Question

我有这种表：

我希望获得一个“a”值，使其具有“b”的确切搜索值。例如，如果我搜索（7,2），我希望查询返回8，如果我搜索（7,2,3），我希望它返回9，否则它必须返回Null：

Search      Result
(7,2)       8
(7,2,3)     9
(7,3)       Null
(7,2,4)     Null
...

是否可以在不使用“分组依据”连接的情况下执行此操作？

[编辑]对于“通过连接分组”，我的意思是像MySql中的GROUP_CONCAT（）或任何字符串函数。

Answer 1

group by /连接方法有效。另一种方法是group by，带有having子句：

select a
from table t
group by a
having sum(case when b = 7 then 1 else 0 end) = 1 and
       sum(case when b = 2 then 1 else 0 end) = 1 and
       sum(case when b not in (7, 2) then 1 else 0 end) = 0;

我们的想法是计算每个值中匹配的值。实际上，您可以使用字符串操作执行此操作，但这些操作在数这是一个＆＃34;将军＆＃34;使用in执行此操作的方式以及您要查找的值的数量：

select a
from table t
group by a
having count(*) = 2 and
       count(distinct (case when b in (2, 7) then b end));

Answer 2

在SQL Server中

SELECT a FROM tableName
WHERE a NOT IN ( SELECT a FROM (   
                               SELECT a,
                               CASE WHEN CHARINDEX(b,@searchTerm)=0 
                               THEN 0 ELSE 1 END as Result
                               FROM tableName
                               )z
                  WHERE z.Result=0
               )

我正在使用CHARINDEX（）函数来检查搜索词是否存在。

Answer 3

不确定你的意思＆＃34;通过连接分组＆＃34; - 我不认为可以在没有 group by 的情况下完成

您没有指定DBMS，因此这是ANSI SQL：

with search_values (val) as ( values (7), (2) ) -- first part gets all those that do have the search values -- but will also include those rows that have more than the searched ones select a from data where b in (select val from search_values) group by a having count(distinct b) = (select count(*) from search_values) intersect -- the intersect then filters out those that have exactly the search values select a from data group by a having count(distinct b) = (select count(*) from search_values);

SQLFiddle示例：http://sqlfiddle.com/#!15/dae93/1

将CTE用于＆＃34;搜索值＆＃34;避免重复它们并避免＆＃34;硬编码＆＃34;要搜索的项目数。如果您查找7,2,3，则只需向CTE添加另一个值

不是使用intersect，而是可以使用共同相关的子查询重写

with search_values (val) as ( values (7), (2) ) select d1.a from data d1 where d1.b in (select val from search_values) group by d1.a having count(distinct d1.b) = (select count(*) from search_values) and count(distinct d1.b) = (select count(*) from data d2 where d2.a = d1.a);

Answer 4

另一种方法是使用 INTERSECT 。

然而，第一种情况似乎仍然需要破解（返回8,9而不是8）。

这是SQL代码：

--(7,2) => 8,9 (instead of 8) :(
   SELECT  distinct(a)
    FROM   t1
    WHERE  b = 7
   INTERSECT
   SELECT  distinct(a)
    FROM   t1
    WHERE  b = 2

--(7,2,3) => 9
   SELECT  distinct(a)
    FROM   t1
    WHERE  b = 7
   INTERSECT
   SELECT  distinct(a)
    FROM   t1
    WHERE  b = 2
   INTERSECT
   SELECT  distinct(a)
    FROM   t1
    WHERE  b = 3

--(7,4) => NULL
   SELECT  distinct(a)
    FROM   t1
    WHERE  b = 7
   INTERSECT
   SELECT  distinct(a)
    FROM   t1
    WHERE  b = 4

--(7,2,4) => NULL
   SELECT  distinct(a)
    FROM   t1
    WHERE  b = 7
   INTERSECT
   SELECT  distinct(a)
    FROM   t1
    WHERE  b = 2
   INTERSECT
   SELECT  distinct(a)
    FROM   t1
    WHERE  b = 4

Fiddle Demo

使用SQL搜索确切的值

4 个答案: