22

I need to filter by calculated column in postgres. It's easy with MySQL but how to implement with Postgres SQL ?

pseudocode:

select id, (cos(id) + cos(id)) as op from myTable WHERE op > 1;

Any SQL tricks ?

2
  • MySQL doesn't have CTEs but that's one way to get around repeating the expression. You can also use an inline view/derived table/nested table or whatever they call it in MySQL Commented Aug 24, 2015 at 16:56
  • 1
    Related: stackoverflow.com/questions/8370114/… Commented Aug 24, 2015 at 17:56

2 Answers 2

27

If you don't want to repeat the expression, you can use a derived table:

select *
from (
   select id, cos(id) + cos(id) as op 
   from myTable 
) as t 
WHERE op > 1;

This won't have any impact on the performance, it is merely syntactic sugar required by the SQL standard.

Alternatively you could rewrite the above to a common table expression:

with t as (
  select id, cos(id) + cos(id) as op 
  from myTable 
)
select *
from t 
where op > 1;

Which one you prefer is largely a matter of taste. CTEs are optimized in the same way as derived tables are, so the first one might be faster especially if there is an index on the expression cos(id) + cos(id)

Sign up to request clarification or add additional context in comments.

2 Comments

If indexes are important it might be a good idea to just use to an explicit range of id values between +/-1.0471975511965977461542144610932 radians.
Is this more efficient than the below answer where the calculation is in the where clause?
8
select id, (cos(id) + cos(id)) as op 
from selfies 
WHERE (cos(id) + cos(id)) > 1

You should specify the calculation in the where clause as you can't use a alias.

1 Comment

For those that are interested, this solution is quite a bit faster than using the derived table above.

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.