English
- # test of problems with having (Reported by Mark Rogers)
- #
- drop table if exists t1;
- create table t1 (a int);
- select count(a) as b from t1 where a=0 having b > 0;
- insert into t1 values (null);
- select count(a) as b from t1 where a=0 having b > 0;
- select count(a) as b from t1 where a=0 having b >=0;
- drop table t1;
