Question

我看了很多线程，但我找不到任何可以回答我的问题（或者我没有搜索到正确的问题：P）我是自学的mysql，现在我被困在这里，无法找到任何溶液

所以我的问题是，如果可以从两个日期之间的表中选择行，只有那些行之间的差异是例如6个月，如果没有显示已经过了多少时间。

表：

+----+--------+--------------+------------+
| id |  name  |    action    |    date    |
+----+--------+--------------+------------+
| 1  | name 1 |   exchange   | 2011-06-15 |
| 2  | name 1 |   exchange   | 2011-12-15 |
| 3  | name 1 |   exchange   | 2012-06-15 |
| 4  | name 1 |   exchange   | 2013-01-15 | -1 month
| 5  | name 2 |   exchange   | 2014-01-15 |
| 6  | name 2 | intervention | 2014-05-15 |
| 7  | name 2 |   exchange   | 2014-06-15 |
| 8  | name 2 |   exchange   | 2015-05-15 | - 11 months
+----+--------+--------------+------------+

所以我卡住了

SELECT * 
FROM (select * from table
        WHERE intervention like '%exchange%'
        AND date between '2011-01-15' and NOW() )
WHERE ....
 - if difference between row 1 and row 2, row 2 and row 3 till no rows? else write difference ?

但是根据我的猜测，我必须做出2个选择，之后加入他们或类似的东西，1选择将采取所有＆＃34;一般＆＃34;行和一个条件介于结果之间我只是不知道我应该如何做到这一点，或者是否只能用sql。

你们有解决方案吗？如果你这样做，你能解释一下你做了什么吗？：d

期望的输出：

+--------+--------------+------------+
|  name  |    action    |    date    |
+--------+--------------+------------+
| name 1 |   exchange   | 2011-06-15 |
| name 1 |   exchange   | 2011-12-15 |
| name 1 | OK exchange  | 2011-12-15 | - generated because row 1 and 2 meet the requirements
| name 1 |   exchange   | 2012-06-15 |
| name 1 | OK exchange  | 2012-06-15 | - generated because row 2 and 3 meet the requirements
| name 1 |   exchange   | 2013-01-15 | 
| name 1 | 1 month late | 2013-12-15 | - generated because row 3 and 4 DONT meet the requirements

提前谢谢你。尊重，iaiu

Answer 1

对我来说，你想要一个混合了源数据和已处理数据的结果，这有点......非正统。让我们从一开始就尝试一下。首先，我们自己加入表格，给他们两个昵称。请注意，t2将每个下一行与t1对配对：

select t1.name as name, t1.date as date, datediff(t2.date, t1.date) as diff_in_days
from table t1, table t2
where
  t1.action like '%exchange%' and
  t1.date between '2011-01-15' and NOW() and
  t1.name = t2.name and
  t1.id = t2.id-1
order by t1.name, t1.date;

我们已经得到了它们之间的时差，但是在几天之内。＆＃34; 6个月＆＃34;的定义由你自己决定，让现在假设它是182天。现在让我们在您的业务逻辑中包含上述内容：

select
  name,
  date,
  if(diff_in_days <=182, "OK", concat(diff_in_days - 182, "days late")) as message
from (
select t1.name as name, t1.date as date, datediff(t2.date, t1.date) as diff_in_days
from table t1, table t2
where
  t1.action like '%exchange%' and
  t1.date between '2011-01-15' and NOW() and
  t1.name = t2.name and
  t1.id = t2.id-1
) res
order by res.name, res.date;

现在，如果您还需要原始数据，请使用UNION并同时放置上述原始数据表和原始数据表。彼此相邻的行。

Mysql - 行之间的条件

1 个答案: