SQL 要求在WHERE子句中满足两个条件

Question

I have table of bird observations. This is an example:

Unique_ID  List_ID  Locality    Common_name
A1         001      Park        Downy Woodpecker
A2         001      Park        Hairy Woodpecker
A3         001      Park        Carolina Chickadee
B1         002      Campground  Blue Jay
B2         002      Campground  Hairy Woodpecker
C1         003      Backyard    Downy Woodpecker
C2         003      Backyard    American Goldfinch
D1         004      School      American Goldfinch
D2         004      School      Hairy Woodpecker
E1         005      Park        Downy Woodpecker
E2         005      Park        Carolina Chickadee

I am trying to write a query for PostgreSQL that will only return the Localities that have occurrences of both woodpeckers, Downy and Hairy, together. In the little example table, that would just be the Park. The other localities only have one or the other of the species.

I tried

SELECT List_ID, LOCALITY, COMMON_NAME FROM table
WHERE COMMON_NAME = 'Downy Woodpecker' and COMMON_NAME = 'Hairy Woodpecker';

But returned no results. My table has 1,000s of observations, it's based on eBird data and those two species are generally common throughout the country, so there has to be at least one list_ID where they occur at the same time. In my example table, only Park (based on list_ID 001) meets the condition for what I'm looking for.

If I understand IN , it will return a row that meets either. Any of the example list_IDs would work for that query, but that's not what I want. How do I write a query that forces the WHERE to meet multiple conditions?

Answer 1

You can join the table with itself to get the locations you want. For example:

select distinct a.locality
from my_table a
join my_table b on a.locality = b.locality
where a.common_name = 'Downy Woodpecker'
  and b.common_name = 'Hairy Woodpecker'

Answer 2

You could first write a query that returns list_ids “having” both types of woodpeckers. (Group by list_id and write a condition in the having clause that will return the ids).

Something like

max(case when common_name like ‘hairy%’ then 1 else 0 end) = 1 
and max(case when common_name like ‘downy%’     then 1 else 0 end) = 1

Then, you can use that query to filter your base table for only the desired list_ids.

Answer 3

First filter the table, group by Locality and return only the localities containing both kinds:

select Locality
from tablename
where Common_name in ('Downy Woodpecker', 'Hairy Woodpecker')
group by Locality
having count(distinct Common_name) = 2

The condition count(distinct Common_name) = 2 in the HAVING clause makes sure that both kinds exist for the same Locality .
If there is no case of duplicate Common_name s in the same Locality you may also omit distinct from count() .
See the demo .
Results:

| locality |
| -------- |
| Park     |