Question

这是sqlfiddle上的代码 - http://sqlfiddle.com/#!6/9596d/8

create table tbla (uid nvarchar(10), name nvarchar(50));
create table tblb (uid nvarchar(10), cd nvarchar(5));

insert into tbla values ('abc','abraham lincoln');
insert into tbla values ('def','david copperfield');
insert into tblb values ('abc','a');
insert into tblb values ('abc','b');
insert into tblb values ('abc','c');
insert into tblb values ('def','a');
insert into tblb values ('def','c');

我需要退回计数。如果没有找到记录则返回0。这就是我所拥有的，但它不会为 uid def

返回0

select a.uid, a.name, count(b.cd) as ct
from tblb b
left join
tbla a
on b.uid = a.uid
where b.cd = 'b'
group by a.uid, a.name

我做错了什么？它使用外连接。

Answer 1

对于左连接，您的表已向后移动，并且您需要对where表中的条件使用join子句而不是left join子句。否则，您将left join变为inner join。

select a.uid, a.name, count(b.cd) as ct
from tbla a
  left join tblb b
    on b.uid = a.uid
   and b.cd = 'b'
group by a.uid, a.name

rextester演示：http://rextester.com/ZXBS56241

返回：

+-----+-------------------+----+
| uid |       name        | ct |
+-----+-------------------+----+
| abc | abraham lincoln   |  1 |
| def | david copperfield |  0 |
+-----+-------------------+----+

Answer 2

恕我直言，这是正确的做法，

create table #tbla (uid nvarchar(10), name nvarchar(50));
create table #tblb (uid nvarchar(10), cd nvarchar(5));

insert into  #tbla values ('abc','abraham lincoln');
insert into  #tbla values ('def','david copperfield');
insert into  #tblb values ('abc','a');
insert into  #tblb values ('abc','b');
insert into  #tblb values ('abc','c');
insert into  #tblb values ('def','a');
insert into  #tblb values ('def','c');

--select * from  #tblb
--select * from  #tbla
;With CTE as
(
select b.uid,  count(b.cd) as ct
from #tblb b
where b.cd = 'b'
group by b.uid
)
select a.uid,isnull(c.ct,0),a.name from #tbla a
left join cte c  on 
a.uid=c.uid 

drop table #tblb
drop table #tbla

如果没有找到行，则Count返回0

2 个答案: