PHP mysql获取两个日期之间的减法

Question

我有这样的表结构：

Id    Name      Rank      Date
-----------------------------------
1     test      1000      2012-1-11

2     test      7000      2012-1-10

3     test2     2000      2012-1-11

4     test2     200       2012-1-10

5     test3     4000      2012-1-10

6     test4     6500      2012-1-11

考虑今天的日期是2012-1/11昨天的日期是2012-1/10

在单个查询中，我得到了今天和昨天的每个用户名之间的差异。即昨天有7000个排名，今天有1000个排名。所以结果是6000，类似于test2的-1800。

我需要如下输出：

Name     Difference (Orderby the difference Desc)
--------------------
test      6000
test2     -1800

如果今天日期或昨天日期的记录不可用，则我们不会使用此记录进行计算。

这在PHP MySQL中是可能的吗？

Answer 1

这个怎么样？(不是很清楚你想要达到什么目的。)请发表意见。

• SQLFIDDLE DEMO

代码：

select b.id, b.name, (b.rank-a.rank) diff
from t1 a
left join t1 b
on b.date < a.date
and b.name = a.name
having not diff is null
;

结果：

ID  NAME    DIFF
2   test    6000
4   test2   -1800

根据OP的评论进行编辑：

请注意，我已经向示例表中添加了额外的几条记录，用于触发条件。

• SQLFIDDLE DEMO2

Code2：

select b.id, b.name,b.rank AS New,
b.Date new_date,
a.Rank as Old, a.date as old_date, 
(b.rank-a.rank) diff
from t1 a
left join t1 b
on b.name = a.name
where b.date > a.date and b.date <= Now()
and datediff(b.date, a.date) = 1
having not diff is null and diff <> 0
order by diff desc
;

结果：

ID  NAME    NEW     NEW_DATE            OLD     OLD_DATE            DIFF
3   test    8000    January, 12 2012    1000    January, 11 2012    7000
4   test2   2000    January, 11 2012    200     January, 10 2012    1800
1   test    1000    January, 11 2012    7000    January, 10 2012    -6000

Answer 2

你所需要的一切都在这一页上

https://dev.mysql.com/doc/refman/5.5/en/date-and-time-functions.html