我在单个表上创建了一个连接索引,当我用where子句查询表时,一个用" =",另一个用"%",但似乎是第一个查询使用主索引,第二个查询不使用它,不明白,对主索引列的两个查询,为什么第二个查询不使用主索引?
单个表上的连接索引如下:
int havetopay(Client x)
{
int freeSMS = x.getFreeSMS();
int sentSMS = x.getSentSMS();
int SMS;
if (freeSMS > sentSMS) SMS = 0;
else SMS = sentSMS - freeSMS;
return (SMS * x.getSMSPrice() + x.getTalkedMinutes() * x.getMinutesPrice());
}
第一个解释,
CREATE JOIN INDEX CustomerService.EMP_JI AS
SELECT employee_number ,
department_number,
employee.last_name,
manager_employee_number
FROM customerservice.employee
PRIMARY INDEX ( last_name );
第二个解释,
explain sel * from customerservice.employee where last_name = 'tony';
1) First, we do a single-AMP RETRIEVE step from
CustomerService.EMP_JI by way of the primary index
"CustomerService.EMP_JI.last_name = 'tony '" with no residual
conditions into Spool 2 (group_amps), which is redistributed by
the hash code of (CustomerService.EMP_JI.employee_number) to few
AMPs. Then we do a SORT to order Spool 2 by row hash. The size
of Spool 2 is estimated with low confidence to be 1 row (45 bytes).
The estimated time for this step is 0.01 seconds.
2) Next, we do a group-AMPs JOIN step from customerservice.employee
by way of a RowHash match scan with no residual conditions, which
is joined to Spool 2 (Last Use) by way of a RowHash match scan.
customerservice.employee and Spool 2 are joined using a merge join,
with a join condition of ("Field_1025 =
customerservice.employee.employee_number"). The result goes into
Spool 1 (group_amps), which is built locally on that AMP. The
size of Spool 1 is estimated with low confidence to be 1 row (143
bytes). The estimated time for this step is 0.03 seconds.
3) Finally, we send out an END TRANSACTION step to all AMPs involved
in processing the request.
-> The contents of Spool 1 are sent back to the user as the result of
statement 1. The total estimated time is 0.04 seconds.
答案 0 :(得分:1)
where子句需要具有用于访问主索引的相等条件(=),因为主索引在主索引的散列值上分配行。