Question

有这样的表：

Table: statuses
ID  | status
----|---------
777 | 1
675 | 1
651 | 0
611 | 1
600 | 0
554 | 1
443 | 0
323 | 0
222 | 1
112 | 1

如何仅选择行中两个（或更多）状态为0的行？（在样本情况下仅为443,323），并按照第一个ID进行分组。

所以输出结果为：

ID  | status | group
----|--------|---------
443 | 0      | 443
323 | 0      | 443

Answer 1

如果无法使用编程方法，您可以尝试此查询（尽管性能会更差）。

 select
    s.id,
    0 as status,
    group_field.id as group
from 
    statuses s,
    (select id from statuses where status = 0 group by id having count(1) > 1 limit 1) group_field
where 
    s.id in(select id from statuses where id = 0 group by id having count(1) > 1)

Answer 2

select 
id, STATUS,
@st_order:=if(@pre_st!=status,0,@st_order+1) as status_order,
@group_leader:=if(@pre_st!=status,id,@group_leader) as group_leader,
@pre_st:=status
from statuses,(select @st_order:=-1, @pre_st:=-1, @group_leader:=-1) val

这个sql为你提供了输出

ID，
  状态，
  行号继续相同的状态，
  同一地位的组长，
  无用的最后一栏

根据您的输入，前四列输出为

ID |状态|状态行号|状态组负责人
  777 | 1 | 0 | 777个
  675 | 1 | 1 | 777个
  651 | 0 | 0 | 651个
  611 | 1 | 0 | 611个
  600 | 0 | 0 | 600个
  554 | 1 | 0 | 554个
  443 | 0 | 0 | 443个
  323 | 0 | 1 | 443个
  222 | 1 | 0 | 222个
  112 | 1 | 1 | 222

所以你可以在外部选择中做任何你想做的事。

Answer 3

您可以通过使用子查询使用少量逻辑来实现此目的

set @cnt=0;
set @id=-1;
SELECT 
    t3.id, t3.status, t3.dup_id AS `group`
FROM
    (SELECT 
        id,
            status,
            IF(status = 0, @cnt:=@cnt + 1, @cnt:=0) AS great,
            IF(@cnt = 1, @id:=id, IF(status = 0, @id, - 1)) AS dup_id
    FROM
        statuses) t3
WHERE
    t3.dup_id IN (SELECT 
            t1.dup_id
        FROM
            (SELECT 
                id,
                    status,
                    IF(status = 0, @cnt:=@cnt + 1, @cnt:=0) AS great,
                    IF(@cnt = 1, @id:=id, IF(status = 0, @id, - 1)) AS dup_id
            FROM
                statuses) t1
        GROUP BY t1.dup_id
        HAVING COUNT(t1.dup_id) > 1
            AND t1.dup_id != - 1);

结果将如预期那样

id  | status | group
----|--------|---------
443 | 0      | 443
323 | 0      | 443

检测表格中的模式

3 个答案: