MySQL-返回在匹配条件之后有n个连续记录的所有记录
MySQL v5.5 Hi-我需要从一个简单的表中创建一个select查询，在该表中，它返回所有在后面有一定数量的连续行且标记为可用的行 示例表数据如下所示：MySQL-返回在匹配条件之后有n个连续记录的所有记录,mysql,Mysql,MySQL v5.5 Hi-我需要从一个简单的表中创建一个select查询，在该表中，它返回所有在后面有一定数量的连续行且标记为可用的行 示例表数据如下所示： id Slot Available 0 5.00pm 1 1 5.10pm 1 2 5.20pm 1 3 5.30pm 0 4 5.40pm 1 5 5.50pm 1 6 6.00pm 1 7 6.10pm 1 我有一个变量，它将决定所需插槽的数量 例如，如
id Slot Available
0 5.00pm 1
1 5.10pm 1
2 5.20pm 1
3 5.30pm 0
4 5.40pm 1
5 5.50pm 1
6 6.00pm 1
7 6.10pm 1
id Slot Available
0 5.00pm 1
4 5.40pm 1
5 5.50pm 1
非常感谢我相信有一种方法可以用SQL实现您想要的功能。我有一个您可以使用的存储过程示例
DELIMITER //
DROP PROCEDURE IF EXISTS GetTimeSlots;
CREATE PROCEDURE GetTimeSlots(NoOfSlots int)
BEGIN
declare finished bool default false;
-- stores id, slot and available from the cursor
declare v_id int;
declare v_slot text;
declare v_available int;
-- stores how many slots were found consecutively
declare v_slotsfound int;
-- loop through all available records
declare c cursor for
select id, slot, available from test where available = 1;
declare continue handler for not found
set finished = true;
-- create empty table to store slots we find
drop temporary table if exists tmp;
create temporary table tmp
select * from test where 1=2;
open c;
looper: loop
fetch c into v_id, v_slot, v_available;
if finished then
leave looper;
end if;
select sum(case
when available then 1
else 0
end) into v_slotsfound
from (
-- list out as many slots you need
-- outer query will total up the # of available slots
select available from test where id >= v_id
order by id
limit NoOfSlots
) t;
-- if # of available slots equals how many you want, store data
-- in temporary table
if v_slotsfound = NoOfSlots then
insert into tmp values (v_id, v_slot, v_available);
end if;
end loop looper;
close c;
-- output the results
select * from tmp;
drop temporary table if exists tmp;
END //
DELIMITER ;
mysql> call gettimeslots(2);
+------+--------+-----------+
| id | slot | available |
+------+--------+-----------+
| 0 | 5.00pm | 1 |
| 1 | 5.10pm | 1 |
| 4 | 5.40pm | 1 |
| 5 | 5.50pm | 1 |
| 6 | 6.00pm | 1 |
+------+--------+-----------+
mysql> call gettimeslots(4);
+------+--------+-----------+
| id | slot | available |
+------+--------+-----------+
| 4 | 5.40pm | 1 |
+------+--------+-----------+
mysql> call gettimeslots(2);
+------+--------+-----------+
| id | slot | available |
+------+--------+-----------+
| 0 | 5.00pm | 1 |
| 1 | 5.10pm | 1 |
| 4 | 5.40pm | 1 |
| 5 | 5.50pm | 1 |
| 6 | 6.00pm | 1 |
+------+--------+-----------+
mysql> call gettimeslots(4);
+------+--------+-----------+
| id | slot | available |
+------+--------+-----------+
| 4 | 5.40pm | 1 |
+------+--------+-----------+
我相信有一种方法可以实现您对SQL的需求。我有一个您可以使用的存储过程示例
DELIMITER //
DROP PROCEDURE IF EXISTS GetTimeSlots;
CREATE PROCEDURE GetTimeSlots(NoOfSlots int)
BEGIN
declare finished bool default false;
-- stores id, slot and available from the cursor
declare v_id int;
declare v_slot text;
declare v_available int;
-- stores how many slots were found consecutively
declare v_slotsfound int;
-- loop through all available records
declare c cursor for
select id, slot, available from test where available = 1;
declare continue handler for not found
set finished = true;
-- create empty table to store slots we find
drop temporary table if exists tmp;
create temporary table tmp
select * from test where 1=2;
open c;
looper: loop
fetch c into v_id, v_slot, v_available;
if finished then
leave looper;
end if;
select sum(case
when available then 1
else 0
end) into v_slotsfound
from (
-- list out as many slots you need
-- outer query will total up the # of available slots
select available from test where id >= v_id
order by id
limit NoOfSlots
) t;
-- if # of available slots equals how many you want, store data
-- in temporary table
if v_slotsfound = NoOfSlots then
insert into tmp values (v_id, v_slot, v_available);
end if;
end loop looper;
close c;
-- output the results
select * from tmp;
drop temporary table if exists tmp;
END //
DELIMITER ;
mysql> call gettimeslots(2);
+------+--------+-----------+
| id | slot | available |
+------+--------+-----------+
| 0 | 5.00pm | 1 |
| 1 | 5.10pm | 1 |
| 4 | 5.40pm | 1 |
| 5 | 5.50pm | 1 |
| 6 | 6.00pm | 1 |
+------+--------+-----------+
mysql> call gettimeslots(4);
+------+--------+-----------+
| id | slot | available |
+------+--------+-----------+
| 4 | 5.40pm | 1 |
+------+--------+-----------+
mysql> call gettimeslots(2);
+------+--------+-----------+
| id | slot | available |
+------+--------+-----------+
| 0 | 5.00pm | 1 |
| 1 | 5.10pm | 1 |
| 4 | 5.40pm | 1 |
| 5 | 5.50pm | 1 |
| 6 | 6.00pm | 1 |
+------+--------+-----------+
mysql> call gettimeslots(4);
+------+--------+-----------+
| id | slot | available |
+------+--------+-----------+
| 4 | 5.40pm | 1 |
+------+--------+-----------+
你能澄清或提供更多的例子吗？1） “连续两个插槽可用”是什么意思？我看到集合中的所有插槽都是连续的，并且排序顺序中有提供。2） 为什么正好是两个连续的插槽，什么是“连续”呢？3） 为什么在所需的解决方案中返回0,4,5而不是0,1,7？区别是什么？你能澄清或提供更多的例子吗？1） “连续两个插槽可用”是什么意思？我看到集合中的所有插槽都是连续的，并且排序顺序中有提供。2） 为什么正好是两个连续的插槽，什么是“连续”呢？3） 为什么在所需的解决方案中返回0,4,5而不是0,1,7？有什么区别吗？太棒了，谢谢！最后一件事，因为我完全不知道它是如何工作的，我只需要始终包括第一个插槽，而不考虑连续的插槽。请你给我指一下正确的方向好吗。再次感谢你，你能举例说明@user1830274是什么意思吗？您可以编辑问题并添加示例。告诉我你想要什么。你提供的代码工作得很好，我只需要始终返回第一行（id 0），只要它标记为可用。哦，我明白了。就在上面
select*from tmp行，类型<代码>插入tmp中，从测试中选择*如果可用=1限值1将在所有其他行上方添加第一个可用的时间段。为了更好地测量，您还应该更改select*fromtmp至按id从tmp订单中选择*
谢谢，真的非常感谢，谢谢！最后一件事，因为我完全不知道它是如何工作的，我只需要始终包括第一个插槽，而不考虑连续的插槽。请你给我指一下正确的方向好吗。再次感谢你，你能举例说明@user1830274是什么意思吗？您可以编辑问题并添加示例。告诉我你想要什么。你提供的代码工作得很好，我只需要始终返回第一行（id 0），只要它标记为可用。哦，我明白了。就在上面select*from tmp行，类型<代码>插入tmp中，从测试中选择*如果可用=1限值1将在所有其他行上方添加第一个可用的时间段。为了更好地测量，您还应该更改select*fromtmp至按id从tmp订单中选择*
谢谢，非常感谢