CODEWITHSUNDEEP
sqlpostgresql
Nicholas Riley
Nicholas Riley

Reputation: 44321

Outer join with conditions?

I'm trying to look up two pieces of metadata (volume and issue) on items which may have either a volume, issue, or both. The metadata is stored in a table with item ID, key (metadata field ID) and value.

This does work, but it seems overly complex and repetitive:

select volume.text_value as volume_value, issue.text_value as issue_value
    from metadatavalue item
    left outer join (select item_id, text_value from metadatavalue
                     where metadata_field_id = 90) volume
                    on item.item_id = volume.item_id
    left outer join (select item_id, text_value from metadatavalue
                     where metadata_field_id = 91) issue
                    on item.item_id = issue.item_id
    where item.metadata_field_id in (90, 91)

Is there a simpler way to write this query?

Thanks.

Upvotes: 2

Views: 1570

Answers (5)

Quassnoi
Quassnoi

Reputation: 425573

SELECT  DISTINCT ON (item_id)
        item_id,
        CASE metadata_field_id
        WHEN 90 THEN
                text_value
        ELSE    (
                SELECT  text_value
                FROM    metadatavalue m
                WHERE   m.metadata_field_id = 90
                        AND m.item_id = i.item_id
                )
        END AS volume,
        CASE metadata_field_id
        WHEN 91 THEN
                text_value
        ELSE    (
                SELECT  text_value
                FROM    metadatavalue m
                WHERE   m.metadata_field_id = 91
                        AND m.item_id = i.item_id
                )
        END AS issue
FROM    metadatavalue
WHERE   metadata_field_id IN (90, 91)
ORDER BY
        item_id

Having an index on (item_id, metadata_field) will improve this query.

This will work better if there are few items with metadata of 90 and 91, compared to overall number of items.

If almost all items have these metadata, just use:

SELECT  *
FROM    (
        SELECT  item_id,
                (
                SELECT  text_value
                FROM    metadatavalue m
                WHERE   m.metadata_field_id = 90
                        AND m.item_id = i.item_id
                ) volume,
                (
                SELECT  text_value
                FROM    metadatavalue m
                WHERE   m.metadata_field_id = 91
                        AND m.item_id = i.item_id
                ) issue
        FROM    items
        ) q
WHERE   issue IS NOT NULL OR volume IS NOT NULL

Upvotes: 2

Paul Morgan
Paul Morgan

Reputation: 32548

PostgreSQL supports a full outer join which can simplify the query:

  select v.text_value as volume_value, i.text_value as issue_value
      from ( select item_id, text_value
               from metadatavalue
              where metadata_field_id = 90) v
           full join
           ( select item_id, text_value
               from metadatavalue
              where metadata_field_id = 91) i
           using (item_id)

Upvotes: 2

araqnid
araqnid

Reputation: 133622

You can take the predicate out of your inner table into the join condition:

select volume.text_value as volume_value, issue.text_value as issue_value
from metadatavalue item
left outer join metadatavalue volume
                on volume.metadata_field_id = 90 and volume.item_id = item.item_id
left outer join metadatavalue issue
                on issue.metadata_field_id = 91 and issue.item_id = item.item_id
where item.metadata_field_id in (90, 91)

Although I'm not sure I'm making sense of your schema-- it seems to me that item should be referring to a different table than volume and issue? sth more like:

select item.*, volume.text_value as volume_value, issue.text_value as issue_value
from item
left outer join metadatavalue volume
                on volume.metadata_field_id = 90 and volume.item_id = item.item_id
left outer join metadatavalue issue
                on issue.metadata_field_id = 91 and issue.item_id = item.item_id

This also makes the rather odd where condition go away. You could put the above query into a view ("item_volume_issue") and it would appear that volume and issue are just additional columns in item_volume_issue instead of separate tables (which is what makes more sense to me).

Upvotes: 0

tom
tom

Reputation: 1639

Using the case statement to segregate the values into buckets, the max function will allow the text_value data to float to the top.

select
   item_id
  ,max(case metadata_field_id when 90 then text_value else null end) as volume_value
  ,max(case metadata_field_id when 91 then text_value else null end) as issue_value
from
  metadatavalue
group by
   item_id

Here's what I used for testing purposes:

select
   item_id
  ,max(case metadata_field_id when 90 then text_value else null end) as volume_value
  ,max(case metadata_field_id when 91 then text_value else null end) as issue_value
from
  (
   select 1 as item_id, 90 as metadata_field_id, '90-I am here' as text_value
   union
   select 1 as item_id, 91 as metadata_field_id, '91-Me too' as text_value
   union
   select 2 as item_id, 90 as metadata_field_id, null as text_value
   union
   select 2 as item_id, 91 as metadata_field_id, '91-funky' as text_value
   union
   select 3 as item_id, 90 as metadata_field_id, '90-fresh' as text_value
  ) metadatavalue
group by
   item_id

Results:

item_id  volume_value  issue_value
1        90-I am here  91-Me too
2        NULL          91-funky
3        90-fresh      NULL

Note: I used SQL Server to model & test and then changed the syntax to match Postgres.

Upvotes: 0

John MacIntyre
John MacIntyre

Reputation: 13031

Try this

select  volume.text_value as volume_value, 
        issue.text_value as issue_value    
from    metadatavalue item    
        left outer join metadatavalue volume                    
            on item.item_id = volume.item_id    
        left outer join metadatavalue issue                    
            on item.item_id = issue.item_id    
where   volume.metadata_field_id = 90
and     issue.metadata_field_id = 91

Upvotes: 2

Related Questions