结合两列mysql的条件
我想将2个不同列的条件组合起来进行查询。这是我原来的查询。您可以在sqlfiddle.com中进行测试。
-- creating database first for test data
create table attendance(Id int, DateTime datetime, Door char(20));
INSERT INTO attendance VALUES
( 1, '2016-01-01 08:00:00', 'In'),
( 2, '2016-01-01 09:00:00', 'Out'),
( 3, '2016-01-01 09:15:00', 'In'),
( 4, '2016-01-01 09:30:00', 'In'),
( 5, '2016-01-01 10:00:00', 'Out'),
( 6, '2016-01-01 15:00:00', 'In');
SELECT * FROM attendance;
SELECT
@id:=@id+1 Id,
MAX(IF(Door = 'In', DateTime, NULL)) `Check In`,
MAX(IF(Door = 'Out', DateTime, NULL)) `Check Out`
FROM
(SELECT
*,
CASE
WHEN
(Door != 'Out' AND @last_door = 'Out')
THEN @group_num:=@group_num+1
ELSE @group_num END door_group,
@last_door:=Door
FROM attendance
JOIN (SELECT @group_num:=1,@last_door := NULL) a
) t JOIN (SELECT @id:=0) b
GROUP BY t.door_group
HAVING SUM(Door = 'In') > 0 AND SUM(Door = 'Out') > 0;
//output
+------+---------------------+---------------------+
| Id | Check In | Check Out |
+------+---------------------+---------------------+
| 1 | 2016-01-01 08:00:00 | 2016-01-01 09:00:00 |
| 2 | 2016-01-01 09:30:00 | 2016-01-01 10:00:00 |
+------+---------------------+---------------------+
从上面的查询中，我想再添加一列。
-- creating database first for test data
create table attendance(Id int, DateTime datetime, Door char(20), Active_door char(20));
INSERT INTO attendance VALUES
( 1, '2016-01-01 08:00:00', 'In', ''),
( 2, '2016-01-01 09:00:00', 'Out', ''),
( 3, '2016-01-01 09:15:00', 'In', ''),
( 4, '2016-01-01 09:30:00', 'In', ''),
( 5, '2016-01-01 09:35:00', '', 'On'),
( 6, '2016-01-01 10:00:00', 'Out', ''),
( 7, '2016-01-01 16:00:00', '', 'Off');
这是我对查询所做的更改，但它无效。
SELECT * FROM attendance;
SELECT
@id:=@id+1 Id,
MAX(IF(Door = 'In' OR Active_door = "On", DateTime, NULL)) `Check In`,
MAX(IF(Door = 'Out' OR Active_door = "Off", DateTime, NULL)) `Check Out`
FROM
(SELECT
*,
CASE
WHEN
((Door != 'Out' OR Active_door != "Off") AND (@last_door = 'Out' OR @last_door = 'Off'))
THEN @group_num:=@group_num+1
ELSE @group_num END door_group,
@last_door:=Door
FROM attendance
JOIN (SELECT @group_num:=1,@last_door := NULL) a
) t JOIN (SELECT @id:=0) b
GROUP BY t.door_group
HAVING SUM(Door = 'In') > 0 OR SUM(Active_door = 'On') > 0 AND SUM(Door = 'Out') > 0 OR SUM(Active_door = 'Off') > 0;
//output
+------+---------------------+---------------------+
| Id | Check In | Check Out |
+------+---------------------+---------------------+
| 1 | 2016-01-01 08:00:00 | 2016-01-01 09:00:00 |
| 2 | 2016-01-01 09:35:00 | 2016-01-01 10:00:00 |
| 3 | NULL | 2016-01-01 16:00:00 |
+------+---------------------+---------------------+
//my desire output
+------+---------------------+---------------------+
| Id | Check In | Check Out |
+------+---------------------+---------------------+
| 1 | 2016-01-01 08:00:00 | 2016-01-01 09:00:00 |
| 2 | 2016-01-01 09:35:00 | 2016-01-01 16:00:00 |
+------+---------------------+---------------------+
请帮助我们如何获得所需的输出。我想从两个列中获得最后一个并且最后一个。提前谢谢。
1 个答案:
答案 0 :(得分：1)
这样可以保持解决方案易于维护，而无需一次性完成最终查询，这将使其大小几乎翻倍（在我看来）。这是因为结果需要匹配并在具有匹配的In和Out事件的一行上表示。所以最后，我使用了一些工作表。它在存储过程中实现。
存储过程使用带有cross join的几个变量。将交叉连接视为初始化变量的机制。变量是安全维护的，所以我相信，在变量查询中经常引用的document的精神。参考的重要部分是在线上安全处理变量，迫使它们在使用它们的其他列之前设置。这是通过greatest()和least()函数实现的，这些函数的优先级高于不使用这些函数而设置的变量。另请注意，coalesce()通常用于同一目的。如果它们的使用看起来很奇怪，例如将已知数大于0或0的最大值，那么这是故意的。故意强制设置变量的优先顺序。
查询中名为dummy2等内容的列是未使用输出的列，但它们用于设置greatest()或其他内部的变量。这是上面提到的。像7777这样的输出是第3个插槽中的占位符，因为使用的if()需要一些值。所以忽略这一切。
我已经包含了代码的几个屏幕截图，因为它逐层进行，以帮助您可视化输出。这些迭代的开发如何慢慢地进入下一阶段，以扩展先前的阶段。
我相信我的同行可以在一个查询中改进这一点。我可以这样完成它。但是我相信这会导致一个令人困惑的混乱，如果被触动就会破裂。
<强>架构：
create table attendance2(Id int, DateTime datetime, Door char(20), Active_door char(20));
INSERT INTO attendance2 VALUES
( 1, '2016-01-01 08:00:00', 'In', ''),
( 2, '2016-01-01 09:00:00', 'Out', ''),
( 3, '2016-01-01 09:15:00', 'In', ''),
( 4, '2016-01-01 09:30:00', 'In', ''),
( 5, '2016-01-01 09:35:00', '', 'On'),
( 6, '2016-01-01 10:00:00', 'Out', ''),
( 7, '2016-01-01 16:00:00', '', 'Off');
drop table if exists oneLinersDetail;
create table oneLinersDetail
( -- architect this depending on multi-user concurrency
id int not null,
dt datetime not null,
door int not null,
grpIn int not null,
grpInSeq int not null,
grpOut int not null,
grpOutSeq int not null
);
drop table if exists oneLinersSummary;
create table oneLinersSummary
( -- architect this depending on multi-user concurrency
id int not null,
grpInSeq int null,
grpOutSeq int null,
checkIn datetime null, -- we are hoping in the end it is not null
checkOut datetime null -- ditto
);
存储过程：
DROP PROCEDURE IF EXISTS fetchOneLiners;
DELIMITER $$
CREATE PROCEDURE fetchOneLiners()
BEGIN
truncate table oneLinersDetail; -- architect this depending on multi-user concurrency
insert oneLinersDetail(id,dt,door,grpIn,grpInSeq,grpOut,grpOutSeq)
select id,dt,door,grpIn,grpInSeq,grpOut,grpOutSeq
from
( select id,dt,door,
if(@lastEvt!=door and door=1,
greatest(@grpIn:=@grpIn+1,0),
7777) as dummy2, -- this output column we don't care about (we care about the variable being set)
if(@lastEvt!=door and door=2,
greatest(@grpOut:=@grpOut+1,0),
7777) as dummy3, -- this output column we don't care about (we care about the variable being set)
if (@lastEvt!=door,greatest(@flip:=1,0),least(@flip:=0,1)) as flip,
if (door=1 and @flip=1,least(@grpOutSeq:=0,1),7777) as dummy4,
if (door=1 and @flip=1,greatest(@grpInSeq:=1,0),7777) as dummy5,
if (door=1 and @flip!=1,greatest(@grpInSeq:=@grpInSeq+1,0),7777) as dummy6,
if (door=2 and @flip=1,least(@grpInSeq:=0,1),7777) as dummy7,
if (door=2 and @flip=1,greatest(@grpOutSeq:=1,0),7777) as dummy8,
if (door=2 and @flip!=1,greatest(@grpOutSeq:=@grpOutSeq+1,0),7777) as dummy9,
@grpIn as grpIn,
@grpInSeq as grpInSeq,
@grpOut as grpOut,
@grpOutSeq as grpOutSeq,
@lastEvt:=door as lastEvt
from
( select id,`datetime` as dt,
CASE
WHEN Door='in' or Active_door='on' THEN 1
ELSE 2
END as door
from attendance2
order by id
) xD1 -- derived table #1
cross join (select @grpIn:=0,@grpInSeq:=0,@grpOut:=0,@grpOutSeq:=0,@lastEvt:=-1,@flip:=0) xParams
order by id
) xD2 -- derived table #2
order by id;
-- select * from oneLinersDetail;
truncate table oneLinersSummary; -- architect this depending on multi-user concurrency
insert oneLinersSummary (id,grpInSeq,grpOutSeq,checkIn,checkOut)
select distinct grpIn,null,null,null,null
from oneLinersDetail
order by grpIn;
-- select * from oneLinersSummary;
update oneLinersSummary ols
join
( select grpIn,max(grpInSeq) m
from oneLinersDetail
where door=1
group by grpIn
) d1
on d1.grpIn=ols.id
set ols.grpInSeq=d1.m;
-- select * from oneLinersSummary;
update oneLinersSummary ols
join
( select grpOut,max(grpOutSeq) m
from oneLinersDetail
where door=2
group by grpOut
) d1
on d1.grpOut=ols.id
set ols.grpOutSeq=d1.m;
-- select * from oneLinersSummary;
update oneLinersSummary ols
join oneLinersDetail old
on old.door=1 and old.grpIn=ols.id and old.grpInSeq=ols.grpInSeq
set ols.checkIn=old.dt;
-- select * from oneLinersSummary;
update oneLinersSummary ols
join oneLinersDetail old
on old.door=2 and old.grpOut=ols.id and old.grpOutSeq=ols.grpOutSeq
set ols.checkOut=old.dt;
-- select * from oneLinersSummary;
-- dump out the results
select id,checkIn,checkOut
from oneLinersSummary
order by id;
-- rows are left in those two tables (oneLinersDetail,oneLinersSummary)
END$$
DELIMITER ;
<强>测试
call fetchOneLiners();
+----+---------------------+---------------------+
| id | checkIn | checkOut |
+----+---------------------+---------------------+
| 1 | 2016-01-01 08:00:00 | 2016-01-01 09:00:00 |
| 2 | 2016-01-01 09:35:00 | 2016-01-01 16:00:00 |
+----+---------------------+---------------------+
这是答案的结尾。以下是开发人员可视化完成存储过程的步骤。
直到最后导致的开发版本。希望这有助于可视化，而不是仅仅丢弃中等大小令人困惑的代码块。
步骤A
步骤B
步骤B输出
第C步
步骤C输出
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