CODEWITHSUNDEEP
mysqlsqlhavinghaving-clauseaggregates
Spencer
Spencer

Reputation: 870

Reference a FROM alias in HAVING

Newbie to sql here. I am trying to select only tuples whose values in one column are larger than the average of that column. The problem is, the relation I am working with is created in the FROM clause, because I need to add a transformation (and also take a subset).

I was originally using a WHERE clause, but this SO answer led me to believe that I could fix the problem by using a HAVING clause instead. I am still receiving this error, however:

Table 'msan692db.l' doesn't exist

How can I make the necessary comparison in a way that is not blind to the alias I created in my FROM clause? I think I could just explicitly recreate the needed relation in the WHERE/HAVING clause, but I will be adding more later and this seems terribly inefficient.

This is my code as it stands:

select *
from 
    (select *, ascii(substring(CREDIT_Grade,1,1)) + cast(substring(CREDIT_Grade,2,1) as unsigned) * .2 as CREDIT_Grade_num
    from loanstats
    where Loan_duration = 36) L
group by loan_ID
having
    Interest_Rate > (select avg(Interest_Rate) from L)

EDIT: I have confirmed, by the way, that the FROM clause by itself returns a valid relation.

Upvotes: 0

Views: 93

Answers (1)

StanislavL
StanislavL

Reputation: 57381

Can't you move the subquery in the FROM section? Like this?

select *
from 
    (select *, ascii(substring(CREDIT_Grade,1,1)) + cast(substring(CREDIT_Grade,2,1) as unsigned) * .2 as CREDIT_Grade_num
    from loanstats
    where Loan_duration = 36) L,
    (select avg(Interest_Rate) avg_rate from L_COPY_HERE) iRateAvg
group by loan_ID
having
    Interest_Rate > iRateAvg.avg_rate

Upvotes: 1

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