Обсуждение: =ANY bug

Hi all,

Because IN is semantically identical to =ANY
here the same error using ANY:

select * from emp where ename =ANY (    select ename from emp group by ename having 1 < count(ename)    );
ename |empno|job       |  hiredate|sal      |comm|deptno|level| mgr
------+-----+----------+----------+---------+----+------+-----+----
ALLEN | 7499|SALESMAN  |1981-02-20|$1,600.00| 300|    20|    4|7782
BLAKE | 7698|MANAGER   |1981-05-01|$2,850.00|    |    30|    3|7782
JONES | 7900|CLERK     |1981-12-03|$950.00  |    |    30|    2|7782
MILLER| 7654|SALESMAN  |1981-09-28|$1,250.00| 400|    30|    3|7839
CLARK | 7844|SALESMAN  |1981-09-08|$1,500.00|    |    10|    2|7839
KING  | 7521|SALESMAN  |1981-02-22|$1,250.00| 500|    10|    1|7782
ALLEN | 7499|SALESMAN  |1981-02-20|$1,600.00| 300|    20|    4|7782
BLAKE | 7698|MANAGER   |1981-05-01|$2,850.00|    |    30|    3|7782
(8 rows)


but I expect the following instead:

ename |empno|job       |  hiredate|sal      |comm|deptno|level| mgr
------+-----+----------+----------+---------+----+------+-----+----
ALLEN | 7499|SALESMAN  |1981-02-20|$1,600.00| 300|    20|    4|7782
BLAKE | 7698|MANAGER   |1981-05-01|$2,850.00|    |    30|    3|7782
ALLEN | 7499|SALESMAN  |1981-02-20|$1,600.00| 300|    20|    4|7782
BLAKE | 7698|MANAGER   |1981-05-01|$2,850.00|    |    30|    3|7782
(4 rows)

-Jose'-

> Hi all,
> 
> Because IN is semantically identical to =ANY
> here the same error using ANY:
> 
> select * from emp where ename =ANY (
>      select ename from emp group by ename having 1 < count(ename)
>      );

Can I just say that using HAVING inside a subquery is buggy?  Vadim, was
this already fixed?


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