3

我有两个查询，我需要找出它们的结果集之间的差异。我的查询如下。 

select star_ident,transition_ident,fix_ident,min(sequence_num)
  from corept.std_star_leg c
  where airport_ident='KLAS' and data_supplier='J'
  group by star_ident,transition_ident;

select name,trans
  from skyplan_deploy.deploy_stars
  where apt='KLAS';

这是我的两个查询。我最初使用左连接但未能得到结果。

select star_ident,transition_ident,fix_ident,min(sequence_num)
  from corept.std_star_leg c
  left join
  (
    select name,trans
    from skyplan_deploy.deploy_stars
    where apt='KLAS' and name != trans
  ) a
 on star_ident=a.name and fix_ident=a.trans
 where airport_ident='KLAS' and data_supplier='J' and a.name is null
 group by star_ident,transition_ident;

我尝试了上面的查询，但它完全给出了错误的结果集。谁能帮我做这个？

谢谢你。

4

1 回答 1

2

没有实际的表格很难测试，但一种方法是这样的；

SELECT t1.*
FROM (
  select star_ident,transition_ident,fix_ident,min(sequence_num)
  from std_star_leg c
  where airport_ident='KLAS' and data_supplier='J'
  group by star_ident,transition_ident) t1
LEFT JOIN (select name,trans
  from deploy_stars
  where apt='KLAS') t2
ON t1.star_ident = t2.name AND t1.fix_ident = t2.trans
WHERE t2.trans IS NULL;

也就是说，将您的选择包装到命名的子选择中，并LEFT JOIN在它们之间做一个标准。这将显示第一个结果集中而不是第二个结果集中的行。

编辑：要查找第二个但不是第一个子选择中的行，您只需将其更改为 a RIGHT JOIN， selectt2.*和 null 检查即可t1.fix_ident；

SELECT t2.*
FROM (
  select star_ident,transition_ident,fix_ident,min(sequence_num)
  from std_star_leg c
  where airport_ident='KLAS' and data_supplier='J'
  group by star_ident,transition_ident) t1
RIGHT JOIN (select name,trans
  from deploy_stars
  where apt='KLAS') t2
ON t1.star_ident = t2.name AND t1.fix_ident = t2.trans
WHERE t1.fix_ident IS NULL;
于 2013-05-22T11:36:51.213 回答