Mysql 联合两个查询并标记重复项

我有两张桌子。一个包含所有选项，一个包含用户选择的选项以及他们创建的一些选项，这些选项可能不在选项表中

我需要合并数据，以便返回一个结果集，其中包括所有选项，以及用户选项，并以某种方式标记这些选项仅为用户，并且与主选项数据重叠

例如

选项

`COLOR`   | `ANIMAL`
---------------------
RED       | DOG
BLUE      | DOG
GREEN     | DOG
YELLOW    | CAT
PINK      | CAT
ORANGE    | CAT

`COLOR` | `ANIMAL`
------------------
GREEN   | SNAKE
BLUE    | DOG
PINK    | CAT
PURPLE  | CAT

用户选择的选项

`COLOR`   | `ANIMAL`
---------------------
RED       | DOG
BLUE      | DOG
GREEN     | DOG
YELLOW    | CAT
PINK      | CAT
ORANGE    | CAT

`COLOR` | `ANIMAL`
------------------
GREEN   | SNAKE
BLUE    | DOG
PINK    | CAT
PURPLE  | CAT

我的结果需要看起来像…

`COLOR` | `ANIMAL`| `DUPLICATE_OR_NEW`
----------------------------------------
RED     | DOG     | 0
BLUE    | DOG     | 1
GREEN   | DOG     | 0
YELLOW  | CAT     | 0
PINK    | CAT     | 1
ORANGE  | CAT     | 0
PURPLE  | CAT     | 1
GREEN   | SNAKE   | 1

---

在这种情况下，排序顺序并不重要。我曾尝试与工会合作，但我认为我需要将两张桌子连接在一起。到目前为止，我还没有想出一个解决办法。

这可能被称为作弊，但它会奏效

SELECT COLOR, ANIMAL, sum(DUPLICATE_OR_NEW)
FROM
(
SELECT COLOR, ANIMAL, 1 as DUPLICATE_OR_NEW FROM options
UNION ALL
SELECT COLOR, ANIMAL, 2 as DUPLICATE_OR_NEW FROM UserSelection
) as UTable
GROUP BY  COLOR, ANIMAL

-- 1 = unchosen option
-- 2 = new user added option
-- 3 = option exsisted chosen by user

---

请参阅SQL Fiddle

另一种方法：

select color, animal, 1 as duplicate_or_new
from UserSelected
union all
select color, animal, 0 as duplicate_or_new
from options o
where not exists (select 1 from UserSelected us where us.color = o.color and us.animal = o.animal)

使用

联合所有

/组的正确方法是：

 select color, animal, max(which) as duplicate_or_new
 from (select color, animal, 1 as which
       from UserSelected
       union all
       select color, animal, 0 as which
       from options
      ) t
group by color, animal

以下查询创建两个单独的标志：

 select color, animal, max(isUser) as IsUser, max(isOption) as IsOption
 from (select color, animal, 1 as IsUser, 0 as IsOption
       from UserSelected
       union all
       select color, animal, 0 as IsUser, 1 as IsOption
       from options
      ) t
group by color, animal

您可以将它们放在案例陈述中以格式化信息：

(case when max(isUser) = 1 and max(isOption) = 1 then 'both'
      when max(isUser) = 1 then 'user'
      when max(isOption) = 1 then 'option'
      else 'impossible'
 end)

---

但问题是什么？如果作弊有效，我喜欢作弊。你得到的答案大约是95%。由于负1，表2中不存在的项显示为零。简单的解决方法是将选择颜色，动物，“1”更改为表\u名称2中的重复\u或\u新****以选择颜色，动物，“2”更改为表\u名称2中的重复\u或\u新。这将在列中为重叠的项添加2，为新项添加1。这对我来说是很好的，因为我可以说大于零的地方。谢谢您如何判断这些选项是完全“新”的还是从现有列表中选择的？