多表的SQL查询

Question

I'm new in Pure Sql I want to write this as Query

select items.* 
from items 
LEFT OUTER JOIN 
    (select sum(purchase_details.quantity) as total 
    from purchase_details 
    where (purchase_details.item_id=items.id)  
GROUP BY purchase_details.item_id) ABC 

but this gives an error

 You have an error in your SQL syntax; check the manual that corresponds to 
    your MariaDB server version for the right syntax to use near 'LIMIT 0, 25' 
    at line 1

i don't know why it's not working

Answer 1

The syntax error here is that you need an on -clause for your left join . But the underlying conceptual problem is different: you cannot join with a dependent subquery .

You can fix your query like this:

select items.* 
from items 
LEFT OUTER JOIN (
  select item_id, sum(purchase_details.quantity) as total
  from purchase_details 
  GROUP BY purchase_details.item_id
) ABC on ABC.item_id = items.id;

This moved your inner where -condition (that would depend on items.id , which is not allowed, as it is out of scope) to the on -clause. Thus item_id is also added in the inner select (as it is needed outside).

A different way to write this would be

select items.*, 
   (select sum(purchase_details.quantity) 
    from purchase_details 
    where purchase_details.item_id=items.id) as total
from items;

Here you have a dependent subquery : the inner where -clause depends on the outer items.id . You do not need a group by anymore, as the where -condition already uses just the rows for that item. (And you can also only return at most one row in this context anyway.)

Both queries are equivalent, and can (if the optimizer finds that execution plan) internally actually be executed in exactly the same way (which is nothing you have to care about much though, as long as you provide appropriate indexes).

So in your case you can use both (and maybe check which one is faster); if you want to get additional information for that item, you should prefer the left join -version though, eg use

...
LEFT OUTER JOIN (
  select item_id, 
    sum(purchase_details.quantity) as total,
    count(purchase_details.item_id) as cnt,
    max(purchase_details.quantity) as max_quantity_per_order,
    max(purchase_details.date) as latest_order,
    ...
  from purchase_details 
  GROUP BY purchase_details.item_id
) ABC on ABC.item_id = items.id;