Question

我在LMS平台上对练习进行评分有问题。

我们有两个查询：

select
    distinct c.city_name,'city' obj_type
from
    shipping.city c
union all
select
    distinct c.state,'state' obj_type
from
    shipping.city c
union all
select
    distinct d.first_name,'driver' obj_type
from
    shipping.driver d
union all
select
    distinct t.Make,'truck' obj_type
from
    shipping.truck t
order by 1 desc

这：

select
    c.city_name,'city' obj_type
from
    shipping.city c
union
select
    c.state,'state' obj_type
from
    shipping.city c
union
select
    d.first_name,'driver' obj_type
from
    shipping.driver d
union
select
    t.Make,'truck' obj_type
from
    shipping.truck t
order by 1 desc

两者都能获得相同的结果，但第二点在平台上没有得分。同时，我在Metabase中使用除外检查结果，它使我“无结果！” （空）

platfrom的工作原理：

它发送2个查询作为基础，并检查学生和参考答案之间的差异。

我认为在Metabase中，“联合”和“全部联合”的数据可能参差不齐，但是比较，除了什么也没给我。谢谢！

UPD：解决问题。在两个查询中有不同的排序，也从platfrom检查，查询也不同。谢谢大家！

Answer 1

如果同一表中的行具有相同的值，或者NULL或city_name或state中的first_name make的结果可能有所不同。

编辑：尝试使用EXCEPT ALL而不是EXCEPT检查结果，否则您将看不到所有差异。

看看这个例子：

DECLARE @T1 TABLE (name VARCHAR(10))
DECLARE @T2 TABLE (surname VARCHAR(10))

DECLARE @R1 TABLE (descr VARCHAR(10), typ VARCHAR(10))
DECLARE @R2 TABLE (descr VARCHAR(10), typ VARCHAR(10))

INSERT INTO @T1
SELECT * 
FROM( VALUES ('Abigail'), ('Luke'), ('Mat'), ('Mat'), ('Zoe')) x (d)

INSERT INTO @T2
SELECT * 
FROM( VALUES ('Brown'), ('Doe'), ('Doe'), ('Smith'), ('Wilkinson')) x (d)

insert into @r1
select name descr, 'name' descr_type from @t1
union 
select surname descr, 'surname' descr_type from @t2

insert into @r2
select name descr, 'name' descr_type from @t1
union all
select surname descr, 'surname' descr_type from @t2

select * from @r1
select * from @r2

如果检查结果：

select * from @r1
except
select * from @r2

将产生“没有结果！” （空）

select * from @r1
except all
select * from @r2

会产生以下结果：

descr   typ
'Doe'   'surname'
'Mat'   'name'

如果您想查看差异详细信息：

select * 
    from 
        (select ROW_NUMBER() over (partition by descr order by descr) n, * from @r1) r1
    full join 
        (select ROW_NUMBER() over (partition by descr order by descr) n, * from @r2) r2 
    on r1.n= r2.n and r1.descr = r2.descr

Answer 2

问题是，由于{city，state，driver，truck}文字的常量不同，UNION的三个腿总是不同的。因此，将没有重复项，并且UNION的结果与UNION ALL相同。

示例：

        -- some data ...
CREATE TABLE one
        ( num integer not null
        , nam text
        );
INSERT INTO one ( num , nam ) values (1,'one'), (2, 'two'), (3, 'three');

        -- QUERY#1
SELECT 'from_first' AS "origin"
        , num, nam
FROM one
UNION
SELECT 'from_second' AS "origin"
        , num, nam
FROM one
ORDER BY 1,2
        ;

        -- QUERY#2
SELECT 'from_first' AS "origin"
        , num, nam
FROM one
UNION ALL
SELECT 'from_second' AS "origin"
        , num, nam
FROM one
ORDER BY 1,2
        ;

结果：

CREATE TABLE
INSERT 0 3
   origin    | num |  nam  
-------------+-----+-------
 from_first  |   1 | one
 from_first  |   2 | two
 from_first  |   3 | three
 from_second |   1 | one
 from_second |   2 | two
 from_second |   3 | three
(6 rows)

   origin    | num |  nam  
-------------+-----+-------
 from_first  |   1 | one
 from_first  |   2 | two
 from_first  |   3 | three
 from_second |   1 | one
 from_second |   2 | two
 from_second |   3 | three
(6 rows)

当一个或两个子选择本身产生重复项时，情况就不同了，如：

        -- some more data ...
CREATE TABLE two
        ( num integer not null
        , nam text
        );
INSERT INTO two ( num , nam ) values (1,'one'), (2, 'two'), (2, 'two');

        -- QUERY#1a
SELECT 'from_first' AS "origin"
        , num, nam
FROM two
UNION
SELECT 'from_second' AS "origin"
        , num, nam
FROM two
ORDER BY 1,2
        ;

        -- QUERY#2b
SELECT 'from_first' AS "origin"
        , num, nam
FROM two
UNION ALL
SELECT 'from_second' AS "origin"
        , num, nam
FROM two
ORDER BY 1,2
        ;

结果2：

CREATE TABLE
INSERT 0 3
   origin    | num | nam 
-------------+-----+-----
 from_first  |   1 | one
 from_first  |   2 | two
 from_second |   1 | one
 from_second |   2 | two
(4 rows)

   origin    | num | nam 
-------------+-----+-----
 from_first  |   1 | one
 from_first  |   2 | two
 from_first  |   2 | two
 from_second |   1 | one
 from_second |   2 | two
 from_second |   2 | two
(6 rows)

LMS平台上的SQL代码分级时，“联合”和“全部联合”之间的差异

2 个答案: