使用多个用户 ID 对 table 进行复杂查询
Complex querying on table with multiple userids
我有一个 table 这样的:
score
id week status
1 1 0
2 1 1
3 1 0
4 1 0
1 2 0
2 2 1
3 2 0
4 2 0
1 3 1
2 3 1
3 3 1
4 3 0
我想获取除第 3 周外所有周状态都为零的人的 ID。像这样:
结果:
result:
id w1.status w2.status w3.status
1 0 0 1
3 0 0 1
我有这个查询,但它在较大的数据集上效率极低。
SELECT w1.id, w1.status, w2.status, w3.status
FROM
(SELECT s.id, s.status
FROM score s
WHERE s.week = 1) w1
LEFT JOIN
(SELECT s.id, s.status
FROM score s
WHERE s.week = 2) w2 ON w1.id=w2.id
LEFT JOIN
(SELECT s.id, s.status
FROM score s
WHERE s.week = 3) w3 ON w1.id=w3.id
WHERE w1.status=0 AND w2.status=0 AND w3.status=1
我正在寻找一种更有效的方法来计算上述内容。
您可以使用 not exists
作为
select
t1.id,
'0' as `w1_status` ,
'0' as `w2_status`,
'1' as `w3_status`
from score t1
where
t1.week = 3
and t1.status = 1
and not exists(
select 1 from score t2
where t1.id = t2.id and t1.week <> t2.week and t2.status = 1
);
为了获得更好的性能,您可以在 table 中添加索引,如
alter table score add index week_status_idx (week,status);
如果周数是静态的 (1-3),group_concat 可能会被用作 hack..
概念:
SELECT
id,
group_concat(status) as totalStatus
/*(w1,w2=0,w3=1 always!)*/
FROM
tableName
WHERE
totalStatus = '(0,0,1)' /* w1=0,w2=1,w3=1 */
GROUP BY
id
ORDER BY
week ASC
(随手写。未测试)
SELECT p1.id, p1.status, p2.status, p3.status
FROM score p1
JOIN score p2 ON p1.id = p2.id
JOIN score p3 ON p2.id = p3.id
WHERE p1.week = 1
AND p1.status = 0
AND p2.week = 2
AND p2.status = 0
AND p3.week = 3
AND p3.status = 1
试试这个,应该有用
select id
from score
where week in (1, 2, 3)
group by id
having sum(
case
when week in (1, 2) and status = 0 then 1
when week = 3 and status = 1 then 1
else 0
end
) = 3
或者更笼统地说...
select id
from score
group by id
having
sum(case when status = 0 then 1 else 0 end) = count(*) - 1
and min(case when status = 1 then week else null end) = max(week)
我有一个 table 这样的:
score
id week status
1 1 0
2 1 1
3 1 0
4 1 0
1 2 0
2 2 1
3 2 0
4 2 0
1 3 1
2 3 1
3 3 1
4 3 0
我想获取除第 3 周外所有周状态都为零的人的 ID。像这样:
结果:
result:
id w1.status w2.status w3.status
1 0 0 1
3 0 0 1
我有这个查询,但它在较大的数据集上效率极低。
SELECT w1.id, w1.status, w2.status, w3.status
FROM
(SELECT s.id, s.status
FROM score s
WHERE s.week = 1) w1
LEFT JOIN
(SELECT s.id, s.status
FROM score s
WHERE s.week = 2) w2 ON w1.id=w2.id
LEFT JOIN
(SELECT s.id, s.status
FROM score s
WHERE s.week = 3) w3 ON w1.id=w3.id
WHERE w1.status=0 AND w2.status=0 AND w3.status=1
我正在寻找一种更有效的方法来计算上述内容。
您可以使用 not exists
作为
select
t1.id,
'0' as `w1_status` ,
'0' as `w2_status`,
'1' as `w3_status`
from score t1
where
t1.week = 3
and t1.status = 1
and not exists(
select 1 from score t2
where t1.id = t2.id and t1.week <> t2.week and t2.status = 1
);
为了获得更好的性能,您可以在 table 中添加索引,如
alter table score add index week_status_idx (week,status);
如果周数是静态的 (1-3),group_concat 可能会被用作 hack..
概念:
SELECT
id,
group_concat(status) as totalStatus
/*(w1,w2=0,w3=1 always!)*/
FROM
tableName
WHERE
totalStatus = '(0,0,1)' /* w1=0,w2=1,w3=1 */
GROUP BY
id
ORDER BY
week ASC
(随手写。未测试)
SELECT p1.id, p1.status, p2.status, p3.status
FROM score p1
JOIN score p2 ON p1.id = p2.id
JOIN score p3 ON p2.id = p3.id
WHERE p1.week = 1
AND p1.status = 0
AND p2.week = 2
AND p2.status = 0
AND p3.week = 3
AND p3.status = 1
试试这个,应该有用
select id
from score
where week in (1, 2, 3)
group by id
having sum(
case
when week in (1, 2) and status = 0 then 1
when week = 3 and status = 1 then 1
else 0
end
) = 3
或者更笼统地说...
select id
from score
group by id
having
sum(case when status = 0 then 1 else 0 end) = count(*) - 1
and min(case when status = 1 then week else null end) = max(week)