Question

我正在尝试从Membership_benefit表中获取NUMSESSION值。我的函数有一个福利ID和一个成员ID作为参数，而Membership_benefit需要这两个值，否则它将返回多行。

这是功能的要旨

`CREATE OR REPLACE FUNCTION benefit_when(_benefitid integer, _membershipid 
 integer) RETURNS text AS $$
 DECLARE
     bfrec record;
 BEGIN
     if _membershipid is not null then
         select 
             benefit.*,
             membership_benefit.numsession as numsession
         from benefit 
             left join membershiptype on benefit.membershiptypeid=membershiptype.id
             left join membership_benefit on _membershipid = membership_benefit.membershipid
         where benefit.id=_benefitid
             AND (membership_benefit.membershipid is null or membership_benefit.membershipid = _membershipid);`

未正确返回numsession值。如果我打印两个参数的值并简单地查询数据库，它将返回正确的参数，但不会在此函数中返回。您是否可以像我在这里一样使用变量进行联接，如果没有，是否有更好的方法通过我拥有的两个参数来获取正在寻找的数字？

谢谢。

Answer 1

更改查询：

select 
         benefit.*,
         membership_benefit.numsession as numsession
     from benefit 
         left join membershiptype on benefit.membershiptypeid=membershiptype.id
         left join membership_benefit on _membershipid = membership_benefit.membershipid
     where benefit.id=_benefitid

变成这样：
select benefit.*, membership_benefit.numsession as numsession from benefit left join membershiptype on benefit.membershiptypeid=membershiptype.id where benefit.id=_benefitid and _membershipid = membership_benefit.membershipid

Answer 2

如果您只想使用numsession，则不确定为什么要进行外部联接。这是您真正想要的吗？

select mb.numsession
from membership_benefit mb
where mb.membershipid = _membershipid and
      mb.benefitid = _benefitid;

如果您总是希望获得好处，请与以下信息一同显示：

select b.*, mb.numsession as numsession
from benefit b join
     membership_benefit mb
     on mb.benefitid = b.id and
        mb.membershipid = _membershipid and
        b.id =_benefitid
where b.id = _benefitid;

我推测您想要这两件事之一。

连接表的正确技术？

2 个答案: