I have a WITH statement which returns a table like followings:
id
parent_id
1
null
3
2
4
null
5
4
The desired output of my query is where parent_id is null or parent_id is present in the id column:
id
parent_id
1
null
4
null
5
4
I wrote the following query:
SELECT * FROM items
WHERE parent_id IS NULL OR parent_id = ANY(SELECT id from items)
As far I have understood, lateral joins are faster than the ANY operator so my idea was to rewrite the above query using them. I started with:
SELECT * FROM items i1
JOIN LATERAL (SELECT * FROM items i2 WHERE i2.parent_id = i1.id ) t ON true
But where do I add the condition to take the items where parent_id is null?
Use a self join:
SELECT t1.*
FROM items t1
LEFT JOIN items t2 ON t1.parent_id = t2.id
WHERE t1.parent_id IS NULL
OR t2.id IS NOT NULL
This is the best performing approach (assuming you have an index on the id column, which is almost certainly the case).
I want to group the records by relation.
products table:
id
price
1
100
2
200
3
300
4
400
product_properties table:
id
productId
propertyId
1
1
2
2
1
3
3
2
2
4
2
3
5
3
4
6
4
4
The query should select lowest price group by product_properties. I mean, If products have same properties in product_properties, query should return product that has lowest price.
So, For these tables query should return products that have ids 1,3.
I use TypeORM, I tried join the relation and distinct on relation alias name but its not worked.
How can I achieve this?
I wrote two variants query for you:
-- variant 1
select distinct t1.product_id from (
select
pr.price, pp.product_id, pp.property_id, min(pr.price) OVER(PARTITION BY pp.property_id) as min_price
from
test.product_properties pp
inner join
test.products pr on pp.product_id = pr.id
) t1
where
t1.price = t1.min_price;
-- variant 2
select distinct t1.product_id from test.product_properties t1
inner join test.products t2 on t1.product_id = t2.id
inner join (
select
pp.property_id, min(pr.price) as min_price
from
test.product_properties pp
inner join
test.products pr on pp.product_id = pr.id
group by pp.property_id
) t3 on t3.property_id = t1.property_id and t3.min_price = t2.price;
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;
I'm working with a database that have a poor design that does not constraint duplicates rows as long as they have a different unique-identifier.
Within one of the table, a given user can have an attribute and a value for the attribute. Normally, a user would only a have a single time the attribute but because of the poor design, I'm getting a lot of duplicates in the table and now I need to clean that mess. This is due to the CRM software not always checking if the row exists when we modify the employee profile but instead it creates a bunch of new rows with duplicates values.
The following query returns the duplicates values:
SELECT ua.ID AS LineID
,ua.Modified AS LineLastModifiedDate
,u.FullName AS EmployeeName
,a.Name AS AttributeName
,ua.value AS AttributeValue
FROM UserAttributes AS ua
INNER JOIN Users AS u ON ua.userid = u.id
INNER JOIN Attributes AS a ON ua.AttributeID = a.ID
WHERE EXISTS (
SELECT NULL
FROM UserAttributes as ua2
WHERE ua2.UserID = ua.UserID
AND ua2.AttributeID = ua.AttributeID
AND ua2.ID != ua.ID
)
And produces results as this:
LineID LineLastModifiedDate EmployeeName AttributeName AttributeValue
------ ----------------------- ------------- --------------- ---------------
15 2016-01-01 Employee1 EmployeeNumber 15
19 2016-07-20 Employee1 EmployeeNumber 15
35 2016-01-01 Employee2 EmployeeSex M
96 2016-07-20 Employee2 EmployeeSex M
21 2016-03-03 Employee1 SickDays 3
99 2016-07-10 Employee1 SickDays 5
What I need to accomplish starting from this query is : ForEach grouping of the same EmployeeName and AttributeName, give me the last modified line expecting results like this :
LineID LineLastModifiedDate EmployeeName AttributeName AttributeValue
------ ----------------------- ------------- --------------- ---------------
19 2016-07-20 Employee1 EmployeeNumber 15
96 2016-07-20 Employee2 EmployeeSex M
99 2016-07-10 Employee1 SickDays 5
How can I modify my query to accomplish this ?
Thank you
-M
;WITH CTE
AS
(
SELECT ua.ID AS LineID
,ua.Modified AS LineLastModifiedDate
,u.FullName AS EmployeeName
,a.Name AS AttributeName
,ua.value AS AttributeValue
,ROW_NUMBER() OVER (PARTITION BY EMPLOYEENAME,EMPLOYEESEX ORDER BY UA.Modified DESC) AS RN
FROM UserAttributes AS ua
INNER JOIN Users AS u ON ua.userid = u.id
INNER JOIN Attributes AS a ON ua.AttributeID = a.ID
WHERE EXISTS (
SELECT NULL
FROM UserAttributes as ua2
WHERE ua2.UserID = ua.UserID
AND ua2.AttributeID = ua.AttributeID
AND ua2.ID != ua.ID
)
)
SELECT * FROM cte where rn=1
You can use row numbering or a scheme as below where you pull out the max value and then use a join. Presumably you can't have ties by date.
select ...
from
UserAttributes as ua
inner join
(
select
UserID, AttributeID,
max(LineLastModifiedDate) as MaxLineLastModifiedDate
fromUserAttributes
group by UserId
) as max_ua
on max_ua.UserID = ua.UserID
and max_ua.AttributeID = max_ua.AttributeID
and max_ua.MaxLineLastModifiedDate = ua.LineLastModifiedDate
...
I have 2 tables. Customers and Orders.
My requirement is...
I would like to get the result like the following
Customer Detail + HasOrders + Count(Orders)
I wrote
SELECT Customers.*
, CASE WHEN o.CustomerID IS NOT NULL THEN 1 ELSE 0 END HasOrders
FROM Customers c
LEFT JOIN Orders o
ON c.CustomerID = o.CustomersID
But it returns many rows. If the customer has 5 orders, it returns 5 rows for each Customer.
Could you please advise me? Thanks.
You need to do the counting in derived table.
SELECT c.*
, case when o.CustomerID is not null
then 1
else 0
end HasOrders
, o.NumberOfOrders
FROM Customers c
LEFT JOIN
(
SELECT CustomerID
, count(*) NumberOfOrders
FROM Orders
GROUP BY CustomerID
) o
ON c.CustomerID = o.CustomersID