如何计算相同事​​件的日差

Question

我的目标是GROUP (eid) SORT BY dayatime()，在每组中选择前两次，计算一天的差异。

我知道这个想法，但我怎么能把它翻译成真正的MySQL查询语法？

mysql> select * from events;
+------+------------+
| eid  | dt         |
+------+------------+
|    1 | 2013-01-01 | -> 3
|    1 | 2013-01-04 | -> 1
|    1 | 2013-01-05 |
|    2 | 2013-04-01 | -> 7
|    2 | 2013-04-08 |
+------+------------+
5 rows in set (0.00 sec)


GOAL: query that gives this result: 
+------+------------+-----------------+
| eid  | dt         | days_until_next |
+------+------------+-----------------+
|    1 | 2013-01-01 |               3 |
|    1 | 2013-01-04 |               1 |
|    2 | 2013-04-01 |               7 |

Answer 1

基本上，您需要$0 ~ text { # run these commands if we find your text getline stn <"inp1" # read a new input line from the file "inp1" sub(/insert_text/,stn) # do the substitution. } 1 # shorthand for "print the current line". 功能，但MySQL不支持此功能。相反，您可以使用相关子查询：

Schedule(() => Console.WriteLine("This task will run at last day of every month."))
        .ToRunEvery(1)
        .Months()
        .OnTheLastDay();