Question

我加入这张桌子时有三张桌子

declare @tbl_1 table (pid int, loc varchar(100), avaId int,xpId int,qf varchar(100));
declare @tbl_2 table (soid int,pid int,sid int,name2 varchar(100), nrt2 int);
declare @tbl_3 table (woid int,pid int,wid int,name3 varchar(100), nrt3 int);

insert into @tbl_1 values (1000,'Bangalore',30,9,'ABC');

insert into @tbl_2 values(0,1000,1,'name1',8);
insert into @tbl_2 values(1,1000,8,'name2',5);
insert into @tbl_2 values(2,1000,7,'name3',6);

insert into @tbl_3 values(0,1000,2,'D1',9);
insert into @tbl_3 values(1,1000,1,'D2',2);
insert into @tbl_3 values(2,1000,3,'D3',0);
insert into @tbl_3 values(3,1000,4,'D4',5);

加入查询：

select 
    t.pid, t.loc, t.avaId, t.xpId, t.qf,
    tt.sid, tt.name2, 
    ttt.wid, ttt.name3
from 
    @tbl_1 t
left join 
    @tbl_2 tt on t.pid = tt.pid
right outer join
    @tbl_3 ttt on t.pid = ttt.pid 
               and tt.soid = ttt.woid

虽然我正在使用此联接，但我得到的结果是

pid    loc    avaId    xpId    qf    sid    name2    wid    name3
1000    Bangalore    30    9    ABC    1    name1    2    D1
1000    Bangalore    30    9    ABC    8    name2    1    D2
1000    Bangalore    30    9    ABC    7    name3    3    D3
NULL    NULL    NULL    NULL    NULL    NULL    NULL    4    D4

我怎样才能得到这个结果：

pid    loc    avaId    xpId    qf    sid    name2    wid    name3
1000    Bangalore    30    9    ABC    1    name1    2    D1
1000    Bangalore    30    9    ABC    8    name2    1    D2
1000    Bangalore    30    9    ABC    7    name3    3    D3
1000    Bangalore    30    9    NULL    NULL    4    D4

由于

Answer 1

如果我理解您的意图，您可以先加入"1,23,45,448,00"和@tbl_2，然后加入@tbl_3。

@tbl_1

的 LiveDemo

输出：

WITH cte AS ( SELECT COALESCE(tt.pid, ttt.pid) AS pid, tt.sid, tt.name2, ttt.wid, ttt.name3 FROM @tbl_2 tt FULL OUTER JOIN @tbl_3 ttt ON tt.pid = ttt.pid AND tt.soid = ttt.woid ) select t.pid, t.loc, t.avaId, t.xpId, t.qf, c.sid, c.name2, c.wid, c.name3 from @tbl_1 t JOIN cte c ON t.pid = c.pid;

如何摆脱最后的记录空

1 个答案: