如何从第一行中减去第二行并在mysql中生成输出列

时间:2015-10-20 23:39:49

标签: mysql sql mysql-workbench

我有一种情况,请考虑以下问题:

select 
    A.cakename, A.status, B.ordertime
from
    Cake as A
        inner join
    Orders as B on A.cakeid = B.cakeid

此查询的输出

cakename,   status,      ordertime
Apple Pie, available,    2014-03-20 22:34:46
Apple Pie, available,    2014-10-10 04:00:00
Apple Pie, available,    2015-03-20 22:34:46
Brownie,   available,    2014-03-20 22:35:07
Brownie,   available,    2015-05-20 22:35:07
Brownie,   available,    2014-03-26 22:36:07
Brownie,   available,    2015-05-20 14:35:48
Brownie,   available,    2015-05-20 22:35:07
Brownie,   available,    2015-05-10 22:35:07

现在我想减去ordertime row n+1 - ordertime row n并创建一个单独的列来存储结果。现在你可以忽略最后一行,因为没有下一行。

例如:2014-10-10 04:00:00 - 2014-03-20 22:34:46此等式将用于第一行,其结果将存储在新列中,我们可以使用difference of consecutive ordertime。谁能给我一些解决这种情况的指示?

如果你可以帮助创建一个很棒的auto increment column,那么我可以自己加入并采取不同的行动。

2 个答案:

答案 0 :(得分:2)

您可以添加额外的计算字段:

select 
A.cakename, A.status, B.ordertime,
(SELECT MAX(ordertime) FROM Orders AS C WHERE C.ordertime < B.ordertime) AS prev
from
Cake as A
    inner join
Orders as B on A.cakeid = B.cakeid
ORDER BY ordertime;

但您最好在ordertime上找到索引:

CREATE INDEX ord_ndx ON Orders(ordertime, cakeid);

实际上,您可以直接计算the difference from the previous time(假设它不是NULL;您必须提供带有IF的默认值)。

另一种方式

但是,根据您的体系结构,您可以通过添加列(previous_order)来获得更好的服务。当您插入订单时,您就知道自己需要什么。使用TRIGGER

查看此示例
CREATE TABLE test ( itemid integer, ordertime timestamp, previous datetime );
CREATE INDEX test_ndx ON test (itemid, ordertime);

CREATE TRIGGER test_set_prev 
    BEFORE INSERT ON test 
    FOR EACH ROW
        SET NEW.previous = (
            SELECT MAX(ordertime) FROM test WHERE itemid = NEW.itemid
        );

现在您的表格会自动生效。 TIMESTAMP列将自动更新,previous列也将自动更新:

insert into test (itemid) values (2);
select sleep(2);
insert into test (itemid) values (2);

+--------+---------------------+---------------------+
| itemid | ordertime           | previous            |
+--------+---------------------+---------------------+
|      2 | 2015-10-21 09:30:44 | NULL                |
|      2 | 2015-10-21 09:30:46 | 2015-10-21 09:30:44 |
+--------+---------------------+---------------------+

它也适用于多个插入:

insert into test (itemid) values (3), (2), (3);

答案 1 :(得分:2)

SELECT t.*,
  IF(@prev IS NULL ,0 , TIMESTAMPDIFF(SECOND,t.ordertime,@prev)) diff,
  @prev = t.ordertime
FROM (
  SELECT 
    A.cakename, A.status, B.ordertime
  FROM Cake as A
  INNER JOIN Orders as B 
  ON A.cakeid = B.cakeid
  ORDER BY ordertime 
) t