Question

我正在尝试编写一个有效的查询，如下所示

＆＃34;获取所有已删除且没有其他行具有相同值且未删除的行＆＃34;

实施例

------------------------
| 2  | "bar" |    1
------------------------
| 4  | "baz" |    1
------------------------
| 6  | "baz" |    1
------------------------

======＆GT;

foo

结果集中deleted = 0行未包含的原因是因为其中一行put returns between paragraphs for linebreak add 2 spaces at end indent code by 4 spaces。

Answer 1

检查此查询

select
    a.*
from
    myTable a
where
    not exists (
        select 1
        from mytable b
        where
            a.val = b.val
            and b.deleted = 0
    )

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Answer 2

为了获取已删除的值而不是其他值，您可以使用subquery形式，其中子查询仅返回已删除的值为1的值。

SELECT * FROM table 
WHERE val IN (
    SELECT val FROM table
    GROUP BY val
    HAVING COUNT(DISTINCT deleted) = 1 AND
           COUNT(DISTINCT case when deleted = 1 then deleted end) = 1
)

相关的子查询可能很有用

SELECT * FROM table t
    WHERE EXISTS (
        SELECT val FROM table
        WHERE val = t.val
        GROUP BY val
        HAVING COUNT(DISTINCT deleted) = 1 AND
               COUNT(DISTINCT case when deleted = 1 then deleted end) = 1
    )

Answer 3

如果您喜欢我喜欢加入条款，这是一个选项：）

declare @t table(id int, val varchar(50) ,deleted bit) 
insert into @t(id,val,deleted)
values 
(1,'foo',1)
,(2,'bar',1)
,(3,'foo',0)
,(4,'baz',1)
,(5,'qux',0)
,(6,'baz',1)

select *
from @t t1
left join(select distinct t2t1.val from @t t2t1 where deleted=0) t2 on t1.val=t2.val
where t2.val is null

如何编写有效的查询来执行此操作？

3 个答案: