Mysql 如何根据“时间”字段中的值添加新列`

我有一张桌子：

+--------+--------------+-----------+   
| person |.....DAY......| ....hours |  
| Name1 ....2014-01-08... 03:53:23  |   
| Name2 ....2014-01-08... 05:30:54  |  
+--------|--------------|-----------+

如何根据时间字段中的值添加新列

标签

？ 如果时间从05:00:00到11:00:00，则值将为

'1'

，否则

'0'

。它应该是这样的：

+---------+--------------+-----------+---------+  
| person  |.....DAY......| ..hours  |  label   |   
| Name1 ....2014-01-08... 03:53:23  |  0 ......|   
| Name2 ....2014-01-08... 05:30:54  |  1.......|   
+---------+--------------+----------+----------+

---

你可以用这样的东西

update table set label = case when hour(hours) between 5 and 10 then 0 else 1 end;

SELECT person, 
       DATE( dt ) AS DAY , 
       case when hour(from_unixtime(sec_to_time(SUM( IF( source LIKE '%вход' OR source LIKE '%въезд', -1, 1 ) * UNIX_TIMESTAMP( dt ) )))) between 5 and 11 
            then 1
            else 0 
       end AS hours
from your_table

---

你可以用这样的东西

update table set label = case when hour(hours) between 5 and 10 then 0 else 1 end;

SELECT person, 
       DATE( dt ) AS DAY , 
       case when hour(from_unixtime(sec_to_time(SUM( IF( source LIKE '%вход' OR source LIKE '%въезд', -1, 1 ) * UNIX_TIMESTAMP( dt ) )))) between 5 and 11 
            then 1
            else 0 
       end AS hours
from your_table

---

因为这是一个计算列，所以您可以像这样在查询中动态生成它

update table set label = case when hour(hours) between 5 and 10 then 0 else 1 end;

SELECT person, 
       DATE( dt ) AS DAY , 
       case when hour(from_unixtime(sec_to_time(SUM( IF( source LIKE '%вход' OR source LIKE '%въезд', -1, 1 ) * UNIX_TIMESTAMP( dt ) )))) between 5 and 11 
            then 1
            else 0 
       end AS hours
from your_table

---

因为这是一个计算列，所以您可以像这样在查询中动态生成它

update table set label = case when hour(hours) between 5 and 10 then 0 else 1 end;

SELECT person, 
       DATE( dt ) AS DAY , 
       case when hour(from_unixtime(sec_to_time(SUM( IF( source LIKE '%вход' OR source LIKE '%въезд', -1, 1 ) * UNIX_TIMESTAMP( dt ) )))) between 5 and 11 
            then 1
            else 0 
       end AS hours
from your_table

---

如果我的列

hours

在SELECT中声明，该怎么办：选择person，DATE（dt）作为DAY，sec_to_time（SUM）（如果（

source

LIKE“%oföззззз”或

source

LIKE“%ofъзззз”，“1）*UNIXзu时间戳（dt））作为小时）如果我的列

hours在SELECT:SELECT中声明，该，日期（dt）为天，秒到时间（总和（如果（
源代码
如“%БСз”或
源代码
如“%Съззз”，-1，1）*UNIX_时间戳（dt））为小时