There is an example request in which there are several joins.
SELECT DISTINCT ON(a.id_1) 1, a.name, b.task, c.created_at
FROM a
INNER JOIN b ON a.id_2 = b.id
INNER JOIN c ON a.ID_2 = c.id
WHERE a.deleted_at IS NULL
ORDER BY a.id_1 desc
In this case, the query will work, sorting by unique values of id_1 will take place. But I need to sort by the column a.name. In this case, postresql will swear with the words ERROR: SELECT DISTINCT ON expressions must match initial ORDER BY expressions.
The following query can serve as a solution to the problem:
SELECT *
FROM(
SELECT DISTINCT ON(a.id_1) a.name, b.task, c.created_at
FROM a
INNER JOIN b ON a.id_2 = b.id
INNER JOIN c ON a.ID_2 = c.id
WHERE a.deleted_at IS NULL
)
ORDER_BY a.name desc
But in reality the database is very large and such a query is not optimal. Are there other ways to sort by the selected column while keeping one uniqueness?
Related
Given the below query
SELECT * FROM A
INNER JOIN B ON (A.b_id = B.id)
WHERE (A.user_id = 'XXX' AND B.provider_id = 'XXX' AND A.type = 'PENDING')
ORDER BY A.created_at DESC LIMIT 1;
The variable values in the query are A.user_id and B.provider_id, the type is always queried on 'PENDING'.
I am planning to add a compound + partial index on A
A(user_id, created_at) where type = 'PENDING'
Also the number of records in A >> B.
Given A.user_id, B.provider_id, A.b_id all are foreign keys. Is there any way I can optimize the query?
Given that you are doing an inner join, I would first express the query as follows, with the join in the opposite direction:
SELECT *
FROM B
INNER JOIN A ON A.b_id = B.id
WHERE A.user_id = 'XXX' AND A.type = 'PENDING' AND
B.provider_id = 'XXX'
ORDER BY
A.created_at DESC
LIMIT 1;
Then I would add the following index to the A table:
CREATE INDEX idx_a ON A (user_id, type, created_at, b_id);
This four column index should cover the join from B to A, as well as the WHERE clause and also the ORDER BY sort at the end of the query. Note that we could probably also have left the query with the join order as you originally wrote above, and this index could still be used.
I have a many to many relation with three columns, (owner_id,property_id,ownership_perc) and for this table applies (many owners have many properties).
So I would like to find all the owner_id who has many properties (property_id) and connect them with other three tables (Table 1,3,4) in order to get further information for the requested result.
All the tables that I'm using are
Table 1: owner (id_owner,name)
Table 2: owner_property (owner_id,property_id,ownership_perc)
Table 3: property(id_property,building_id)
Table 4: building(id_building,address,region)
So, when I'm trying it like this, the query runs but it returns empty.
SELECT address,region,name
FROM owner_property
JOIN property ON owner_property.property_id = property.id_property
JOIN owner ON owner.id_owner = owner_property.owner_id
JOIN building ON property.building_id=building.id_building
GROUP BY owner_id,address,region,name
HAVING count(owner_id) > 1
ORDER BY owner_id;
Only when I'm trying the code below, it returns the owner_id who has many properties (see image below) but without joining it with the other three tables:
SELECT a.*
FROM owner_property a
JOIN (SELECT owner_id, COUNT(owner_id)
FROM owner_property
GROUP BY owner_id
HAVING COUNT(owner_id)>1) b
ON a.owner_id = b.owner_id
ORDER BY a.owner_id,property_id ASC;
So, is there any suggestion on what I'm doing wrong when I'm joining the tables? Thank you!
This query:
SELECT owner_id
FROM owner_property
GROUP BY owner_id
HAVING COUNT(property_id) > 1
returns all the owner_ids with more than 1 property_ids.
If there is a case of duplicates in the combination of owner_id and property_id then instead of COUNT(property_id) use COUNT(DISTINCT property_id) in the HAVING clause.
So join it to the other tables:
SELECT b.address, b.region, o.name
FROM (
SELECT owner_id
FROM owner_property
GROUP BY owner_id
HAVING COUNT(property_id) > 1
) t
INNER JOIN owner_property op ON op.owner_id = t.owner_id
INNER JOIN property p ON op.property_id = p.id_property
INNER JOIN owner o ON o.id_owner = op.owner_id
INNER JOIN building b ON p.building_id = b.id_building
ORDER BY op.owner_id, op.property_id ASC;
Always qualify the column names with the table name/alias.
You can try to use a correlated subquery that counts the ownerships with EXISTS in the WHERE clause.
SELECT b1.address,
b1.region,
o1.name
FROM owner_property op1
INNER JOIN owner o1
ON o1.id_owner = op1.owner_id
INNER JOIN property p1
ON p1.id_property = op1.property_id
INNER JOIN building b1
ON b1.id_building = p1.building_id
WHERE EXISTS (SELECT ''
FROM owner_property op2
WHERE op2.owner_id = op1.owner_id
HAVING count(*) > 1);
I have this SQL:
select * from `posts`
where `posts`.`deleted_at` is null
and `expire_at` >= '2017-03-26 21:23:42.000000'
and (
select count(distinct tags.id) from `tags`
inner join `post_tag` on `tags`.`id` = `post_tag`.`tag_id`
where `post_tag`.`post_id` = `posts`.`id`
and (`tags`.`tag` like 'PHP' or `tags`.`tag` like 'pop' or `tags`.`tag` like 'UI')
) >= 1
Is it possible order the results by number of tags in posts?
Maybe add there alias?
Any information can help me.
Convert your correlated subquery into a join:
select p.*
from posts p
join (
select pt.post_id,
count(distinct t.id) as tag_count
from tags t
inner join post_tag pt on t.id = pt.tag_id
where t.tag in ('PHP', 'pop', 'UI')
group by pt.post_id
) pt on p.id = pt.post_id
where p.deleted_at is null
and p.expire_at >= '2017-03-26 21:23:42.000000'
order by pt.tag_count desc;
Also, note that I changed the bunch of like and or to single IN because you are not matching any pattern i.e. there is no % in the string. So, better using single IN instead.
Also, if you have defined your table names, column names etc keeping keywords etc in mind, you shouldn't have the need to use the backticks. They make reading a query difficult.
I have a difficulty dealing with a SQL query. I use PostgreSQL.
The query says: Show the customers that have done at least an order that contains products from 3 different categories. The result will be 2 columns, CustomerID, and the amount of orders. I have written this code but I don't think it's correct.
select SalesOrderHeader.CustomerID,
count(SalesOrderHeader.SalesOrderID) AS amount_of_orders
from SalesOrderHeader
inner join SalesOrderDetail on
(SalesOrderHeader.SalesOrderID=SalesOrderDetail.SalesOrderID)
inner join Product on
(SalesOrderDetail.ProductID=Product.ProductID)
where SalesOrderDetail.SalesOrderDetailID in
(select DISTINCT count(ProductCategoryID)
from Product
group by ProductCategoryID
having count(DISTINCT ProductCategoryID)>=3)
group by SalesOrderHeader.CustomerID;
Here are the database tables needed for the query:
where SalesOrderDetail.SalesOrderDetailID in
(select DISTINCT count(ProductCategoryID)
Is never going to give you a result as an ID (SalesOrderDetailID) will never logically match a COUNT (count(ProductCategoryID)).
This should get you the output I think you want.
SELECT soh.CustomerID, COUNT(soh.SalesOrderID) AS amount_of_orders
FROM SalesOrderHeader soh
INNER JOIN SalesOrderDetail sod ON soh.SalesOrderID = sod.SalesOrderID
INNER JOIN Product p ON sod.ProductID = p.ProductID
HAVING COUNT(DISTINCT p.ProductCategoryID) >= 3
GROUP BY soh.CustomerID
Try this :
select CustomerID,count(*) as amount_of_order from
SalesOrder join
(
select SalesOrderID,count(distinct ProductCategoryID) CategoryCount
from SalesOrderDetail JOIN Product using (ProductId)
group by 1
) CatCount using (SalesOrderId)
group by 1
having bool_or(CategoryCount>=3) -- At least on CategoryCount>=3
I have a simple query:
SELECT * FROM Products p
LEFT JOIN SomeTable st ON st.SomeId = p.SomeId AND st.SomeOtherId = p.SomeOtherId
So far so good.
But the first join to SomeId can be NULL, In that case the check should be IS NULL, and that's where the join fails. I tried to use a CASE, but can't get that to work also.
Am I missing something simple here?
From Undocumented Query Plans: Equality Comparisons.
SELECT *
FROM Products p
LEFT JOIN SomeTable st
ON st.SomeOtherId = p.SomeOtherId
AND EXISTS (SELECT st.SomeId INTERSECT SELECT p.SomeId)