Question

我有3张表如下。

salesman(sid,sname)
location(lid,lname)
sales_loc(sid,lid)

现在我要打印已经访问过所有地点的销售员的sid和sname。我想要一个SQL查询，我不想要PL / SQL代码。

Answer 1

select sid, sname from salesman
where not exists 
        (select 1 from location
          where not exists
                  (select 1 from sales_loc
                    where sid=saleman.sid 
                     and lid = location.lid));

Answer 2

另一种方法：

select sid, sname from salesman
where
  (select count(*) from location) =
  (select count(*) from sales_loc where sales_loc.sid = salesman.sid)

修改如果sales_loc(sid,lid)对不是键，则以下查询更合适，
正如ammoQ建议：

select sid, sname from salesman where (select count(*) from location) = (select count(distinct lid) from sales_loc where sales_loc.sid = salesman.sid)

Answer 3

获取所有销售人员记录，其中访问的位置数等于位置数。

select sm.* from salesman as sm
where (select count(sl.*) from sales_loc as sl where sl.sid = sm.sid)
 = (select count(l.*) from location as l);

Answer 4

还有一个锅！

鉴于这些推销员及其领土......

SQL> select s.sname, l.lname
  2  from   salesman s
  3         , location l
  4         , sales_loc sl
  5  where  sl.sid = s.sid
  6  and sl.lid = l.lid
  7  order by s.sid, l.lid
  8  /

SNAME      LNAME
---------- ----------
FOX        TRAIN
FOX        BOAT
KNOX       BOAT
KNOX       HOUSE
SAM        TRAIN
SAM        BOAT
SAM        HOUSE

7 rows selected.

SQL>

...此查询提取已访问过所有人的人...

SQL> select s.sname
  2  from   salesman s
  3  where s.sid not in  (
  4      select sid from (
  5          select cjs.sid, cjl.lid
  6          from   salesman cjs
  7                 cross join location cjl
  8          minus
  9          select sl.sid, sl.lid
 10          from   sales_loc sl
 11      )
 12  )
 13  /

SNAME  
------ ----
SAM

SQL>

Answer 5

这应该有效。我试过了。

select a.sid, a.salesman, count(a.sid) as total from salesman a
inner join sales_loc b on b.sid = a.sid
inner join location c on c.lid = b.lid
group by a.sid, a.salesman
having count(a.sid) = (select count(*) from location)

它使用要比较的位置总数。

Answer 6

推销员（SID，SNAME）位置（盖，L-NAME） sales_loc（SID，盖子）

从salesman s，location l，sales_loc sl中选择s.sid，s.sname 其中s.sid = sl.sid 和l.lid = sl.lid

简单的SQL查询

6 个答案: