 | 500
4 | 200
You can do a conditional sum() for only the root row:
with recursive tree as (
select id, parent_id, name, id as root_id
from expense
where parent_id is null
union all
select c.id, c.parent_id, c.name, p.root_id
from expense c
join tree p on c.parent_id = p.id
)
select e.id,
e.name,
e.root_id,
case
when e.id = e.root_id then sum(ea.amount) over (partition by root_id)
else amount
end as amount
from tree e
left join expense_amount ea on e.id = ea.ref_id
order by id;
I prefer doing the recursive part first, then join the related tables to the result of the recursive query, but you could do the join to the expense_amount also inside the CTE.
Online example: http://rextester.com/TGQUX53703
However, the above only aggregates on the top-level parent, not for any intermediate non-leaf rows.
If you want to see intermediate aggregates as well, this gets a bit more complicated (and is probably not very scalable for large results, but you said your tables aren't that big)
with recursive tree as (
select id, parent_id, name, 1 as level, concat('/', id) as path, null::numeric as amount
from expense
where parent_id is null
union all
select c.id, c.parent_id, c.name, p.level + 1, concat(p.path, '/', c.id), ea.amount
from expense c
join tree p on c.parent_id = p.id
left join expense_amount ea on ea.ref_id = c.id
)
select e.id,
lpad(' ', (e.level - 1) * 2, ' ')||e.name as name,
e.amount as element_amount,
(select sum(amount)
from tree t
where t.path like e.path||'%') as sub_tree_amount,
e.path
from tree e
order by path;
Online example: http://rextester.com/MCE96740
The query builds up a path of all IDs belonging to a (sub)tree and then uses a scalar sub-select to get all child rows belonging to a node. That sub-select is what will make this quite slow as soon as the result of the recursive query can't be kept in memory.
I used the level column to create a "visual" display of the tree structure - this helps me debugging the statement and understanding the result better. If you need the real name of an element in your program you would obviously only use e.name instead of pre-pending it with blanks.
I could not get your query to work for some reason. Here's my attempt that works for the particular table you provided (parent-child, no grandchild) without recursion. SQL Fiddle
--- step 1: get parent-child data together
with parent_child as(
select t.*, amount
from
(select e.id, f.name as name,
coalesce(f.name, e.name) as pname
from expense e
left join expense f
on e.parent_id = f.id) t
left join expense_amount ea
on ea.ref_id = t.id
)
--- final step is to group by id, name
select id, pname, sum(amount)
from
(-- step 2: group by parent name and find corresponding amount
-- returns A, B
select e.id, t.pname, t.amount
from expense e
join (select pname, sum(amount) as amount
from parent_child
group by 1) t
on t.pname = e.name
-- step 3: to get C, D we union and get corresponding columns
-- results in all rows and corresponding value
union
select id, name, amount
from expense e
left join expense_amount ea
on e.id = ea.ref_id
) t
group by 1, 2
order by 1;
Can someone help me with this query?
SELECT p.OwnerName, SUM(ru.MonthlyRent) AS PotentinalRent, SUM(
(SELECT COUNT(t.ID) * ru.MonthlyRent FROM tblTenant t
WHERE t.UnitID = ru.ID)
) AS ExpectedRent
FROM tblRentalUnit ru
LEFT JOIN tblProperty p ON p.ID = ru.PropertyID
GROUP BY p.OwnerName
I'm having problems with the second sum, it won't let me do it. Evidently SUM won't work on subqueries, but I need to calculate the expected rent (MonthlyRent if there is a tenant assigned to the RentalUnit's id, 0 of they're not). How can I make this work?
SELECT p.OwnerName, SUM(ru.MonthlyRent) AS PotentialRent, SUM(cnt) AS ExpectedRent
FROM tblRentalUnit ru
LEFT JOIN
tblProperty p
ON p.ID = ru.PropertyID
OUTER APPLY
(
SELECT COUNT(t.id) * ru.MonthlyRent AS cnt
FROM tblTenant t
WHERE t.UnitID = ru.ID
) td
GROUP BY p.OwnerName
Here's a test script to check:
WITH tblRentalUnit AS
(
SELECT 1 AS id, 100 AS MonthlyRent, 1 AS PropertyID
UNION ALL
SELECT 2 AS id, 300 AS MonthlyRent, 2 AS PropertyID
),
tblProperty AS
(
SELECT 1 AS id, 'Owner 1' AS OwnerName
UNION ALL
SELECT 2 AS id, 'Owner 2' AS OwnerName
),
tblTenant AS
(
SELECT 1 AS id, 1 AS UnitID
UNION ALL
SELECT 2 AS id, 1 AS UnitID
)
SELECT p.OwnerName, SUM(ru.MonthlyRent) AS PotentialRent, SUM(cnt) AS ExpectedRent
FROM tblRentalUnit ru
LEFT JOIN
tblProperty p
ON p.ID = ru.PropertyID
OUTER APPLY
(
SELECT COUNT(t.id) * ru.MonthlyRent AS cnt
FROM tblTenant t
WHERE t.UnitID = ru.ID
) td
GROUP BY p.OwnerName
What is the meaning of the sum of the unitMonthlyRent times the number of tenants, for some partiicular rental unit (COUNT(t.ID) * ru.MonthlyRent )?
Is it the case that all you are trying to do is see the difference between the total potential rent from all untis versus the expected rent (From only occcupied units) ? If so, then try this
Select p.OwnerName,
Sum(r.MonthlyRent) AS PotentinalRent,
Sum(Case t.Id When Null Then 0
Else r.MonthlyRent End) ExpectedRent
From tblRentalUnit r
Left Join tblTenant t
On t.UnitID = r.ID
left Join tblProperty p
On p.ID = r.PropertyID)
Group By p.OwnerName