如何从第一行中减去第二行并在mysql中生成输出列

Question

我有一个情况，考虑以下查询：

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

这个查询的输出

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 这个等式将用于第一行，其结果将存储在新列中让我们采用difference of consecutive ordertime。谁能给我一些解决这种情况的指示？

如果您能帮助创建一个 auto increment column，那就太好了，因为这样我就可以自行加入并从中受益。

Answer 1

您可以添加一个额外的计算字段：

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;

但你最好在ordertime 上有一个索引：

CREATE INDEX ord_ndx ON Orders(ordertime, cakeid);

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

另一种方式

但是，根据您的架构，添加列 (previous_order) 可能会更好地为您服务。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
        );

现在您的桌​​子会自动工作。 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);

Answer 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